Chapter 5: Oscillations
Sometime in the 1580s, a young medical student named Galileo Galilei sat in the cathedral at Pisa. A hanging lamp above him had been set swinging, and instead of following the service, he found himself watching the lamp. The swings were slowly dying down — each arc was shorter than the one before it. That was expected. What was not expected was what Galileo noticed next.
He had no watch. Mechanical clocks of that time could not measure short intervals reliably, so he used the one steady timekeeper available to him: his own pulse. Counting beats, he timed the wide swings and then the narrow ones. The count came out very nearly the same. The lamp was covering less distance, yet taking the same time to complete each swing.
This single observation — that a pendulum’s period barely depends on how far it swings — became the foundation of accurate timekeeping. Decades later it led to the pendulum clock, and for nearly three centuries pendulums were the most precise clocks humanity possessed.
The lamp in Pisa was doing something that a guitar string, a tuning fork, a sewing-machine needle, a floating ball bobbing in water, and the quartz crystal in a wristwatch all do: repeating the same motion, about the same central position, again and again. Physics calls this oscillation, and it follows a surprisingly simple set of rules.
Diagram
Fig. 5.0 – A lamp swinging on a long chain inside a tall cathedral, with a faint arc showing wide and narrow swings, and a student below with fingers on his wrist counting his pulse.
This chapter will answer each of these questions as you work through it.
5.1 Introduction
Oscillations Are All Around You
Oscillation is one of the most common and interesting phenomena in the world of Physics. Once you learn to look for it, you find it almost everywhere in ordinary life.
Think of the rocking of a cradle, the swinging of a swing in a park, or the steady motion of the pendulum of a wall clock. Look at the vibrations of a guitar or violin string when it is plucked, the rapid up and down motion of the needle of a sewing machine, the motion of the prongs of a tuning fork after it is struck, or the bouncing of a block attached to a spring.
Different as these look, they share one feature. Each of them repeats the same movement again and again after a certain fixed interval of time. Motion that repeats itself in this way is called periodic motion, and you will meet its precise definition in the very next section.
In all the examples listed above, there is a second common feature as well. The object does not travel away and keep going — it moves mostly to and fro, or up and down, about one central position and returns to it repeatedly. This is what makes them oscillations rather than merely repeating motions of some general kind.
Diagram
Fig. 5.0b – A montage of everyday oscillations: a rocking cradle, a clock pendulum, a plucked guitar string, a sewing-machine needle and a block on a spring, each with a double-headed arrow showing the to-and-fro direction.
Why Sine and Cosine Enter the Picture
Every oscillatory motion is a periodic motion. What makes oscillations especially convenient to study is that the quantities describing them can be written down as neat mathematical functions of time.
In this chapter you will see that for such motion, the displacement (how far the body is from its central position), the velocity and the acceleration can all be represented by sine and cosine functions of time. These two functions are together known as harmonic functions. They are the natural mathematical language for anything that rises, falls and returns in a regular cycle.
Because of this, an oscillatory motion that obeys such functions is called harmonic motion. The word “harmonic” here is not about music — it simply records the fact that sines and cosines describe the motion exactly.
What This Chapter Will Give You
By the time you finish this chapter, you will be able to use the correct terminology for describing any oscillation — amplitude, period, frequency and phase.
You will study simple harmonic motion (S.H.M.) in detail, both as a force law and as a differential equation, and you will learn its graphical representations, showing how displacement, velocity and acceleration vary with time.
You will also examine the energy changes that take place during S.H.M., how oscillations gradually die down through damping, and how an oscillator can be driven by an outside agency to give the important effect called resonance.
These ideas do not stay inside this chapter. They return when you study waves, sound, alternating current circuits and the vibrations of atoms in solids, so the effort you invest here pays back repeatedly.
5.2 Explanation of Periodic Motion
What Makes a Motion Periodic
Some motions in nature simply repeat. A body performing such motion goes through one complete set of movements, and then goes through exactly the same set again, and again, without any change in the pattern.
The time taken to complete one such full set of movements is called the period of the motion, or its periodic time. Its SI unit is the second (s).
There is a neat way to check whether a set of movements is truly complete. At the end of one set, the state of the body — its position and the direction in which it is moving — must be exactly the same as it was at the beginning. If anything about the state is different, the set is not yet complete.
Examples of Periodic Motion
The Moon going round the Earth is periodic: it returns to the same position relative to the Earth after a definite time, and then starts the same journey again.
The same is true for the planets moving around the Sun, each with its own period, and for the motion of electrons around the nucleus of an atom.
As you saw in Chapter 1, the uniform circular motion of any object is also a periodic motion. The object keeps returning to the same point on the circle, moving in the same direction, after a fixed interval of time.
Diagram
Fig. 5.0c – Two panels side by side: the left shows the Moon circling the Earth with an arrow marking one complete revolution; the right shows a bob moving to and fro along a straight path about a central point.
Oscillatory Motion: A Special Kind of Periodic Motion
Now consider a different type of periodic motion, one in which a particle repeatedly moves to and fro along the same path. Such motion is called oscillatory motion or vibratory motion.
Notice the extra condition here. It is not enough that the motion repeats; the particle must retrace the same path forward and backward about a central position.
This gives us an important one-way relationship. Every oscillatory motion is periodic, because the to-and-fro journey repeats after a definite time. But every periodic motion need not be oscillatory.
Circular motion is the standard example that proves the point. It is certainly periodic, since it repeats after a fixed time, yet it is not oscillatory, because the particle keeps going round in one sense and never retraces its path backwards.
The Simplest Oscillation of All
Among all oscillatory motions, the simplest form is simple harmonic motion. In this motion, every particle of the oscillating body moves to and fro about its mean position — the central position it settles at when left undisturbed — along a certain fixed path.
The shape of that fixed path gives us two categories. If the path is a straight line, the motion is called linear simple harmonic motion. If the path is an arc of a circle, it is called angular simple harmonic motion.
Period and Frequency of the Oscillation
For an oscillation, the period is defined through the to-and-fro journey. The smallest interval of time after which the to and fro motion is repeated is called the period of the periodic motion, written as \(T\) and measured in seconds (s).
The word smallest matters. If a motion repeats every 2 s, it also repeats every 4 s and every 6 s, but only the shortest of these — 2 s — is called the period.
The companion quantity counts oscillations instead of timing them. The number of oscillations completed per unit time is called the frequency of the periodic motion, written here as \(n\). Its SI unit is the hertz (Hz), where 1 Hz means one oscillation per second.
Period and frequency are two ways of describing the same rhythm — one measures time per oscillation, the other counts oscillations per unit time. Section 5.6.3 will make the exact relation between them explicit.
5.3 Linear Simple Harmonic Motion (S.H.M.)
Setting Up the Experiment
To see linear S.H.M. in the cleanest possible way, we set up a simple arrangement on a laboratory table.
Place a rectangular block on a smooth frictionless horizontal surface. Attach one end of a spring to a rigid wall and the other end to the block, as shown in Fig. 5.1. The surface is taken to be frictionless so that the only horizontal force acting on the block is the force from the spring.
Diagram
Fig. 5.1 (a), (b) and (c) – Spring mass oscillator: (a) spring stretched with block displaced by \(+x\) to the right and force \(F = -kx\) pointing left, (b) relaxed spring at equilibrium with \(F = 0\), (c) spring compressed with block displaced by \(-x\) to the left and force pointing right, each on a frictionless surface.
Now pull the block of mass \(m\) towards the right and release it. The block will not simply return and stop. It will begin a to and fro motion on either side of its equilibrium position, moving right, left, right again, and so on.
This to and fro motion of the block along a straight line is linear simple harmonic motion, and the rest of this section explains exactly why it happens.
The Equilibrium Position and the Restoring Force
Fig. 5.1(b) shows the equilibrium position, in which the spring is relaxed and exerts no force on the block. This is the mean position of the oscillation, and it is the position we measure all displacements from.
Suppose the block is displaced towards the right from its equilibrium position. The spring is now stretched, and the force exerted by the spring on the block is directed towards the left, as shown in Fig. 5.1(a).
The reason lies in the elastic properties of the spring. On being stretched, the spring tends to regain its original shape and size, and therefore it exerts a restoring force on the block. This force is responsible for bringing the block back to its original position.
Two features of this force matter, and both are needed for S.H.M. First, the force is proportional to the displacement — pull the block twice as far and the spring pulls back twice as hard. Second, its direction is opposite to that of the displacement.
If \(x\) is the displacement of the block from the mean position, the restoring force \(f\) is given by
\[f = -kx \qquad \text{--- (5.1)}\]
Here \(f\) is the restoring force in newtons (N) and \(x\) is the displacement from the mean position in metres (m).
The quantity \(k\) is a constant that depends upon the elastic properties of the spring. It is called the force constant of the spring, and its SI unit is newton per metre (N/m). A stiff spring has a large \(k\); a loose, easily stretched spring has a small \(k\).
The negative sign in Eq. (5.1) is not decoration. It indicates that the force and the displacement are oppositely directed, so that the force always acts to undo the displacement.
The Law Works on Both Sides
The same rule holds if the block is pushed the other way. If the block is displaced towards the left from its equilibrium position, the spring is compressed, and the force exerted by the spring on the block is now directed towards the right. Its magnitude is again proportional to the displacement from the mean position, as shown in Fig. 5.1(c).
This is worth pausing on. The single equation \(f = -kx\) already contains both cases automatically. When the block is on the right, \(x\) is positive and \(f\) comes out negative, meaning a leftward force. When the block is on the left, \(x\) is negative and \(f\) comes out positive, meaning a rightward force.
Thus \(f = -kx\) can be used as the equation of motion of the block for the entire oscillation, without needing separate treatment for the two sides.
From Force to Acceleration
Now suppose the block is released from the rightmost position. The restoring force exerted by the spring accelerates it towards its equilibrium position.
We can find this acceleration directly from Newton’s second law of motion, which tells us that acceleration equals force divided by mass. The acceleration \(a\) of the block is therefore given by
\[a = \frac{f}{m} = -\left(\frac{k}{m}\right)x \qquad \text{--- (5.2)}\]
where \(m\) is the mass of the block in kilograms (kg) and \(a\) is its acceleration in metres per second squared (m/s²).
Read the physical content of Eq. (5.2) carefully. Since \(k\) and \(m\) are both constants for a given block and spring, the whole factor \(k/m\) is a constant. This shows that the acceleration is also proportional to the displacement, and that its direction is opposite to that of the displacement.
In other words, the force and the acceleration are both directed towards the mean or equilibrium position, whichever side the block happens to be on. This is the single most important structural fact about S.H.M., and everything in the following sections is built on it.
Following the Block Through One Full Oscillation
It helps to trace the whole journey once, slowly, because every later result about S.H.M. is contained in this description.
As the block moves towards the mean position, its speed starts increasing due to its acceleration, but its displacement from the mean position goes on decreasing. Notice that these two quantities change in opposite senses — the block speeds up precisely as it runs out of displacement.
When the block returns to its mean position, the displacement is zero, and hence the force and the acceleration are also zero. There is nothing left to pull it back at this instant.
At the same time, the speed of the block at the mean position becomes maximum, and hence its kinetic energy attains its maximum value.
This is why the block does not stop at the mean position. Having no force on it there, it simply keeps moving, and continues beyond the mean position towards the left.
During this process the spring is compressed, and it exerts a restoring force on the block towards the right. Once again, the force and the displacement are oppositely directed.
This opposing force now retards the motion of the block, so that the speed goes on reducing and finally becomes zero. This position is shown in Fig. 5.1(c). In this position, the displacement from the mean position and the restoring force are both maximum.
The block is at rest here, but only for an instant. This force now accelerates the block towards the right, towards the equilibrium position, and the entire process begins again in the opposite sense.
The process goes on repeating, and this causes the block to oscillate on either side of its equilibrium (mean) position.
Defining Linear S.H.M.
Such oscillatory motion along a straight path is called linear simple harmonic motion (S.H.M.). We can now state its definition precisely, because every part of the definition has been demonstrated above.
Both halves of this definition are exam-critical. A motion in which the force is directed towards a fixed point but is not proportional to the displacement is still oscillatory, but it is not simple harmonic.
Conditions for S.H.M., and Harmonic Versus Non-Harmonic Oscillations
The conditions required for simple harmonic motion are: 1. Oscillation of the particle is about a fixed point. 2. The net force or acceleration is always directed towards the fixed point. 3. The particle comes back to the fixed point due to the restoring force.
Harmonic oscillation is that oscillation which can be expressed in terms of a single harmonic function, such as \(x = a \sin \omega t\) or \(x = a \cos \omega t\).
Non-harmonic oscillation is that oscillation which cannot be expressed in terms of a single harmonic function. It may be a combination of two or more harmonic oscillations, such as \(x = a \sin \omega t + b \sin 2\omega t\), etc.
Two Experiments That Demonstrate Linear S.H.M.
Two simple arrangements show linear S.H.M. without needing a spring at all, which is a useful reminder that S.H.M. is about the form of the restoring force, not about springs in particular.
In the first, a hydrometer is immersed in a glass jar filled with water. In its equilibrium position, it floats vertically in the water, as shown in Fig. 5.1(d). If it is slightly depressed and released, it bobs up and down, performing linear S.H.M. Here the restoring force comes from the extra upthrust of the displaced water, which grows in proportion to how far the hydrometer is pushed in.
Diagram
Fig. 5.1(d) – A hydrometer floating vertically in a glass jar of water, with a double-headed vertical arrow showing the up and down bobbing about the equilibrium level.
In the second, a U-tube is filled with a sufficiently long column of mercury. Initially, when both arms of the U-tube are exposed to the atmosphere, the level of mercury in both the arms is the same, as shown in Fig. 5.1(e). Now, if the level of mercury in one of the arms is depressed slightly and released, the level of mercury in each arm starts moving up and down about the equilibrium position, performing linear S.H.M. The restoring force here is supplied by the weight of the extra column of mercury that stands above the level in the other arm.
Diagram
Fig. 5.1(e) – A U-tube containing mercury with the level in one arm depressed by \(x\) below and raised by \(x\) above the common equilibrium level, marked with arrows.
5.4 Differential Equation of S.H.M.
Turning the Force Law into an Equation of Motion
We already have a complete description of linear S.H.M. in words. In a linear S.H.M., the force is directed towards the mean position, and its magnitude is directly proportional to the displacement of the body from the mean position.
Written as an equation, this is exactly what we obtained in Eq. (5.1),
\[f = -kx\]
where \(k\) is the force constant in N/m and \(x\) is the displacement from the mean position in metres.
A statement about force, however, does not by itself tell us where the body will be at a given instant. To get that, we must bring in Newton’s second law, which connects force to acceleration.
According to Newton’s second law of motion, \(f = ma\), and therefore
\[ma = -kx \qquad \text{--- (5.3)}\]
Here \(m\) is the mass of the particle in kg and \(a\) is its acceleration in m/s².
Writing Acceleration in Terms of Displacement
Equation (5.3) still contains two different quantities, \(a\) and \(x\). To make progress we express the acceleration itself in terms of the displacement, using the definitions of velocity and acceleration from calculus.
The velocity of the particle is the rate at which its displacement changes with time,
\[\mathrm{v} = \frac{dx}{dt}\]
and its acceleration is the rate at which the velocity changes with time,
\[a = \frac{d\mathrm{v}}{dt} = \frac{d^{2}x}{dt^{2}}\]
The symbol \(\dfrac{d^{2}x}{dt^{2}}\) simply means that the displacement \(x\) has been differentiated twice with respect to time — first to get velocity, then again to get acceleration. Its SI unit is m/s².
Substituting this expression for acceleration in Eq. (5.3), we get
\[m\frac{d^{2}x}{dt^{2}} = -kx\]
Bringing every term to one side and dividing throughout by \(m\),
\[\therefore \quad \frac{d^{2}x}{dt^{2}} + \frac{k}{m}x = 0 \qquad \text{--- (5.4)}\]
Introducing the Angular Frequency
The combination \(k/m\) appears so often, and is such a fixed property of the given oscillator, that it is given a name and a symbol of its own.
Substituting \(\dfrac{k}{m} = \omega^{2}\), where \(\omega\) is the angular frequency, Eq. (5.4) becomes
\[\frac{d^{2}x}{dt^{2}} + \omega^{2}x = 0 \qquad \text{--- (5.5)}\]
Equation (5.5) is the differential equation of linear S.H.M. An equation of this kind is called a differential equation because it links a quantity (\(x\)) with its own rate of change — here, its second rate of change with time.
The physical meaning of Eq. (5.5) is worth stating in plain words. It says that at every instant, the acceleration of the particle is proportional to its displacement and directed opposite to it. Any motion whose displacement obeys this equation is a linear S.H.M., whatever the physical system may be — a block on a spring, a bobbing hydrometer, or a mercury column in a U-tube.
Note that \(\omega\) is fixed entirely by \(k\) and \(m\), that is, by the stiffness of the spring and the mass of the body. It has nothing to do with how far you pulled the block before releasing it. Its SI unit is radian per second (rad/s).
5.5 Acceleration (a), Velocity (v) and Displacement (x) of S.H.M.
We can obtain expressions for the acceleration, velocity and displacement of a particle performing S.H.M. by solving the differential equation of S.H.M. in terms of displacement \(x\) and time \(t\).
The plan is a staircase. The differential equation gives us acceleration at once. Integrating it once gives velocity. Integrating once more gives displacement. Each step trades one derivative for one constant of integration, and each constant is fixed by a physical fact about the oscillation.
Acceleration in Terms of Displacement
From Eq. (5.5), we have
\[\frac{d^{2}x}{dt^{2}} + \omega^{2}x = 0\]
Taking the second term to the right-hand side,
\[\therefore \quad \frac{d^{2}x}{dt^{2}} = -\omega^{2}x \qquad \text{--- (5.6)}\]
But \(a = \dfrac{d^{2}x}{dt^{2}}\) is the acceleration of the particle performing S.H.M. Therefore
\[\therefore \quad a = -\omega^{2}x \qquad \text{--- (5.7)}\]
This is the expression for acceleration in terms of displacement \(x\). Here \(a\) is in m/s², \(x\) is in m and \(\omega\) is in rad/s.
Equation (5.7) is compact but rich. It says the acceleration is largest where the displacement is largest, vanishes where the displacement vanishes, and always carries the opposite sign to the displacement.
Velocity in Terms of Displacement
To get velocity we must integrate, but there is an obstacle: Eq. (5.6) involves \(t\), while we want an answer in terms of \(x\). The standard way past this obstacle is to rewrite the acceleration using the chain rule.
From Eq. (5.6), we have \(\dfrac{d^{2}x}{dt^{2}} = -\omega^{2}x\)
\[\therefore \quad \frac{d}{dt}\left(\frac{dx}{dt}\right) = -\omega^{2}x\]
\[\therefore \quad \frac{d\mathrm{v}}{dt} = -\omega^{2}x\]
\[\therefore \quad \frac{d\mathrm{v}}{dx} \cdot \frac{dx}{dt} = -\omega^{2}x\]
\[\therefore \quad \mathrm{v}\frac{d\mathrm{v}}{dx} = -\omega^{2}x\]
\[\therefore \quad \mathrm{v}\,d\mathrm{v} = -\omega^{2}x\,dx\]
Integrating both the sides, we get
\[\int \mathrm{v}\,d\mathrm{v} = -\omega^{2}\int x\,dx\]
\[\therefore \quad \frac{\mathrm{v}^{2}}{2} = -\frac{\omega^{2}x^{2}}{2} + C \qquad \text{--- (5.8)}\]
where \(C\) is the constant of integration.
To evaluate \(C\) we need one situation in which both \(x\) and \(\mathrm{v}\) are known. The extreme position supplies exactly that.
Let \(A\) be the maximum displacement (amplitude) of the particle in S.H.M. — the farthest distance it reaches from the mean position on either side. Its formal definition is taken up in Section 5.6.1.
When the particle is at the extreme position, velocity (\(\mathrm{v}\)) is zero, because the particle is momentarily at rest there before turning back.
Thus, at \(x = \pm A\), \(\mathrm{v} = 0\). Substituting in Eq. (5.8), we get
\[0 = -\frac{\omega^{2}A^{2}}{2} + C\]
\[\therefore \quad C = +\frac{\omega^{2}A^{2}}{2} \qquad \text{--- (5.9)}\]
Using \(C\) in Eq. (5.8), we get
\[\frac{\mathrm{v}^{2}}{2} = -\frac{\omega^{2}x^{2}}{2} + \frac{\omega^{2}A^{2}}{2}\]
\[\therefore \quad \mathrm{v}^{2} = \omega^{2}\left(A^{2} - x^{2}\right)\]
\[\therefore \quad \mathrm{v} = \pm\omega\sqrt{A^{2} - x^{2}} \qquad \text{--- (5.10)}\]
This is the expression for the velocity of a particle performing linear S.H.M. in terms of displacement \(x\). Here \(\mathrm{v}\) is in m/s, \(A\) and \(x\) are in m, and \(\omega\) is in rad/s.
The \(\pm\) sign is physically necessary, not a mathematical loose end. At any position other than the extremes, the particle may be travelling either outward or back towards the mean position, and the two cases differ only in the sign of the velocity.
Displacement in Terms of Time
Substituting \(\mathrm{v} = \dfrac{dx}{dt}\) in Eq. (5.10), we get
\[\frac{dx}{dt} = \pm\omega\sqrt{A^{2} - x^{2}}\]
\[\therefore \quad \frac{dx}{\sqrt{A^{2} - x^{2}}} = \pm\omega\,dt\]
Integrating both the sides, we get
\[\int \frac{dx}{\sqrt{A^{2} - x^{2}}} = \pm\omega\int dt\]
\[\sin^{-1}\left(\frac{x}{A}\right) = \omega t + \phi \qquad \text{--- (5.11)}\]
Here \(\phi\) is the constant of integration. To know \(\phi\), we need to know the value of \(x\) at any instance of time \(t\), the most convenient being \(t = 0\).
Taking the sine of both sides of Eq. (5.11),
\[\therefore \quad x = A\sin\left(\omega t + \phi\right) \qquad \text{--- (5.12)}\]
This is the general expression for the displacement (\(x\)) of a particle performing linear S.H.M. at time \(t\).
Here at last is the sine curve promised in the introduction. It has arrived not by assumption but as the forced consequence of a restoring force proportional to displacement.
Let us find expressions for displacement for two particular cases.
Case (i): Starting from the Mean Position
If the particle starts S.H.M. from the mean position, \(x = 0\) at \(t = 0\).
Using Eq. (5.11), we get
\[\phi = \sin^{-1}\left(\frac{0}{A}\right) = 0 \ \text{or} \ \pi\]
Substituting in Eq. (5.12), we get
\[\therefore \quad x = \pm A\sin\omega t \qquad \text{--- (5.13)}\]
This is the expression for displacement at any instant if the particle starts S.H.M. from the mean position. The positive sign is to be chosen if it starts towards the positive side, and the negative sign for starting towards the negative side.
Both choices, \(\phi = 0\) and \(\phi = \pi\), describe a particle at the mean position; they differ only in which way it is heading at that instant.
Case (ii): Starting from the Extreme Position
If the particle starts S.H.M. from the extreme position, \(x = \pm A\) at \(t = 0\).
\[\therefore \quad \phi = \sin^{-1}\left(\pm 1\right) = \frac{\pi}{2} \ \text{or} \ \frac{3\pi}{2}\]
Substituting in Eq. (5.12), we get
\[x = A\sin\left(\omega t + \frac{\pi}{2}\right) \quad \text{or} \quad x = A\sin\left(\omega t + \frac{3\pi}{2}\right)\]
Using the standard trigonometric result that \(\sin\left(\theta + \dfrac{\pi}{2}\right) = \cos\theta\) and \(\sin\left(\theta + \dfrac{3\pi}{2}\right) = -\cos\theta\),
\[\therefore \quad x = \pm A\cos\omega t \qquad \text{--- (5.14)}\]
This is the expression for displacement at any instant if the particle starts S.H.M. from the extreme position. The positive sign is for starting from the positive extreme position and the negative sign for starting from the negative extreme position.
So the same physical motion is described by a sine or by a cosine, depending only on where we decided to begin timing it. Neither form is more correct than the other.
Expressions of Displacement, Velocity and Acceleration at Time \(t\)
Having the displacement as a function of time, the other two quantities follow by simple differentiation.
From Eq. (5.12),
\[x = A\sin\left(\omega t + \phi\right)\]
\[\therefore \quad \mathrm{v} = \frac{dx}{dt} = A\omega\cos\left(\omega t + \phi\right)\]
\[\therefore \quad a = \frac{d\mathrm{v}}{dt} = -A\omega^{2}\sin\left(\omega t + \phi\right)\]
Observe the pattern. Each differentiation multiplies the amplitude by \(\omega\) and shifts the curve by a quarter cycle, which is why velocity is a cosine when displacement is a sine. It also explains why \(a = -\omega^{2}x\), since the last expression is just \(-\omega^{2}\) times the first.
Extreme Values of Displacement (x), Velocity (v) and Acceleration (a)
We now ask where in the oscillation each of the three quantities reaches its largest and smallest value. These results are used constantly in numerical problems.
1) Displacement: The general expression for displacement \(x\) in S.H.M. is \(x = A\sin\left(\omega t + \phi\right)\).
At the mean position, \(\left(\omega t + \phi\right) = 0\) or \(\pi\)
\[\therefore \quad x_{min} = 0\]
Thus, at the mean position, the displacement of the particle performing S.H.M. is minimum (i.e. zero).
At the extreme position, \(\left(\omega t + \phi\right) = \dfrac{\pi}{2}\) or \(\dfrac{3\pi}{2}\)
\[\therefore \quad x = A\sin\left(\omega t + \phi\right)\]
\[\therefore \quad x = \pm A\sin\frac{\pi}{2} \quad \therefore \quad x_{max} = \pm A\]
Thus, at the extreme position the displacement of the particle performing S.H.M. is maximum.
2) Velocity: According to Eq. (5.10) the magnitude of velocity of the particle performing S.H.M. is \(\mathrm{v} = \pm\omega\sqrt{A^{2} - x^{2}}\)
At the mean position, \(x = 0\) \(\therefore\) \(\mathrm{v}_{max} = \pm A\omega\).
Thus, the velocity of the particle in S.H.M. is maximum at the mean position.
At the extreme position, \(x = \pm A\) \(\therefore\) \(\mathrm{v}_{min} = 0\).
Thus, the velocity of the particle in S.H.M. is minimum at the extreme positions.
3) Acceleration: The magnitude of the acceleration of the particle in S.H.M. is \(\omega^{2}x\).
At the mean position \(x = 0\), so that the acceleration is minimum. \(\therefore\) \(a_{min} = 0\).
At the extreme positions \(x = \pm A\), so that the acceleration is maximum, \(a_{max} = \mp\omega^{2}A\).
The opposite sign in \(\mp\omega^{2}A\) is worth reading carefully: at the positive extreme the acceleration points in the negative direction, and at the negative extreme it points in the positive direction. In both cases it points back towards the mean position.
Notice the neat complementarity running through all three results, as summarised in Fig. 5.0d. Where displacement is greatest, velocity is zero and acceleration is greatest. Where displacement is zero, velocity is greatest and acceleration is zero. Velocity and acceleration never reach their peaks at the same place.
Diagram
Fig. 5.0d – A straight-line path with mean position O and extremes at \(\pm A\), showing at each of the three points the values of \(x\), \(\mathrm{v}\) and \(a\) with arrows marking their directions.
5.6 Amplitude (A), Period (T) and Frequency (n) of S.H.M.
Every S.H.M., no matter what is oscillating, can be described completely by three numbers: how far the body swings, how long each swing takes, and how many swings happen per second. These are the amplitude, the period and the frequency, and we take them one at a time.
5.6.1 Amplitude of S.H.M.
Consider a particle P performing S.H.M. along the straight line MN, as shown in Fig. 5.2. The centre O of MN is the mean position of the particle, and the particle travels back and forth between the two ends M and N.
Diagram
Fig. 5.2 – S.H.M. of a particle: a straight line MN with centre O marked as the mean position and a particle P shown between O and M, with the displacement measured from O.
The displacement of the particle, as given by Eq. (5.12), is
\[x = A\sin\left(\omega t + \phi\right)\]
A sine can never exceed 1 in magnitude, and this single fact sets the limit on how far the particle can go.
The particle will have its maximum displacement when \(\sin\left(\omega t + \phi\right) = \pm 1\), i.e., when \(x = \pm A\). This distance \(A\) is called the amplitude of S.H.M., measured in metres (m).
Two cautions are worth noting for numerical problems. The amplitude is measured from the mean position, not from one extreme to the other, so the total length of the path travelled between the extremes is \(2A\). Also, the amplitude is fixed by how the oscillation was started — by how far the body was initially pulled aside — and not by the properties of the spring.
5.6.2 Period of S.H.M.
The time taken by the particle performing S.H.M. to complete one oscillation is called the period of S.H.M.
We can find it from the displacement equation itself, by asking after how much time the particle returns to exactly the same state.
Displacement of the particle at time \(t\) is given by
\[x = A\sin\left(\omega t + \phi\right)\]
After a time \(t' = t + \dfrac{2\pi}{\omega}\) the displacement will be
\[x' = A\sin\left[\omega\left(t + \frac{2\pi}{\omega}\right) + \phi\right]\]
\[\therefore \quad x' = A\sin\left(\omega t + 2\pi + \phi\right)\]
Since adding \(2\pi\) to the angle leaves a sine unchanged,
\[\therefore \quad x' = A\sin\left(\omega t + \phi\right)\]
This result shows that the particle is at the same position after a time \(\dfrac{2\pi}{\omega}\). That means the particle completes one oscillation in time \(\dfrac{2\pi}{\omega}\). It can be shown that \(t = \dfrac{2\pi}{\omega}\) is the minimum time after which it repeats.
Hence its period \(T\) is given by
\[T = \frac{2\pi}{\omega}\]
Here \(T\) is in seconds (s) and \(\omega\) in rad/s.
Expressing the Period Through the Physical Properties
The relation \(T = 2\pi/\omega\) is correct but not yet practical, because \(\omega\) is not something we measure directly. We therefore go back to where \(\omega\) came from.
From Eq. (5.4) and Eq. (5.5),
\[\omega^{2} = \frac{k}{m} = \frac{\text{force per unit displacement}}{\text{mass}} = \text{acceleration per unit displacement}\]
\[\therefore \quad T = \frac{2\pi}{\sqrt{\text{acceleration per unit displacement}}}\]
This form is worth remembering, because it applies to any S.H.M., including ones with no spring in sight. Whenever you can show that acceleration is proportional to displacement, the constant of proportionality immediately gives you the period.
For the spring-and-block oscillator in particular, substituting \(\omega = \sqrt{k/m}\),
\[T = 2\pi\sqrt{\frac{m}{k}} \qquad \text{--- (5.15)}\]
where \(m\) is the mass of the body in kg and \(k\) is the force constant in N/m.
Read Eq. (5.15) physically. A heavier block has more inertia, resists being turned around, and so takes longer per oscillation — hence \(m\) in the numerator. A stiffer spring pulls back harder for the same displacement and hurries the block along — hence \(k\) in the denominator. Because both appear under a square root, quadrupling the mass only doubles the period.
Most importantly, the amplitude \(A\) does not appear anywhere in Eq. (5.15). Pull the block far or only a little, and it still takes exactly the same time per oscillation. When pulled further it has more distance to cover, but it also travels correspondingly faster, and the two effects cancel exactly.
5.6.3 Frequency of S.H.M.
The number of oscillations performed by a particle performing S.H.M. per unit time is called the frequency of S.H.M.
The relation between frequency and period follows from counting. In time \(T\), the particle performs one oscillation. Hence in unit time it performs \(\dfrac{1}{T}\) oscillations.
Hence, frequency \(n\) of S.H.M. is given by
\[n = \frac{1}{T} = \frac{\omega}{2\pi} = \frac{1}{2\pi}\sqrt{\frac{k}{m}} \qquad \text{--- (5.16)}\]
The SI unit of frequency is the hertz (Hz), where 1 Hz is one oscillation per second.
Do not confuse \(n\) with \(\omega\). The frequency \(n\) counts complete oscillations per second, while the angular frequency \(\omega\) counts radians per second, and one complete oscillation corresponds to \(2\pi\) radians. This is why the two are related by \(\omega = 2\pi n\), a relation used constantly in numerical problems and one that students frequently drop.
Combination of Springs
Series combination (Fig. A): In this case, all the springs are connected one after the other, forming a single chain.
Diagram
Fig. A – Two springs of force constants \(k_1\) and \(k_2\) joined end to end in a vertical chain from a rigid support, with a stretching force \(f\) applied at the lower end.
Consider an arrangement of two such springs of spring constants \(k_1\) and \(k_2\). If the springs are massless, each will have the same stretching force \(f\). For a vertical arrangement, it will be the weight \(mg\).
If \(e_1\) and \(e_2\) are the respective extensions, we can write
\[f = k_1 e_1 = k_2 e_2 \quad \text{and} \quad e_1 = \frac{f}{k_1}, \quad e_2 = \frac{f}{k_2}\]
The total extension is the sum of the individual extensions, since the springs are joined end to end:
\[e = e_1 + e_2 = f\left(\frac{1}{k_1} + \frac{1}{k_2}\right)\]
If \(k_s\) is the effective spring constant (as if there is a single spring that gives the same total extension for the same force), we can write
\[e = \frac{f}{k_s} \quad \therefore \quad \frac{f}{k_s} = f\left(\frac{1}{k_1} + \frac{1}{k_2}\right) \quad \therefore \quad \frac{1}{k_s} = \frac{1}{k_1} + \frac{1}{k_2}\]
For a number of such (massless) springs in series,
\[\frac{1}{k_s} = \frac{1}{k_1} + \frac{1}{k_2} + \cdots = \sum_{i}\frac{1}{k_i}\]
For only two massless springs in series,
\[k_s = \frac{k_1 k_2}{k_1 + k_2} = \frac{\text{Product}}{\text{Sum}}\]
For \(n\) such identical massless springs of spring constant \(k\) each, in series,
\[k_s = \frac{k}{n}\]
A series combination is therefore always softer than any single spring in it, because the same force now stretches several springs one after another.
Parallel combination (Fig. B): In such a combination, all the springs are connected between the same two points, one of them being the support, and at the other end the stretching force \(f\) is applied at a suitable point.
Diagram
Fig. B – Two springs of force constants \(k_1\) and \(k_2\) hanging side by side from the same rigid support and joined at the bottom to a common bar, with a stretching force \(f\) applied at the bar.
Irrespective of their spring constants, each spring will now have the same extension \(e\). The springs now share the force such that in the equilibrium position, the total restoring force is equal and opposite to the stretching force \(f\).
Let \(f_1 = -k_1 e\), \(f_2 = -k_2 e\), … be the individual restoring forces.
If \(k_p\) is the effective spring constant, a single spring of this spring constant will be stretched by the same extension \(e\) by the same stretching force \(f\).
\[\therefore \quad -f = k_p e = -\left(f_1 + f_2\right) = k_1 e + k_2 e\]
\[\therefore \quad k_p = k_1 + k_2 = \sum_{i} k_i\]
For \(m\) such identical massless springs of spring constant \(k\) each, in parallel,
\[k_p = mk\]
A parallel combination is therefore always stiffer than any single spring in it, because several springs now pull back at once.
5.7 Reference Circle Method
A Shadow That Oscillates
There is a second, entirely geometrical way of arriving at S.H.M., and it explains several things that the algebra of the previous sections left unexplained.
Figure 5.3 shows a rod rotating along a vertical circle in the x-y plane. If the rod is illuminated parallel to the x-axis from either side by a linear source parallel to the rod, as shown in Fig. 5.3, the shadow (projection) of the rod will be produced on the y-axis.
Diagram
Fig. 5.3 – Projection of a rotating rod: a rod rotating in a vertical circle in the x-y plane, illuminated by a linear source parallel to the x-axis, with its shadow falling along the y-axis.
Watch the tip of that shadow. As the rod goes round steadily, the tip of this shadow can be seen to be oscillating about the origin, along the y-axis. It runs up, slows, turns, runs down, slows, turns again — exactly the behaviour of an oscillator, produced by a rod that is doing nothing but rotating uniformly.
We shall now prove that the motion of the tip of the projection is an S.H.M. if the corresponding motion of the tip of the rod is a U.C.M. (uniform circular motion). For this, we should take the projections of displacement, velocity and acceleration on any reference diameter, and confirm that in each case we get the corresponding quantity for a linear S.H.M.
Setting Up the Reference Circle
Figure 5.4 shows the anticlockwise uniform circular motion of a particle P, with centre at the origin O.
Diagram
Fig. 5.4 – S.H.M. as projection of a U.C.M.: a particle P moving anticlockwise on a circle of radius r centred at O, with reference points E, F, G, H on the circle, the initial position \(P_0\) at angle \(\phi\), the reference angle \((\omega t + \phi)\) marked from OX, and M the foot of the projection of OP on the y-axis.
Its angular positions are decided with the reference OX. It means that if the particle is at E, the angular position is zero; at F it is \(90^{\circ} = \dfrac{\pi}{2}\) rad; at G it is \(180^{\circ} = \pi\) rad, and so on.
If it comes to E again, the angular position will be \(360^{\circ} = 2\pi\) rad, and not zero. This is a deliberate choice and an important one: the angle is allowed to keep growing past one full turn, so that it never repeats and always tells us which round the particle is on.
Let \(\vec{r} = \text{OP}\) be the position vector of this particle, of magnitude \(r\) in metres.
At \(t = 0\), let the particle be at \(P_0\) with reference angle \(\phi\). During time \(t\), it has angular displacement \(\omega t\). Thus, the reference angle at time \(t\) is \(\left(\omega t + \phi\right)\).
Let us choose the diameter FH along the y-axis as the reference diameter, and label OM as the projection of \(\vec{r} = \text{OP}\) on this.
Projection of Displacement
At time \(t\), we get the projection, or the position vector,
\[\text{OM} = \text{OP}\sin\left(\omega t + \phi\right) \quad \Rightarrow \quad y = r\sin\left(\omega t + \phi\right)\]
Compare this with Eq. (5.12). This is exactly the equation of linear S.H.M., of amplitude \(r\). The radius of the reference circle plays the part of the amplitude, which makes sense — the shadow can never stray further from O than the radius itself.
This result also settles a question raised earlier. The term \(\omega\) can now be understood as the angular velocity of the reference circular motion, measured in rad/s. For linear S.H.M. we may call it the angular frequency, as it decides the periodicity of the S.H.M.
So the strange-looking name “angular frequency”, used for a motion along a straight line, is borrowed from the circle hiding behind that motion. In the next section you will come to know that the phase angle \(\left(\omega t + \phi\right)\) of the circular motion can be used as the phase of the corresponding S.H.M.
Projection of Velocity
Instantaneous velocity of the particle P in the circular motion is the tangential velocity of magnitude \(r\omega\), directed along the tangent to the circle, as shown in Fig. 5.5.
Diagram
Fig. 5.5 – Projection of velocity: the tangential velocity \(r\omega\) of particle P on the reference circle, resolved to show its component along the reference diameter.
Its projection on the reference diameter will be
\[\mathrm{v}_y = r\omega\cos\left(\omega t + \phi\right)\]
The cosine appears here rather than a sine because the tangent to a circle is always perpendicular to the radius, so resolving the tangential velocity brings in the angle’s complement.
This is the expression for the velocity of a particle performing a linear S.H.M., matching what we obtained by differentiating Eq. (5.12).
Projection of Acceleration
Instantaneous acceleration of the particle P in circular motion is the radial or centripetal acceleration of magnitude \(r\omega^{2}\), directed towards O.
Its projection on the reference diameter will be
\[a_y = -r\omega^{2}\sin\left(\omega t + \phi\right) = -\omega^{2}y\]
The negative sign appears automatically because the centripetal acceleration points inward, towards the centre, and so its projection always points back towards the mean position of the shadow.
Again, this is the corresponding acceleration for the linear S.H.M., in agreement with Eq. (5.7).
The General Result
From this analogy it is clear that the projection of any quantity for a uniform circular motion gives us the corresponding quantity of linear S.H.M.
Nothing in the argument depended on our having chosen the y-axis. This analogy can be verified for any diameter as the reference diameter, since a different choice merely changes the value of the initial angle \(\phi\).
This result is more than an elegant curiosity. It converts every question about an oscillation into a question about a rotating radius, and rotating radii are easy to visualise and easy to add together. Section 5.8 uses this to give a precise meaning to phase, and Section 5.10 uses it again when two S.H.M.s have to be combined.
5.8 Phase in S.H.M.
What “State of Oscillation” Means
Phase in S.H.M. — or for any motion — is basically the state of oscillation. It is a single number that tells us everything about where the oscillator has reached in its cycle.
To know the state of oscillation in S.H.M., we need to know three things at that instant of time: the displacement (position), the direction of velocity, and the oscillation number, that is, during which oscillation the particle is.
It is worth seeing why nothing simpler will do.
Knowing only the displacement is not enough. At a given position there are two possible directions of velocity — the particle may be going outward or returning — and this is true everywhere except at the extreme positions. Besides, the same displacement repeats in every successive oscillation.
Knowing only the velocity is not enough either. There are two different positions for the same velocity, one on each side of the mean position, and this is true everywhere except at the mean position itself. And again, both of these repeat for the successive oscillations.
So neither quantity, taken alone, can identify the state of the oscillator. Worse, both of them keep coming back to the same values, so neither can tell us which oscillation we are in.
The Quantity That Never Repeats
Hence, to know the phase, we need a quantity that is continuously changing with time, and that does not return to an earlier value.
The reference circle of the previous section supplies exactly such a quantity. It is clear that all the quantities of linear S.H.M. (\(x\), \(\mathrm{v}\), \(a\), etc.) are the projections, taken on a diameter, of the respective quantities for the reference circular motion.
The angular displacement \(\left(\omega t + \phi\right)\) can thus be used as the phase of S.H.M., as it varies continuously with time. In this case, it will be called the phase angle.
This works because we agreed in Section 5.7 never to reset the reference angle to zero after a full turn. The angle simply keeps increasing, so the count of completed rounds is carried inside the number itself.
Since the phase angle is an angle, it may be quoted in degrees or in radians, and both are used freely in problems.
Diagram
Fig. 5.0e – A reference circle marked with the four key phase angles 0°, 90°, 180° and 270°, each joined by a horizontal arrow to the corresponding point of the straight-line oscillation alongside: mean position moving positive, positive extreme, mean position moving negative, and negative extreme.
Special Cases
The following four values of phase are worth committing to memory, since almost every phase problem is built from them. Each is illustrated in Fig. 5.0e.
(i) Phase \(\theta = 0\) indicates that the particle is at the mean position, moving to the positive, during the beginning of the first oscillation. Phase angle \(360^{\circ}\) or \(2\pi^{c}\) is the beginning of the second oscillation, and so on for the successive oscillations.
(ii) Phase \(\theta = 180^{\circ}\) or \(\pi^{c}\) indicates that during its first oscillation, the particle is at the mean position and moving to the negative. The similar state in the second oscillation will have phase \(\left(360 + 180\right) = 540^{\circ}\), or \(2\pi + \pi = 3\pi^{c}\), and so on for the successive oscillations.
(iii) Phase \(\theta = 90^{\circ}\) or \(\left(\dfrac{\pi}{2}\right)^{c}\) indicates that the particle is at the positive extreme position during the first oscillation. For the second oscillation it will be \(\left(360 + 90\right) = 450^{\circ}\), or \(2\pi + \dfrac{\pi}{2} = \left(\dfrac{5\pi}{2}\right)^{c}\), and so on for the successive oscillations.
(iv) Phase \(\theta = 270^{\circ}\) or \(\left(\dfrac{3\pi}{2}\right)^{c}\) indicates that the particle is at the negative extreme position during the first oscillation. For the second oscillation it will be \(\left(360 + 270\right) = 630^{\circ}\), or \(2\pi + \dfrac{3\pi}{2} = \left(\dfrac{7\pi}{2}\right)^{c}\), and so on for the successive oscillations.
Notice the working rule contained in these four cases. Divide the given phase angle by \(360^{\circ}\) (or \(2\pi^{c}\)). The quotient tells you how many complete oscillations are already over, so the oscillation currently in progress is one more than that. The remainder tells you the position and the direction of motion within that oscillation.
5.9 Graphical Representation of S.H.M.
Equations tell us what happens; graphs let us see it happening. In this section we plot the displacement, velocity and acceleration of a particle in S.H.M. against time, for the two standard starting conditions met in Section 5.5.
(a) Particle Executing S.H.M., Starting from the Mean Position, Towards Positive
As the particle starts from the mean position, as shown in Fig. 5.6, towards the positive side, the initial phase is \(\phi = 0\).
Putting \(\phi = 0\) in the general expressions obtained in Section 5.5,
\[\text{displacement} \quad x = A\sin\omega t\]
\[\text{Velocity} \quad \mathrm{v} = A\omega\cos\omega t\]
\[\text{Acceleration} \quad a = -A\omega^{2}\sin\omega t\]
We now evaluate these at intervals of a quarter period, which is enough to trace out the whole shape of each curve.
| \((t)\) | 0 | \(T/4\) | \(T/2\) | \(3T/4\) | \(T\) | \(5T/4\) |
|---|---|---|---|---|---|---|
| \((\theta)^{*}\) | 0 | \(\dfrac{\pi}{2}\) | \(\pi\) | \(\dfrac{3\pi}{2}\) | \(2\pi\) | \(\dfrac{5\pi}{2}\) |
| \((x)\) | 0 | \(A\) | 0 | \(-A\) | 0 | \(A\) |
| \((\mathrm{v})\) | \(A\omega\) | 0 | \(-A\omega\) | 0 | \(A\omega\) | 0 |
| \((a)\) | 0 | \(-A\omega^{2}\) | 0 | \(A\omega^{2}\) | 0 | \(-A\omega^{2}\) |
* phase \(\theta = \omega t + \phi\)
Plotting these values and joining them smoothly gives the three curves of Fig. 5.6.
Diagram
Fig. 5.6 – (a) Variation of displacement with time, a sine curve starting at zero; (b) variation of velocity with time, a cosine curve starting at its maximum; (c) variation of acceleration with time, an inverted sine curve starting at zero; all three drawn one below the other on a common time axis marked in quarters of \(T\).
Read the table across each row and the physics comes out at once. Wherever \(x\) reaches \(\pm A\), the velocity is zero and the acceleration is at its largest. Wherever \(x\) is zero, the velocity is at its largest and the acceleration is zero.
(b) Particle Performing S.H.M., Starting from the Positive Extreme Position
As the particle starts from the positive extreme position, as shown in Fig. 5.7, the initial phase is \(\phi = \dfrac{\pi}{2}\).
\[\therefore \quad \text{displacement}, \quad x = A\sin\left(\omega t + \frac{\pi}{2}\right) = A\cos\omega t\]
Differentiating once with respect to time gives the velocity,
\[\text{Velocity}, \quad \mathrm{v} = \frac{dx}{dt} = \frac{d}{dt}\left(A\cos\omega t\right) = -A\omega\sin\omega t\]
and differentiating once more gives the acceleration,
\[\text{Acceleration}, \quad a = \frac{d\mathrm{v}}{dt} = \frac{d}{dt}\left(-A\omega\sin\omega t\right) = -A\omega^{2}\cos\omega t\]
| \((t)\) | 0 | \(T/4\) | \(T/2\) | \(3T/4\) | \(T\) | \(5T/4\) |
|---|---|---|---|---|---|---|
| \((\theta)^{*}\) | \(\dfrac{\pi}{2}\) | \(\pi\) | \(\dfrac{3\pi}{2}\) | \(2\pi\) | \(\dfrac{5\pi}{2}\) | \(3\pi\) |
| \((x)\) | \(A\) | 0 | \(-A\) | 0 | \(A\) | 0 |
| \((\mathrm{v})\) | 0 | \(-A\omega\) | 0 | \(A\omega\) | 0 | \(-A\omega\) |
| \((a)\) | \(-A\omega^{2}\) | 0 | \(A\omega^{2}\) | 0 | \(-A\omega^{2}\) | 0 |
* phase \(\theta = \omega t + \phi\)
Diagram
Fig. 5.7 – (a) Variation of displacement with time, a cosine curve starting at \(+A\); (b) variation of velocity with time, an inverted sine curve starting at zero; (c) variation of acceleration with time, an inverted cosine curve starting at \(-A\omega^{2}\); all three drawn one below the other on a common time axis marked in quarters of \(T\).
Compare Fig. 5.7 with Fig. 5.6. The curves have identical shapes and identical maximum values; every curve has merely been slid along the time axis by a quarter of a period. This is the graphical statement of what was said in Section 5.5 — the choice between a sine and a cosine records only where we started counting time, not any difference in the physical motion.
Conclusions from the Graphs
- Displacement, velocity and acceleration of S.H.M. are periodic functions of time.
- The displacement-time curve and the acceleration-time curve are sine curves, and the velocity-time curve is a cosine curve.
- There is a phase difference of \(\dfrac{\pi}{2}\) radian between displacement and velocity.
- There is a phase difference of \(\dfrac{\pi}{2}\) radian between velocity and acceleration.
- There is a phase difference of \(\pi\) radian between displacement and acceleration.
- Shapes of all the curves get repeated after \(2\pi\) radian, or after a time \(T\).
The last two conclusions carry the most physics. A phase difference of \(\pi\) radian between displacement and acceleration means the two curves are exact mirror images about the time axis, which is simply the graph of \(a = -\omega^{2}x\): whenever the particle is on one side, its acceleration points to the other.
The phase difference of \(\dfrac{\pi}{2}\) radian between displacement and velocity means the velocity always runs a quarter of a cycle ahead of the displacement, peaking where the displacement crosses zero.
5.10 Composition of two S.H.M.s having same period and along the same path
The Problem
A particle need not be subjected to only one S.H.M. at a time. Two separate agencies may act on it together, each trying to make it perform its own simple harmonic motion.
Consider a particle subjected simultaneously to two S.H.M.s having the same period and along the same path — let it be along the x-axis — but of different amplitudes and initial phases.
The resultant displacement at any instant is equal to the vector sum of its displacements due to both the S.H.M.s at that instant. Since both motions lie along the same straight line, this vector sum reduces to an ordinary algebraic sum, with signs taken care of automatically.
The question we must answer is whether the combined motion is still an S.H.M., and if so, what its amplitude and initial phase are.
Adding the Two Motions
Equations of displacement of the two S.H.M.s along the same straight line (x-axis) are
\[x_1 = A_1\sin\left(\omega t + \phi_1\right) \quad \text{and} \quad x_2 = A_2\sin\left(\omega t + \phi_2\right)\]
where \(A_1\) and \(A_2\) are the two amplitudes in metres, \(\phi_1\) and \(\phi_2\) the two initial phases in radians, and \(\omega\) the common angular frequency in rad/s.
The resultant displacement (\(x\)) at any instant (\(t\)) is given by \(x = x_1 + x_2\)
\[x = A_1\sin\left(\omega t + \phi_1\right) + A_2\sin\left(\omega t + \phi_2\right)\]
Expanding each sine using \(\sin\left(P + Q\right) = \sin P\cos Q + \cos P \sin Q\),
\[\therefore \quad x = A_1\sin\omega t\cos\phi_1 + A_1\cos\omega t\sin\phi_1 + A_2\sin\omega t\cos\phi_2 + A_2\cos\omega t\sin\phi_2\]
Now \(A_1\), \(A_2\), \(\phi_1\) and \(\phi_2\) are constants, and \(\omega t\) is the only variable. Thus, collecting the constants together,
\[x = \left(A_1\cos\phi_1 + A_2\cos\phi_2\right)\sin\omega t + \left(A_1\sin\phi_1 + A_2\sin\phi_2\right)\cos\omega t\]
The expression has now been reduced to a fixed constant multiplying \(\sin\omega t\), plus another fixed constant multiplying \(\cos\omega t\). Anything of that form can be rewritten as a single sine, and the next step performs exactly that conversion.
As \(A_1\), \(A_2\), \(\phi_1\) and \(\phi_2\) are constants, we can combine them in terms of another two convenient constants \(R\) and \(\delta\) as
\[R\cos\delta = A_1\cos\phi_1 + A_2\cos\phi_2 \qquad \text{--- (5.17)}\]
\[\text{and} \quad R\sin\delta = A_1\sin\phi_1 + A_2\sin\phi_2 \qquad \text{--- (5.18)}\]
Substituting these into the expression for \(x\),
\[\therefore \quad x = R\left(\sin\omega t\cos\delta + \cos\omega t\sin\delta\right)\]
\[\therefore \quad x = R\sin\left(\omega t + \delta\right)\]
What the Result Means
This is the equation of an S.H.M. of the same angular frequency — hence the same period — but of amplitude \(R\) and initial phase \(\delta\).
It shows that the combination (superposition) of two linear S.H.M.s of the same period and occurring along the same path is also an S.H.M.
This is a strong statement. The family of S.H.M.s of a given period is closed under addition — combining its members can never produce anything outside the family.
Diagram
Fig. 5.0f – A phasor diagram: two arrows of lengths \(A_1\) and \(A_2\) drawn from a common origin at angles \(\phi_1\) and \(\phi_2\) to the reference axis, completing a parallelogram whose diagonal of length \(R\) makes an angle \(\delta\) with the same axis.
The two constants \(R\) and \(\delta\) have a clear geometrical meaning, shown in Fig. 5.0f. Equations (5.17) and (5.18) are exactly the rules for adding two arrows of lengths \(A_1\) and \(A_2\), drawn at angles \(\phi_1\) and \(\phi_2\). The resultant arrow has length \(R\) and direction \(\delta\). This is the reference circle of Section 5.7 doing useful work: adding oscillations becomes the same problem as adding vectors.
Resultant Amplitude
To find \(R\), we square and add Eqs. (5.17) and (5.18), because \(\sin^{2}\delta + \cos^{2}\delta = 1\) removes \(\delta\) entirely:
\[R = \sqrt{\left(R\sin\delta\right)^{2} + \left(R\cos\delta\right)^{2}}\]
Substituting from Eq. (5.17) and Eq. (5.18), we get
\[R^{2} = A_1^{2} + A_2^{2} + 2A_1A_2\cos\left(\phi_1 - \phi_2\right)\]
\[\therefore \quad R = \sqrt{A_1^{2} + A_2^{2} + 2A_1A_2\cos\left(\phi_1 - \phi_2\right)} \qquad \text{--- (5.19)}\]
Notice that only the difference of the two initial phases appears, never \(\phi_1\) or \(\phi_2\) separately. What matters physically is how far one oscillation lags behind the other, not where either of them was measured from.
Initial Phase of the Resultant Motion
Dividing Eq. (5.18) by Eq. (5.17), the \(R\) cancels and we get
\[\frac{R\sin\delta}{R\cos\delta} = \frac{A_1\sin\phi_1 + A_2\sin\phi_2}{A_1\cos\phi_1 + A_2\cos\phi_2}\]
\[\therefore \quad \tan\delta = \frac{A_1\sin\phi_1 + A_2\sin\phi_2}{A_1\cos\phi_1 + A_2\cos\phi_2}\]
\[\therefore \quad \delta = \tan^{-1}\left(\frac{A_1\sin\phi_1 + A_2\sin\phi_2}{A_1\cos\phi_1 + A_2\cos\phi_2}\right) \qquad \text{--- (5.20)}\]
The initial phase \(\delta\) of the resultant is also called its epoch, a term that appears in some examination questions.
Special Cases
Three particular phase differences occur so often that their results are worth memorising.
(i) If the two S.H.M.s are in phase, \(\left(\phi_1 - \phi_2\right) = 0^{\circ}\), \(\therefore\) \(\cos\left(\phi_1 - \phi_2\right) = 1\).
\[\therefore \quad R = \sqrt{A_1^{2} + A_2^{2} + 2A_1A_2} = \pm\left(A_1 + A_2\right)\]
Further, if \(A_1 = A_2 = A\), we get \(R = 2A\).
This is the case of maximum reinforcement: the two motions pull the particle the same way at every instant, and the amplitudes simply add.
(ii) If the two S.H.M.s are \(90^{\circ}\) out of phase, \(\left(\phi_1 - \phi_2\right) = 90^{\circ}\) \(\therefore\) \(\cos\left(\phi_1 - \phi_2\right) = 0\).
\[\therefore \quad R = \sqrt{A_1^{2} + A_2^{2}}\]
Further, if \(A_1 = A_2 = A\), we get \(R = \sqrt{2}\,A\).
Here the two amplitudes combine exactly as two perpendicular vectors would, which is what the phasor picture of Fig. 5.0f leads you to expect.
(iii) If the two S.H.M.s are \(180^{\circ}\) out of phase, \(\left(\phi_1 - \phi_2\right) = 180^{\circ}\) \(\therefore\) \(\cos\left(\phi_1 - \phi_2\right) = -1\).
\[\therefore \quad R = \sqrt{A_1^{2} + A_2^{2} - 2A_1A_2} \quad \therefore \quad R = \left|A_1 - A_2\right|\]
Further, if \(A_1 = A_2 = A\), we get \(R = 0\).
The last result is striking: two oscillations of equal strength, exactly opposed, leave the particle standing still. Nothing has been destroyed — at every instant, each motion is asking the particle to go one way while the other asks it to go the other way by the same amount.
5.11 Energy of a Particle Performing S.H.M.
Why an Oscillator Has Both Kinds of Energy
While performing an S.H.M., the particle possesses speed, and hence kinetic energy, at all the positions except at the extreme positions.
In spite of the presence of a restoring force — everywhere except at the mean position — the particle nevertheless occupies various positions away from the mean. This is an indication that work is done against that force, and that the system therefore has potential energy stored in it.
The physical form of this stored energy depends on the system: it is elastic in the case of a spring, gravitational for a pendulum, magnetic for a magnet, and so on. The mathematics that follows is the same in every case.
Total energy of the particle performing an S.H.M. is thus the sum of its kinetic and potential energies.
Consider a particle of mass \(m\), performing a linear S.H.M. along the path MN about the mean position O. At a given instant, let the particle be at P, at a distance \(x\) from O, as shown in Fig. 5.8.
Diagram
Fig. 5.8 – Energy in an S.H.M.: a straight path from N through the mean position O to M, with the particle at a point P at distance \(x\) from O, and a small further displacement \(dx\) marked beyond P.
Kinetic Energy
Velocity of the particle in S.H.M. is given as
\[\mathrm{v} = \omega\sqrt{A^{2} - x^{2}} = A\omega\cos\left(\omega t + \phi\right)\]
where \(x\) is the displacement of the particle performing S.H.M. and \(A\) is the amplitude of S.H.M.
Substituting this velocity into \(E_k = \dfrac{1}{2}m\mathrm{v}^{2}\), and using \(k = m\omega^{2}\),
\[E_k = \frac{1}{2}m\omega^{2}\left(A^{2} - x^{2}\right) = \frac{1}{2}k\left(A^{2} - x^{2}\right) \qquad \text{--- (5.21)}\]
This is the kinetic energy at displacement \(x\), measured in joules (J).
At time \(t\), it is
\[E_k = \frac{1}{2}m\mathrm{v}^{2} = \frac{1}{2}mA^{2}\omega^{2}\cos^{2}\left(\omega t + \phi\right)\]
\[= \frac{1}{2}kA^{2}\cos^{2}\left(\omega t + \phi\right) \qquad \text{--- (5.22)}\]
Thus, with time, it varies as \(\cos^{2}\theta\).
Equation (5.21) already tells us where the kinetic energy is large and where it is small: it falls to zero at \(x = \pm A\) and reaches its greatest value at \(x = 0\).
Potential Energy
To find the potential energy we ask how much work an outside agency must do to bring the particle from the mean position out to P, pushing against the restoring force all the way.
The restoring force acting on the particle at point P is given by \(f = -kx\), where \(k\) is the force constant. Suppose that the particle is displaced further by an infinitesimal displacement \(dx\) against the restoring force \(f\).
The external work done (\(dW\)) during this displacement is
\[dW = f\left(-dx\right) = -kx\left(-dx\right) = kx\,dx\]
We must integrate rather than simply multiply force by distance, because the force is not constant — it grows steadily as the particle moves outward.
The total work done on the particle to displace it from O to P is given by
\[W = \int_{0}^{x} dW = \int_{0}^{x} kx\,dx = \frac{1}{2}kx^{2}\]
This should be the potential energy (P.E.) \(E_p\) of the particle at displacement \(x\).
\[\therefore \quad E_P = \frac{1}{2}kx^{2} = \frac{1}{2}m\omega^{2}x^{2} \qquad \text{--- (5.23)}\]
At time \(t\), it is
\[E_P = \frac{1}{2}kx^{2} = \frac{1}{2}kA^{2}\sin^{2}\left(\omega t + \phi\right)\]
\[= \frac{1}{2}mA^{2}\omega^{2}\sin^{2}\left(\omega t + \phi\right) \qquad \text{--- (5.23a)}\]
Thus, with time, it varies as \(\sin^{2}\theta\).
Total Energy and Its Conservation
The total energy of the particle is the sum of its kinetic energy and potential energy.
\[\therefore \quad E = E_k + E_p\]
Using Eq. (5.21) and Eq. (5.23), we get
\[E = \frac{1}{2}m\omega^{2}\left(A^{2} - x^{2}\right) + \frac{1}{2}m\omega^{2}x^{2}\]
The two terms in \(x^{2}\) cancel exactly, and this cancellation is the whole point of the result:
\[E = \frac{1}{2}m\omega^{2}A^{2} = \frac{1}{2}kA^{2} = \frac{1}{2}m\left(\mathrm{v}_{max}\right)^{2} \qquad \text{--- (5.24)}\]
This expression gives the total energy of the particle at point P. As \(m\), \(\omega\) and \(A\) are constant, the total energy of the particle at any point P is constant, that is, independent of \(x\) and \(t\). In other words, the energy is conserved in S.H.M.
If \(n\) is the frequency of S.H.M., \(\omega = 2\pi n\). Using this in Eq. (5.24), we get
\[E = \frac{1}{2}m\left(2\pi n\right)^{2}A^{2} = 2\pi^{2}n^{2}A^{2}m\]
\[= 2\pi^{2}m\frac{A^{2}}{T^{2}} \qquad \text{--- (5.25)}\]
Thus, the total energy in S.H.M. is directly proportional to (a) the mass of the particle, (b) the square of the amplitude, (c) the square of the frequency, (d) the force constant, and inversely proportional to the square of the period.
The dependence on the squares of amplitude and frequency is heavily used in numerical questions. Doubling the amplitude alone multiplies the energy by four; doubling the frequency alone also multiplies it by four; doing both multiplies it by sixteen.
Special Cases
(i) At the mean position, \(x = 0\) and velocity is maximum.
Hence \(E = \left(E_k\right)_{max} = \dfrac{1}{2}m\omega^{2}A^{2}\) and potential energy \(\left(E_p\right)_{min} = 0\)
(ii) At the extreme positions, the velocity of the particle is zero and \(x = \pm A\).
Hence \(E = \left(E_p\right)_{max} = \dfrac{1}{2}m\omega^{2}A^{2}\) and kinetic energy \(\left(E_k\right)_{min} = 0\)
As the particle oscillates, the energy changes between kinetic and potential. At the mean position, the energy is entirely kinetic; while at the extreme positions, it is entirely potential. At other positions the energy is partly kinetic and partly potential. However, the total energy is always conserved.
(iii) If \(K.E. = P.E.\),
\[\frac{1}{2}m\omega^{2}\left(A^{2} - x^{2}\right) = \frac{1}{2}m\omega^{2}x^{2} \quad \therefore \quad x = \frac{\pm A}{\sqrt{2}}\]
Thus at \(x = \dfrac{\pm A}{\sqrt{2}}\), the \(K.E. = P.E. = \dfrac{E}{2}\) for a particle performing linear S.H.M.
Note that this equal division happens at about \(0.707A\), not at the midpoint \(A/2\) — a point on which students are frequently caught out.
(iv) At \(x = \dfrac{\pm A}{2}\),
\[P.E. = \frac{1}{2}kx^{2} = \frac{1}{4}\left(\frac{1}{2}kA^{2}\right) = \frac{E}{4}\]
\[\therefore \quad K.E. = 3\,P.E.\]
Thus, at \(x = \dfrac{\pm A}{2}\), the energy is 25% potential and 75% kinetic.
The variation of K.E. and P.E. with displacement in S.H.M. is shown in Fig. 5.9.
Diagram
Fig. 5.9 – Energy in S.H.M.: potential energy plotted against displacement as an upward parabola with its minimum at \(x = 0\), kinetic energy as a downward parabola with its maximum at \(x = 0\), and the total energy as a horizontal straight line at \(\frac{1}{2}kA^{2}\), the two curves crossing at \(x = \pm A/\sqrt{2}\).
Fig. 5.9 summarises the whole section in one picture. The two parabolas are mirror images, they cross at half the total energy, and their sum is the flat line at the top — the graphical statement of conservation of energy.
5.12 Simple Pendulum
What a Simple Pendulum Is
The pendulum is the oldest and most familiar oscillator, and it is worth defining carefully before we analyse it.
An ideal simple pendulum is a heavy particle suspended by a massless, inextensible, flexible string from a rigid support.
No such object exists, of course, so in the laboratory we use the closest practical approximation to it.
A practical simple pendulum is a small heavy (dense) sphere, called the bob, suspended by a light and inextensible string from a rigid support.
Each word in these definitions matters. The string must be inextensible, or the length would change during the swing; it must be light, or its own mass would take part in the motion; and the support must be rigid, or energy would leak away into the support.
The distance between the point of suspension and the centre of gravity of the bob — the point of oscillation — is called the length of the pendulum, denoted by \(L\) and measured in metres. Note that it is measured to the centre of the bob, not to its top, which is a common source of error in experiments.
Diagram
Fig. 5.10 – Simple pendulum: a bob of mass \(m\) hanging by a string of length \(L\) from a rigid support at O, shown displaced through a small angle \(\theta\), with the tension \(T'\) along the string and the weight \(mg\) resolved into components \(mg\cos\theta\) along the string and \(mg\sin\theta\) perpendicular to it.
Let \(m\) be the mass of the bob and \(T'\) be the tension in the string. The pendulum remains in equilibrium in the position OA, with the centre of gravity of the bob vertically below the point of suspension O.
If now the pendulum is displaced through a small angle \(\theta\), called the angular amplitude, and released, it begins to oscillate on either side of the mean (equilibrium) position in a single vertical plane.
We shall now show that the bob performs S.H.M. about the mean position for small angular amplitude \(\theta\).
Finding the Restoring Force
In the displaced position (extreme position), two forces are acting on the bob, as shown in Fig. 5.10:
- Force \(T'\) due to tension in the string, directed along the string, towards the support, and
- Weight \(mg\), in the vertically downward direction.
At the extreme positions, there should not be any net force along the string, since the string cannot stretch. Only a component of \(mg\) can balance the force due to tension.
Thus, weight \(mg\) is resolved into two components:
- The component \(mg\cos\theta\) along the string, which is balanced by the tension \(T'\), and
- The component \(mg\sin\theta\) perpendicular to the string, which is the restoring force acting on mass \(m\), tending to return it to the equilibrium position.
\[\therefore \quad \text{Restoring force,} \quad F = -mg\sin\theta \qquad \text{--- (5.26)}\]
The negative sign records that this component always acts towards the mean position, opposite to the displacement.
At this stage the motion is not yet simple harmonic, because \(F\) depends on \(\sin\theta\) rather than on \(\theta\) itself. The next step is what turns it into S.H.M., and it is an approximation.
The Small-Angle Approximation
As \(\theta\) is very small (\(\theta < 10^{\circ}\)), we can write \(\sin\theta \approx \theta^{c}\), where \(\theta\) must be expressed in radians.
\[\therefore \quad F = -mg\theta\]
From Fig. 5.10, for a small angle, \(\theta = \dfrac{x}{L}\), where \(x\) is the displacement of the bob from the mean position, measured along the arc.
\[\therefore \quad F = -\frac{mg}{L}x \qquad \text{--- (5.27)}\]
As \(m\), \(g\) and \(L\) are constant, \(F \propto -x\).
Thus, for small displacement, the restoring force is directly proportional to the displacement and is oppositely directed. Hence the bob of a simple pendulum performs linear S.H.M. for small amplitudes.
Comparing Eq. (5.27) with \(f = -kx\), we can see that the combination \(\dfrac{mg}{L}\) plays exactly the role that the force constant \(k\) played for the spring.
Period of a Simple Pendulum
From Eq. (5.15), the period \(T\) of oscillation of a pendulum can be given as
\[T = \frac{2\pi}{\omega} = \frac{2\pi}{\sqrt{\text{acceleration per unit displacement}}}\]
So all we need is the acceleration per unit displacement. Using Eq. (5.27),
\[F = -\frac{mg}{L}x\]
\[\therefore \quad ma = -\frac{mg}{L}x\]
The mass \(m\) cancels from both sides — a cancellation with far-reaching consequences, as we shall see.
\[\therefore \quad a = -\frac{g}{L}x \quad \therefore \quad \frac{a}{x} = \frac{g}{L} \quad \text{(in magnitude)}\]
Substituting in the expression for \(T\), we get
\[T = 2\pi\sqrt{\frac{L}{g}} \qquad \text{--- (5.28)}\]
Here \(L\) is the length of the pendulum in m, \(g\) is the acceleration due to gravity in m/s², and \(T\) is the period in s.
Equation (5.28) gives the expression for the time period of a simple pendulum. However, while deriving the expression, the following assumptions are made:
- The amplitude of oscillations is very small (at least 20 times smaller than the length).
- The length of the string is large, and
- During the oscillations, the bob moves along a single vertical plane.
These conditions are exam-critical. The formula is simply not valid if any of them is violated.
The frequency of oscillation \(n\) of the simple pendulum is
\[n = \frac{1}{T} = \frac{1}{2\pi}\sqrt{\frac{g}{L}} \qquad \text{--- (5.29)}\]
Laws of the Simple Pendulum
From Eq. (5.28), we can conclude the following for a simple pendulum.
- The period of a simple pendulum is directly proportional to the square root of its length.
- The period of a simple pendulum is inversely proportional to the square root of the acceleration due to gravity.
- The period of a simple pendulum does not depend on its mass.
- The period of a simple pendulum does not depend on its amplitude (for small amplitude).
These conclusions are also called the laws of simple pendulum.
Law (c) follows from the cancellation of \(m\) noted above. A heavier bob is pulled back harder, but it also has proportionally more inertia to be moved, and the two effects cancel exactly — the same reason all bodies fall with the same acceleration.
5.12.1 Second’s Pendulum
A simple pendulum whose period is two seconds is called a second’s pendulum.
The name can be misleading at first. Its period is 2 s, which means each single swing from one extreme to the other takes 1 s, and it is this one-second beat that gives the pendulum its name and its usefulness in clocks.
Its length follows from the general formula. From
\[\text{Period} \ T = 2\pi\sqrt{\frac{L}{g}}\]
\[\therefore \quad \text{For a second's pendulum,} \quad 2 = 2\pi\sqrt{\frac{L_s}{g}}\]
where \(L_s\) is the length of the second’s pendulum, having period \(T = 2\) s.
Squaring both sides and rearranging,
\[\therefore \quad L_s = \frac{g}{\pi^{2}} \qquad \text{--- (5.30)}\]
Using this relation, we can find the length of a second’s pendulum at a place, if we know the acceleration due to gravity at that place.
The relation can equally be read the other way round. Experimentally, if \(L_s\) is known, it can be used to determine the acceleration due to gravity \(g\) at that place.
How Small Must the Amplitude Be?
To answer this, it would be better to measure the time period for different angular amplitudes and compare them.
Let \(T_0 = 2\pi\sqrt{\dfrac{L}{g}}\) be the period for (ideally) very small angular amplitude, and let \(T_\theta\) be the period at a higher angular amplitude \(\theta\). Experimentally determined values of the ratio \(\dfrac{T_\theta}{T_0}\) are as shown in the table below.
| \(\theta\) | 20° | 45° | 50° | 70° | 90° |
|---|---|---|---|---|---|
| \(\dfrac{T_\theta}{T_0}\) | 1.02 | 1.04 | 1.05 | 1.10 | 1.18 |
It shows that the error in the time period is about 2% at an amplitude of 20°, 5% at an amplitude of 50°, 10% at an amplitude of 70° and 18% at an amplitude of 90°.
Thus, the recommended maximum angular amplitude is less than 20°. It also helps us in restricting the oscillations in a single vertical plane, which was assumption (iii) of the derivation.
Conical Pendulum Compared with Simple Pendulum
You have already studied the conical pendulum, in which the bob moves in a horizontal circle rather than swinging in a vertical plane. The two arrangements look similar but behave very differently, and the following comparison brings out every point of difference.
| Conical pendulum | Simple pendulum | |
|---|---|---|
| 1 | Trajectory and the plane of the motion of the bob is a horizontal circle. | Trajectory and the plane of motion of the bob is part of a vertical circle. |
| 2 | K.E. and gravitational P.E. are constant. | K.E. and gravitational P.E. are interconverted and their sum is conserved. |
| 3 | Horizontal component of the force due to tension is the necessary centripetal force (governing force). | Tangential component of the weight is the governing force for the energy conversions during the motion. |
| 4 | Period, \(T = 2\pi\sqrt{\dfrac{L\cos\theta}{g}}\) | Period, \(T = 2\pi\sqrt{\dfrac{L}{g}}\) |
| 5 | String always makes a fixed angle with the horizontal and can never be horizontal. | With large amplitude, the string can be horizontal at some instances. |
| 6 | During the discussion for both, we have ignored the stretching of the string and the energy spent for it. However, the string is always stretched, otherwise it will never have tension (except at the extreme positions of the simple pendulum). Also, non-conservative forces like air resistance are neglected. |
Row 2 is the sharpest distinction. In a conical pendulum the bob stays at a fixed height and moves at a fixed speed, so neither form of energy changes. In a simple pendulum the bob rises and falls, and the continuous exchange between the two forms of energy is precisely what the oscillation consists of.
5.13 Angular S.H.M. and its Differential Equation
Oscillations That Twist Instead of Slide
So far, every oscillator we have studied moved along a line. There is a second family in which the body does not travel anywhere at all — it merely turns to and fro about an axis.
Figure 5.11 shows a metallic disc attached centrally to a thin wire, preferably a nylon or metallic wire, hanging from a rigid support.
Diagram
Fig. 5.11 – Torsional (angular) oscillations: a horizontal metallic disc suspended centrally from a rigid support by a thin vertical wire, with a curved double-headed arrow showing the disc turning clockwise and anticlockwise about the wire as axis.
If the disc is slightly twisted about the axis along the wire and released, it performs rotational motion partly in the clockwise and partly in the anticlockwise (or opposite) sense. Such oscillations are called angular oscillations or torsional oscillations.
Notice how closely this parallels the spring-and-block system. There, a stretched spring pulled the block back with a force; here, a twisted wire turns the disc back with a torque. Torque is the rotational counterpart of force, and it is measured in newton metre (N m).
This motion is governed by the restoring torque in the wire, which is always opposite to the angular displacement. If its magnitude happens to be proportional to the corresponding angular displacement, we can call the motion an angular S.H.M.
The word “happens” in the sentence above is deliberate. Not every twisting oscillation is simple harmonic; it becomes so only when the proportionality holds, which for a wire means keeping the twist small.
The Period of Angular S.H.M.
The result of Section 5.6.2 carries over directly, with every linear quantity replaced by its rotational counterpart. The time period \(T\) of angular S.H.M. is given by
\[T = \frac{2\pi}{\omega} = \frac{2\pi}{\sqrt{\text{angular acceleration per unit angular displacement}}}\]
This is the same rule as before — displacement has become angular displacement, and acceleration has become angular acceleration.
Setting Up the Differential Equation
Thus, for the angular S.H.M. of a body, the restoring torque \(\tau\) acting upon it, for angular displacement \(\theta\), is
\[\tau \propto -\theta \quad \text{or} \quad \tau = -c\theta \qquad \text{--- (5.31)}\]
The constant of proportionality \(c\) is the restoring torque per unit angular displacement, measured in N m per radian. It plays exactly the part that the force constant \(k\) played in linear S.H.M., and for a wire it depends on the material of the wire, its length and its thickness.
If \(I\) is the moment of inertia of the body, the torque acting on the body is given by
\[\tau = I\alpha\]
where \(\alpha\) is the angular acceleration in rad/s², and \(I\) is measured in kg m². This relation is the rotational form of Newton’s second law: moment of inertia measures a body’s reluctance to have its rotation changed, just as mass measures its reluctance to have its motion changed.
Using this in Eq. (5.31) we get
\[I\alpha = -c\theta\]
Writing the angular acceleration as the second derivative of the angular displacement with respect to time, \(\alpha = \dfrac{d^{2}\theta}{dt^{2}}\), and bringing both terms to one side,
\[\therefore \quad I\frac{d^{2}\theta}{dt^{2}} + c\theta = 0 \qquad \text{--- (5.32)}\]
This is the differential equation for angular S.H.M. Compare it with Eq. (5.5): the two have exactly the same structure, with \(I\) in place of \(m\), \(c\) in place of \(k\), and \(\theta\) in place of \(x\).
From this equation, the angular acceleration \(\alpha\) can be written as
\[\alpha = \frac{d^{2}\theta}{dt^{2}} = -\frac{c}{I}\theta\]
Since \(c\) and \(I\) are constants, the angular acceleration \(\alpha\) is directly proportional to \(\theta\) and its direction is opposite to that of the angular displacement. Hence, this oscillatory motion is called angular S.H.M.
Comparing with \(\alpha = -\omega^{2}\theta\), we can identify \(\omega^{2} = \dfrac{c}{I}\), and therefore the period of the torsional oscillator is \(T = 2\pi\sqrt{\dfrac{I}{c}}\).
5.13.1 Magnet Vibrating in Uniform Magnetic Field
Why a Suspended Magnet Oscillates
A bar magnet in a magnetic field provides a second, very useful example of angular S.H.M., and one that is used in the laboratory to measure magnetic quantities.
If a bar magnet is freely suspended in the plane of a uniform magnetic field, it remains in equilibrium with its axis parallel to the direction of the field. This is its mean position, in the same sense in which a hanging pendulum bob has a mean position.
If it is given a small angular displacement \(\theta\) — about an axis passing through its centre, perpendicular both to itself and to the field — and released, it performs angular oscillations, as shown in Fig. 5.12.
Diagram
Fig. 5.12 – Magnet vibrating in a uniform magnetic field: a bar magnet suspended at its centre, shown deflected through a small angle \(\theta\) from the field direction, with the uniform field lines drawn horizontally and a curved arrow indicating the restoring torque.
Let \(\mu\) be the magnetic dipole moment of the magnet, measured in A m², and \(B\) the magnetic field, measured in tesla (T) or Wb/m². In the deflected position, a restoring torque acts on the magnet that tends to bring it back to its equilibrium position.
Showing That the Motion Is Angular S.H.M.
The magnitude of this torque is
\[\tau = \mu B\sin\theta\]
This is the standard expression for the torque on a magnetic dipole in a field, and it is largest when the magnet lies across the field and zero when it lies along the field.
If \(\theta\) is small, \(\sin\theta \approx \theta^{c}\)
\[\therefore \quad \tau = \mu B\theta\]
This is the same small-angle approximation used for the simple pendulum, and it plays the same role here: it converts a sine into a plain angle and so makes the torque proportional to the displacement.
For a clockwise angular displacement \(\theta\), the restoring torque is in the anticlockwise direction, which is what makes it a restoring torque and requires the negative sign below.
\[\therefore \quad \tau = I\alpha = -\mu B\theta\]
where \(I\) is the moment of inertia of the bar magnet and \(\alpha\) is its angular acceleration.
\[\therefore \quad \alpha = -\frac{\mu B}{I}\theta \qquad \text{--- (5.33)}\]
Since \(\mu\), \(B\) and \(I\) are constants, Eq. (5.33) shows that the angular acceleration is directly proportional to the angular displacement and directed opposite to the angular displacement. Hence the magnet performs angular S.H.M.
Period of the Vibrating Magnet
The period of vibrations of the magnet is given by
\[T = \frac{2\pi}{\omega} = \frac{2\pi}{\sqrt{\text{angular acceleration per unit angular displacement}}}\]
From Eq. (5.33), the angular acceleration per unit angular displacement is \(\dfrac{\mu B}{I}\), and therefore
\[\therefore \quad T = 2\pi\sqrt{\frac{I}{\mu B}} \qquad \text{--- (5.34)}\]
Equation (5.34) makes physical sense on inspection. A stronger field or a stronger magnet means a larger restoring torque and hence quicker oscillations, so \(\mu\) and \(B\) sit in the denominator. A bulkier magnet is harder to turn, so \(I\) sits in the numerator.
5.14 Damped Oscillations
Why Real Oscillations Die Down
Every oscillation studied so far has been ideal. Once started, it continued for ever with the same amplitude, because the total energy \(\frac{1}{2}kA^{2}\) was conserved. No real oscillation behaves like this.
If the amplitude of oscillations of an oscillator is reduced by the application of an external force, the oscillator and its motion are said to be damped. Periodic oscillations of gradually decreasing amplitude are called damped harmonic oscillations, and the oscillator is called a damped harmonic oscillator.
For example, the motion of a simple pendulum dies out eventually, as air exerts a viscous force on the pendulum and there may also be some friction at the support.
A Controlled Damped Oscillator
To study damping properly, it helps to have an arrangement in which the resistive force can be made large and steady rather than left to stray air currents.
Figure 5.13 shows a block of mass \(m\) that can oscillate vertically on a spring. From the block, a rod extends to a vane that is submerged in a liquid.
Diagram
Fig. 5.13 – A damped oscillator: a block of mass \(m\) hanging from a spring fixed to a rigid support, with a rod extending downward from the block to a flat vane immersed in a vessel of liquid.
As the vane moves up and down, the liquid exerts a drag force on it, and thus on the complete oscillating system. The mechanical energy of the block-spring system decreases with time, as energy is transferred to thermal energy of the liquid and vane.
Nothing is destroyed here. The energy that leaves the oscillation reappears as a very slight warming of the liquid, so the total energy of the wider system is still conserved — it has simply left the oscillator.
The Equation of a Damped Oscillator
The damping force (\(F_d\)) depends on the nature of the surrounding medium and is directly proportional to the speed \(\mathrm{v}\) of the vane and the block.
\[\therefore \quad F_d = -b\mathrm{v}\]
Here \(b\) is the damping constant, measured in N s/m, and the negative sign indicates that \(F_d\) opposes the velocity.
Note the important difference from the restoring force. The restoring force depends on where the block is; the damping force depends on how fast it is moving. A block momentarily at rest at an extreme position feels no damping force at all.
For spring constant \(k\), the force on the block from the spring is
\[F_s = -kx\]
Assuming that the gravitational force on the block is negligible compared to \(F_d\) and \(F_s\), the total force acting on the mass at any time \(t\) is
\[F = F_d + F_s\]
\[\therefore \quad ma = F_d + F_s\]
\[\therefore \quad ma = -b\mathrm{v} - kx\]
\[\therefore \quad ma + b\mathrm{v} + kx = 0\]
Writing \(\mathrm{v} = \dfrac{dx}{dt}\) and \(a = \dfrac{d^{2}x}{dt^{2}}\),
\[\therefore \quad m\frac{d^{2}x}{dt^{2}} + b\frac{dx}{dt} + kx = 0 \qquad \text{--- (5.35)}\]
Compare Eq. (5.35) with Eq. (5.5). One extra term has appeared, the middle one, and it is the only difference between an oscillation that lasts for ever and one that fades away.
The Solution and What It Describes
The solution of Eq. (5.35) describes the motion of the block under the influence of a damping force which is proportional to the speed.
The solution is found to be of the form
\[x = A\,e^{-bt/2m}\cos\left(\omega' t + \phi\right) \qquad \text{--- (5.36)}\]
\(A\,e^{-bt/2m}\) is the amplitude of the damped harmonic oscillations.
Read Eq. (5.36) as two factors doing two separate jobs. The cosine keeps the block oscillating to and fro exactly as before, while the exponential factor steadily shrinks the size of those oscillations.
As shown in the displacement against time graph, Fig. 5.14, the amplitude decreases with time exponentially. The term \(\cos\left(\omega' t + \phi\right)\) shows that the motion is still an S.H.M.
Diagram
Fig. 5.14 – Displacement against time graph for a damped oscillator: an oscillating curve whose successive peaks grow smaller, enclosed between two mirror-image exponential envelope curves \(\pm A e^{-bt/2m}\).
The two smooth curves touching the peaks in Fig. 5.14 are called the envelope of the motion. They are not part of the oscillation itself; they simply record how the amplitude is being squeezed as time passes.
The angular frequency of the damped oscillation is
\[\omega' = \sqrt{\frac{k}{m} - \left(\frac{b}{2m}\right)^{2}}\]
Period of oscillation,
\[T = \frac{2\pi}{\omega'} = \frac{2\pi}{\sqrt{\dfrac{k}{m} - \left(\dfrac{b}{2m}\right)^{2}}}\]
The damping increases the period (slows down the motion) and decreases the amplitude.
5.15 Free Oscillations, Forced Oscillations and Resonance
Free Oscillations and Natural Frequency
If an object is allowed to oscillate or vibrate on its own, it does so with its natural frequency, or with one of its natural frequencies.
For example, if the bob of a simple pendulum of length \(l\) is displaced and released, it will oscillate only with the frequency
\[n = \frac{1}{2\pi}\sqrt{\frac{g}{l}}\]
which is called its natural frequency, and the oscillations are called free oscillations.
The natural frequency is decided entirely by the properties of the oscillator itself — for a pendulum, by its length and the local value of \(g\); for a spring-and-block system, by \(k\) and \(m\). Nothing about how the oscillation was started can change it.
However, by applying a periodic force, the same pendulum can be made to oscillate with a different frequency. The oscillations then will be forced oscillations, and the frequency is called the driver frequency or forced frequency.
The distinction is worth holding on to. In free oscillation the body chooses its own frequency; in forced oscillation the frequency is imposed on it from outside.
An Experiment with Four Pendula
Consider the arrangement shown in Fig. 5.15. There are four pendula tied to a string.
Diagram
Fig. 5.15 – Forced oscillations: four pendula A, B, C and D hanging from a common horizontal string stretched between two supports; A has a solid bob and acts as the driver, C is of the same length as A, B is shorter and D is longer, all three having hollow bobs.
Pendula A and C are of the same length, pendulum B is of shorter length and pendulum D is of longer length.
Pendulum A has a solid rubber ball as its bob and acts as the driver pendulum, or the source pendulum. The other three pendula have hollow rubber balls as their bobs and act as the driven pendula. The bob of A is made heavier deliberately, so that it can go on supplying energy to the others without stopping quickly itself.
As the pendula A and C are of the same lengths, their natural frequencies are the same. Pendulum B has a higher natural frequency, as its length is shorter than that of pendulum A. The natural frequency of pendulum D is less than that of pendula A and C.
Pendulum A is now set into oscillations in a plane perpendicular to the plane of the paper. In the course of time, it will be observed that the other three pendula also start oscillating in the same plane. This happens due to transfer of the vibrational energy through the string.
Oscillations of pendulum A are free oscillations, and those of pendula B, C and D are forced oscillations of the same frequency as that of A. Note carefully that B, C and D all oscillate at A’s frequency, not at their own natural frequencies — this is exactly what “forced” means.
The natural frequency of pendulum C is the same as that of A, as its length is the same as that of A.
Resonance
It can also be seen that among the pendula B, C and D, the pendulum C oscillates with maximum amplitude, and the other two with smaller amplitudes.
As the energy depends upon the amplitude, it is clear that pendulum C has absorbed maximum energy from the source pendulum A, while the other two absorbed less.
This is the whole point of the experiment. It shows that the object C, having the same natural frequency as that of the source, absorbs maximum energy from the source. In such a case, it is said to be in resonance with the source (pendulum A).
For unequal natural frequencies on either side, higher or lower, the energy absorbed — and hence the amplitude — is less.
If the activity is repeated for a set of pendula of different lengths, and the squares of their amplitudes are plotted against their natural frequencies, the plot will be similar to that shown in Fig. 5.16.
Diagram
Fig. 5.16 – Resonant frequency: a graph of \(A^{2}\) on the vertical axis against forced frequency \(f\) on the horizontal axis, showing a single sharp peak at \(f = f_r\), the resonant frequency, and falling away on both sides.
The peak occurs when the forced frequency matches the natural frequency, i.e., at the resonant frequency.
The square of the amplitude is plotted rather than the amplitude itself because, as Section 5.11 showed, the energy of an oscillation is proportional to \(A^{2}\). The height of the curve at any point therefore shows directly how much energy is being absorbed at that frequency.
In the next chapter, on superposition of waves, you will see that most of the traditional musical instruments use the principle of resonance. In the topic of A.C. circuits, the resonance in L.C. circuits is discussed in detail.
5.16 Summary
Periodic and Oscillatory Motion
- Any motion which repeats itself after a definite interval of time is called periodic motion. The time taken for one complete set of movements is called its period.
- Oscillatory (or vibratory) motion is a periodic motion in which a particle repeatedly moves to and fro along the same path about a mean position.
- Every oscillatory motion is periodic, but every periodic motion need not be oscillatory. Circular motion is periodic but not oscillatory.
- Simple harmonic motion is the simplest oscillatory periodic motion. If the path is a straight line it is linear S.H.M.; if the path is an arc of a circle it is angular S.H.M.
Linear S.H.M.
- Linear S.H.M. is the linear periodic motion of a body in which the force (or acceleration) is always directed towards the mean position and its magnitude is proportional to the displacement from the mean position.
- Restoring force: \(f = -kx\), where \(k\) is the force constant in N/m and \(x\) is the displacement from the mean position in m. The negative sign shows that force and displacement are oppositely directed.
- Acceleration: \(a = \dfrac{f}{m} = -\left(\dfrac{k}{m}\right)x\)
- Conditions for S.H.M.: oscillation about a fixed point; net force or acceleration always directed towards that fixed point; the particle returns to it due to the restoring force.
- A harmonic oscillation can be expressed by a single harmonic function such as \(x = a\sin\omega t\) or \(x = a\cos\omega t\). A non-harmonic oscillation cannot, and may be a combination such as \(x = a\sin\omega t + b\sin 2\omega t\).
Differential Equation and Its Solutions
- Differential equation of linear S.H.M.: \(\dfrac{d^{2}x}{dt^{2}} + \omega^{2}x = 0\), with \(\omega^{2} = \dfrac{k}{m}\), where \(\omega\) is the angular frequency in rad/s.
- Acceleration in terms of displacement: \(a = -\omega^{2}x\)
- Velocity in terms of displacement: \(\mathrm{v} = \pm\omega\sqrt{A^{2} - x^{2}}\)
- General displacement at time \(t\): \(x = A\sin\left(\omega t + \phi\right)\)
- Starting from the mean position: \(x = \pm A\sin\omega t\) (positive sign for starting towards the positive side).
- Starting from the extreme position: \(x = \pm A\cos\omega t\) (positive sign for the positive extreme).
- At time \(t\): \(\mathrm{v} = A\omega\cos\left(\omega t + \phi\right)\) and \(a = -A\omega^{2}\sin\left(\omega t + \phi\right)\)
- The instant \(t = 0\) is chosen by the observer; the particle is already in motion.
Extreme Values
| Quantity | At mean position | At extreme position |
|---|---|---|
| Displacement | \(x_{min} = 0\) | \(x_{max} = \pm A\) |
| Velocity | \(\mathrm{v}_{max} = \pm A\omega\) | \(\mathrm{v}_{min} = 0\) |
| Acceleration | \(a_{min} = 0\) | \(a_{max} = \mp\omega^{2}A\) |
Amplitude, Period and Frequency
- The maximum displacement of a particle performing S.H.M. from its mean position is called the amplitude \(A\). The total path length is \(2A\).
- Period: \(T = \dfrac{2\pi}{\omega} = \dfrac{2\pi}{\sqrt{\text{acceleration per unit displacement}}} = 2\pi\sqrt{\dfrac{m}{k}}\)
- Frequency: \(n = \dfrac{1}{T} = \dfrac{\omega}{2\pi} = \dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}\), with SI unit hertz (Hz), and \(\omega = 2\pi n\).
- The period of an S.H.M. does not depend on its amplitude.
Combination of Springs
- Series: \(\dfrac{1}{k_s} = \sum_{i}\dfrac{1}{k_i}\); for two springs \(k_s = \dfrac{k_1k_2}{k_1 + k_2}\); for \(n\) identical springs \(k_s = \dfrac{k}{n}\). A series combination is softer.
- Parallel: \(k_p = \sum_{i}k_i\); for \(m\) identical springs \(k_p = mk\). A parallel combination is stiffer.
Reference Circle and Phase
- The projection of a uniform circular motion on any diameter of its circle is a linear S.H.M., of amplitude equal to the radius and angular frequency equal to the angular velocity of the circular motion.
- Projections: displacement \(y = r\sin\left(\omega t + \phi\right)\), velocity \(\mathrm{v}_y = r\omega\cos\left(\omega t + \phi\right)\), acceleration \(a_y = -r\omega^{2}\sin\left(\omega t + \phi\right) = -\omega^{2}y\).
- Phase is the state of oscillation, given by the phase angle \(\theta = \left(\omega t + \phi\right)\), which fixes the position, the direction of velocity and the oscillation number.
- Key phase values in the first oscillation: \(0\) — mean position moving positive; \(\dfrac{\pi}{2}\) — positive extreme; \(\pi\) — mean position moving negative; \(\dfrac{3\pi}{2}\) — negative extreme. Add \(2\pi\) for each completed oscillation.
Graphical Representation
- Displacement, velocity and acceleration are all periodic functions of time, repeating after \(2\pi\) radian or after a time \(T\).
- Starting from the mean position: \(x = A\sin\omega t\), \(\mathrm{v} = A\omega\cos\omega t\), \(a = -A\omega^{2}\sin\omega t\).
- Starting from the positive extreme position: \(x = A\cos\omega t\), \(\mathrm{v} = -A\omega\sin\omega t\), \(a = -A\omega^{2}\cos\omega t\).
- Phase difference between displacement and velocity is \(\dfrac{\pi}{2}\) radian; between velocity and acceleration is \(\dfrac{\pi}{2}\) radian; between displacement and acceleration is \(\pi\) radian.
Composition of Two S.H.M.s
- Two S.H.M.s of the same period along the same path superpose to give another S.H.M. of the same period: \(x = R\sin\left(\omega t + \delta\right)\).
- Resultant amplitude: \(R = \sqrt{A_1^{2} + A_2^{2} + 2A_1A_2\cos\left(\phi_1 - \phi_2\right)}\)
- Initial phase (epoch): \(\delta = \tan^{-1}\left(\dfrac{A_1\sin\phi_1 + A_2\sin\phi_2}{A_1\cos\phi_1 + A_2\cos\phi_2}\right)\)
- In phase (\(0^{\circ}\)): \(R = A_1 + A_2\), and \(R = 2A\) if \(A_1 = A_2 = A\).
- \(90^{\circ}\) out of phase: \(R = \sqrt{A_1^{2} + A_2^{2}}\), and \(R = \sqrt{2}A\) if \(A_1 = A_2 = A\).
- \(180^{\circ}\) out of phase: \(R = \left|A_1 - A_2\right|\), and \(R = 0\) if \(A_1 = A_2 = A\).
Energy in S.H.M.
- Kinetic energy: \(E_k = \dfrac{1}{2}m\omega^{2}\left(A^{2} - x^{2}\right) = \dfrac{1}{2}k\left(A^{2} - x^{2}\right)\), varying with time as \(\cos^{2}\theta\).
- Potential energy: \(E_p = \dfrac{1}{2}kx^{2} = \dfrac{1}{2}m\omega^{2}x^{2}\), varying with time as \(\sin^{2}\theta\).
- Total energy: \(E = \dfrac{1}{2}m\omega^{2}A^{2} = \dfrac{1}{2}kA^{2} = \dfrac{1}{2}m\left(\mathrm{v}_{max}\right)^{2} = 2\pi^{2}n^{2}A^{2}m = 2\pi^{2}m\dfrac{A^{2}}{T^{2}}\), and is constant — energy is conserved in S.H.M.
- Total energy is directly proportional to the mass, the square of the amplitude, the square of the frequency and the force constant, and inversely proportional to the square of the period.
- At the mean position the energy is entirely kinetic; at the extreme positions it is entirely potential.
- \(K.E. = P.E. = \dfrac{E}{2}\) at \(x = \dfrac{\pm A}{\sqrt{2}}\).
- At \(x = \dfrac{\pm A}{2}\), \(P.E. = \dfrac{E}{4}\) and \(K.E. = 3\,P.E.\), i.e. the energy is 25% potential and 75% kinetic.
Simple Pendulum
- An ideal simple pendulum is a heavy particle suspended by a massless, inextensible, flexible string from a rigid support. Its length is measured from the point of suspension to the centre of gravity of the bob.
- Restoring force: \(F = -mg\sin\theta\); for small \(\theta\) (\(\theta < 10^{\circ}\)), \(\sin\theta \approx \theta\) and \(F = -mg\dfrac{x}{L}\), so the bob performs linear S.H.M.
- Period: \(T = 2\pi\sqrt{\dfrac{L}{g}}\); frequency: \(n = \dfrac{1}{2\pi}\sqrt{\dfrac{g}{L}}\)
- Assumptions: the amplitude is very small (at least 20 times smaller than the length); the length of the string is large; the bob moves in a single vertical plane.
- Laws of simple pendulum: \(T \propto \sqrt{L}\); \(T \propto \dfrac{1}{\sqrt{g}}\); \(T\) is independent of the mass of the bob; \(T\) is independent of the amplitude (for small amplitude).
- The error in period is about 2% at an angular amplitude of 20°, 5% at 50°, 10% at 70° and 18% at 90°; hence the recommended maximum angular amplitude is less than 20°.
- Second’s pendulum: a simple pendulum of period two seconds, with length \(L_s = \dfrac{g}{\pi^{2}}\).
Angular S.H.M.
- Angular S.H.M. is the oscillatory motion of a body in which the torque producing the angular acceleration is directly proportional to the angular displacement and oppositely directed.
- Restoring torque: \(\tau = -c\theta\), where \(c\) is the restoring torque per unit angular displacement.
- Differential equation: \(I\dfrac{d^{2}\theta}{dt^{2}} + c\theta = 0\), giving \(\alpha = -\dfrac{c}{I}\theta\), where \(I\) is the moment of inertia in kg m².
- Period: \(T = \dfrac{2\pi}{\sqrt{\text{angular acceleration per unit angular displacement}}}\)
- Magnet in a uniform magnetic field: \(\tau = \mu B\sin\theta\); for small \(\theta\), \(\alpha = -\dfrac{\mu B}{I}\theta\) and \(T = 2\pi\sqrt{\dfrac{I}{\mu B}}\), where \(\mu\) is the magnetic dipole moment in A m² and \(B\) the magnetic field in T.
Damped Oscillations
- Periodic oscillations of gradually decreasing amplitude are called damped harmonic oscillations.
- Damping force: \(F_d = -b\mathrm{v}\), where \(b\) is the damping constant.
- Equation of motion: \(m\dfrac{d^{2}x}{dt^{2}} + b\dfrac{dx}{dt} + kx = 0\)
- Solution: \(x = A\,e^{-bt/2m}\cos\left(\omega' t + \phi\right)\), with amplitude \(A\,e^{-bt/2m}\) decreasing exponentially with time.
- Angular frequency: \(\omega' = \sqrt{\dfrac{k}{m} - \left(\dfrac{b}{2m}\right)^{2}}\), and period \(T = \dfrac{2\pi}{\omega'}\).
- Damping increases the period and decreases the amplitude.
Free Oscillations, Forced Oscillations and Resonance
- Free oscillations are performed by a body on its own, at its natural frequency, for example \(n = \dfrac{1}{2\pi}\sqrt{\dfrac{g}{l}}\) for a simple pendulum.
- Forced oscillations are performed under an external periodic force, at the driver (forced) frequency.
- Resonance occurs when the forced frequency matches the natural frequency of the body; the body then oscillates with maximum amplitude and absorbs maximum energy from the source.
- A plot of \(A^{2}\) against forced frequency peaks at the resonant frequency.
5.17 Objective Questions
- A particle performs linear S.H.M. starting from the mean position. Its amplitude is \(A\) and time period is \(T\). At the instance when its speed is half the maximum speed, its displacement \(x\) is
- \(\dfrac{\sqrt{3}}{2}A\) (B) \(\dfrac{2}{\sqrt{3}}A\) (C) \(\dfrac{A}{2}\) (D) \(\dfrac{1}{\sqrt{2}}A\)
- A body of mass 1 kg is performing linear S.H.M. Its displacement \(x\) (cm) at \(t\) (second) is given by \(x = 6\sin\left(100t + \dfrac{\pi}{4}\right)\). Maximum kinetic energy of the body is
- 36 J (B) 9 J (C) 27 J (D) 18 J
- The length of second’s pendulum on the surface of earth is nearly 1 m. Its length on the surface of moon should be [Given: acceleration due to gravity (\(g\)) on moon is \(1/6^{\text{th}}\) of that on the earth’s surface]
- 1/6 m (B) 6 m (C) 1/36 m (D) \(\dfrac{1}{\sqrt{6}}\) m
- Two identical springs of constant \(k\) are connected, first in series and then in parallel. A metal block of mass \(m\) is suspended from their combination. The ratio of their frequencies of vertical oscillations will be in a ratio
- 1:4 (B) 1:2 (C) 2:1 (D) 4:1
- The graph shows variation of displacement of a particle performing S.H.M. with time \(t\). Which of the following statements is correct from the graph?
- The acceleration is maximum at time \(T\).
- The force is maximum at time \(3T/4\).
- The velocity is zero at time \(T/2\).
- The kinetic energy is equal to total energy at time \(T/4\).
Diagram
Fig. 5.17 – Displacement-time graph for question 5: a sine curve starting from the origin, with the time axis marked at \(T/4\), \(T/2\), \(3T/4\) and \(T\), and dashed lines dropped from the peaks and troughs.
5.18 Check Your Concepts
Every oscillatory motion is periodic, but every periodic motion need not be oscillatory. Justify this statement with one example of each kind.
Why does a block on a spring not come to rest at the mean position, even though the net force on it there is zero?
The equation \(f = -kx\) contains a negative sign. Explain what would be physically wrong with the motion if the sign were positive instead.
In the differential equation of linear S.H.M., the constant \(k/m\) is written as \(\omega^{2}\) rather than as any ordinary constant. Give one mathematical and one physical reason for this choice.
A student claims that a particle in S.H.M. has its maximum velocity and its maximum acceleration at the same point of the path. Identify the error and state where each maximum actually occurs.
Two identical blocks are attached to identical springs and set oscillating, one with amplitude 2 cm and the other with amplitude 6 cm. Compare their periods, their maximum speeds and their total energies, giving a reason in each case.
Explain why a series combination of two springs is always softer than either spring alone, while a parallel combination is always stiffer.
Why is knowing only the displacement, or only the velocity, insufficient to specify the state of an oscillator? What single quantity conveys the complete state, and why is it allowed to exceed \(2\pi\) radian?
A particle in S.H.M. is described once by a sine function and once by a cosine function. Does this mean the two motions are physically different? Explain.
Sketch, on the same time axis, the displacement and acceleration of a particle performing linear S.H.M., and use the graphs to explain the phase difference of \(\pi\) radian between them.
Two S.H.M.s of equal amplitude and the same period act along the same line on a particle. Under what condition does the particle remain permanently at rest? Explain how this is consistent with conservation of energy.
The potential energy of a particle in S.H.M. varies as \(x^{2}\) and not as \(x\). Explain why, using the way the restoring force changes with displacement.
At what fraction of the amplitude are the kinetic and potential energies of an oscillator equal? Explain why the answer is not \(A/2\), as students often assume.
The mass of the bob cancels out in the derivation of the period of a simple pendulum. Explain the physical reason for this cancellation.
Will a simple pendulum oscillate inside a lift that is (a) at rest, (b) moving up with constant velocity, (c) falling freely? Give a reason for each answer.
A pendulum experiment is performed with an angular amplitude of 70° instead of 10°. State what happens to the measured period and explain why, referring to the approximation used in the derivation.
A second’s pendulum is taken from Mumbai to a high mountain station. Will the clock using it gain or lose time? Justify your answer.
Compare the conical pendulum and the simple pendulum with respect to (a) the plane of motion of the bob, (b) the energy changes taking place, and (c) the force that governs the motion.
State which quantity in angular S.H.M. plays the role of mass, and which plays the role of the force constant, in linear S.H.M. Hence write the period of a torsional oscillator by analogy.
Two identical bar magnets are placed together and suspended in a uniform magnetic field, first with like poles together and then with unlike poles together. In which case is the period longer, and why?
Damping reduces the amplitude of an oscillation. Explain why it also increases the period, rather than leaving the timing unchanged.
A damped oscillator loses mechanical energy with time. Does this violate the law of conservation of energy? Explain where the energy goes.
Why is damping deliberately introduced in the moving coil of an analog ammeter, and what would happen if it were absent?
Distinguish between free oscillations and forced oscillations, and state the frequency with which the body oscillates in each case.
In the experiment with four pendula tied to a common string, explain why only one of the driven pendula oscillates with a large amplitude while the others do not.
In a resonance experiment, why is the square of the amplitude, rather than the amplitude itself, plotted against the forced frequency?
Soldiers are ordered to break step while crossing a bridge. Explain the physics behind this instruction.
Define linear simple harmonic motion.
Using differential equation of linear S.H.M., obtain the expression for (a) velocity in S.H.M., (b) acceleration in S.H.M.
Obtain the expression for the period of a simple pendulum performing S.H.M.
State the laws of simple pendulum.
Prove that under certain conditions a magnet vibrating in uniform magnetic field performs angular S.H.M.
Obtain the expression for the period of a magnet vibrating in a uniform magnetic field and performing S.H.M.
Show that a linear S.H.M. is the projection of a U.C.M. along any of its diameter.
Draw graphs of displacement, velocity and acceleration against phase angle, for a particle performing linear S.H.M. from (a) the mean position (b) the positive extreme position. Deduce your conclusions from the graph.
Deduce the expressions for the kinetic energy and potential energy of a particle executing S.H.M. Hence obtain the expression for total energy of a particle performing S.H.M. and show that the total energy is conserved. State the factors on which total energy depends.
Deduce the expression for period of simple pendulum. Hence state the factors on which its period depends.
5.19 Practice with Numericals
A body of mass 0.4 kg performs linear S.H.M. and experiences a restoring force of 2 N when its displacement from the mean position is 5 cm. Find the force constant, the angular frequency and the period of the motion.
A particle performing linear S.H.M. has an amplitude of 6 cm and an angular frequency of 5 rad/s. Calculate its speed and its acceleration when it is 3.6 cm from the mean position.
The displacement of a particle performing linear S.H.M. is given by \(x = 0.05\sin\left(20t + \dfrac{\pi}{6}\right)\) metre. Write down its amplitude, angular frequency, period, frequency and initial phase, and find its maximum speed and maximum acceleration.
A particle performing linear S.H.M. of period 8 s starts from the positive extreme position. Find the time it takes to reach a point midway between the extreme position and the mean position for the first time.
The speeds of a particle performing linear S.H.M. are 12 cm/s and 5 cm/s at displacements of 5 cm and 12 cm respectively. Find its amplitude and period.
The maximum speed of a particle performing linear S.H.M. is 0.6 m/s and its maximum acceleration is 3.6 m/s². Find the amplitude, the angular frequency and the period of the motion.
A block of mass 0.8 kg is attached to a spring of force constant 200 N/m on a frictionless horizontal surface and is pulled 5 cm aside before release. Find the period, the maximum speed of the block and its speed when it is 3 cm from the mean position.
Two springs of force constants 400 N/m and 600 N/m support a block of mass 3 kg, first in series and then in parallel. Calculate the frequency of vertical oscillations in each arrangement and find the ratio of the two frequencies.
A particle performing linear S.H.M. of amplitude 10 cm is in its third oscillation and is at a displacement of \(+5\) cm, moving towards the positive extreme. Determine its phase angle in radians.
Describe completely the state of oscillation of a particle whose phase angle is \(1470^{\circ}\), stating the oscillation number, the displacement as a fraction of the amplitude, and the direction of motion.
Two S.H.M.s of the same period act along the same straight line on a particle and are represented by \(x_1 = 6\sin\left(3\pi t\right)\) cm and \(x_2 = 8\sin\left(3\pi t + \dfrac{\pi}{2}\right)\) cm. Find the amplitude and the epoch of the resultant S.H.M.
A particle of mass 100 g performs linear S.H.M. of amplitude 8 cm and period 0.4 s. Find its total energy, and its kinetic and potential energies at a displacement of 4 cm.
The total energy of a particle performing linear S.H.M. is 0.08 J and its amplitude is 4 cm. Find the force constant of the oscillator and the displacement at which the kinetic energy is equal to the potential energy.
A body of mass 0.25 kg performs S.H.M. with a frequency of 4 Hz and an amplitude of 5 cm. Calculate its total energy, and state by what factor the energy would change if the frequency alone were doubled.
A simple pendulum has a length of 2.25 m at a place where \(g = 9.8\) m/s². Find its period and frequency, and calculate the new period if the length is reduced to one quarter of its value.
The period of a simple pendulum increases by 20% when its length is increased by 44 cm. Find its original length and its original period. (Take \(g = 9.8\) m/s²)
Find the length of a second’s pendulum at a place where \(g = 9.78\) m/s². If this pendulum is carried to a place where \(g = 9.83\) m/s² without changing its length, find how much time the clock using it would gain or lose per day.
A simple pendulum of length 80 cm carries a bob of mass 40 g. Find the restoring force acting on the bob when its displacement from the mean position is 4 cm. (Take \(g = 9.8\) m/s²)
A metallic disc of moment of inertia \(5 \times 10^{-4}\) kg m² is suspended from a wire and performs torsional oscillations of period 2 s. Calculate the restoring torque per unit angular displacement of the wire, and the maximum restoring torque when the disc is twisted through 30° and released.
A bar magnet of moment of inertia \(2.5 \times 10^{-6}\) kg m² and magnetic moment 1.5 A m² performs angular oscillations in a uniform magnetic field of \(2 \times 10^{-5}\) T. Find its period and the number of oscillations it completes per minute.
Two bar magnets of identical size and mass have magnetic moments of 80 A m² and 20 A m². When suspended together with like poles in the same direction in the Earth’s field, the period of angular oscillation is 4 s. Find the period when their unlike poles are placed together.
A block of mass 0.5 kg oscillates on a spring of force constant 50 N/m in a medium for which the damping constant is 0.10 N s/m. Calculate the angular frequency of the damped oscillation and compare it with the undamped value.
A simple pendulum of length 1.6 m hangs at a place where \(g = 9.8\) m/s². Find its natural frequency. A periodic force is now applied to it, first at 0.4 Hz and then at 0.39 Hz. State which will produce the larger amplitude and give the reason.
At what distance from the mean position is the speed of a particle performing S.H.M. half its maximum speed? Given path length of S.H.M. = 10 cm.
In SI units, the differential equation of an S.H.M. is \(\dfrac{d^{2}x}{dt^{2}} = -36x\). Find its frequency and period.
A needle of a sewing machine moves along a path of amplitude 4 cm with frequency 5 Hz. Find its acceleration \(\dfrac{1}{30}\) s after it has crossed the mean position.
Potential energy of a particle performing linear S.H.M. is \(0.1\pi^{2}x^{2}\) joule. If mass of the particle is 20 g, find the frequency of S.H.M.
The total energy of a body of mass 2 kg performing S.H.M. is 40 J. Find its speed while crossing the centre of the path.
A simple pendulum performs S.H.M. of period 4 seconds. How much time after crossing the mean position will the displacement of the bob be one third of its amplitude?
A simple pendulum of length 100 cm performs S.H.M. Find the restoring force acting on its bob of mass 50 g when the displacement from the mean position is 3 cm.
Find the change in length of a second’s pendulum, if the acceleration due to gravity at the place changes from 9.75 m/s² to 9.80 m/s².
At what distance from the mean position is the kinetic energy of a particle performing S.H.M. of amplitude 8 cm, three times its potential energy?
A particle performing linear S.H.M. of period \(2\pi\) seconds about the mean position O is observed to have a speed of \(b\sqrt{3}\) m/s when at a distance \(b\) (metre) from O. If the particle is moving away from O at that instant, find the time required by the particle to travel a further distance \(b\).
The period of oscillation of a body of mass \(m_1\) suspended from a light spring is \(T\). When a body of mass \(m_2\) is tied to the first body and the system is made to oscillate, the period is \(2T\). Compare the masses \(m_1\) and \(m_2\).
The displacement of an oscillating particle is given by \(x = a\sin\omega t + b\cos\omega t\), where \(a\), \(b\) and \(\omega\) are constants. Prove that the particle performs a linear S.H.M. with amplitude \(A = \sqrt{a^{2} + b^{2}}\).
Two parallel S.H.M.s represented by \(x_1 = 5\sin\left(4\pi t + \dfrac{\pi}{3}\right)\) cm and \(x_2 = 3\sin\left(4\pi t + \dfrac{\pi}{4}\right)\) cm are superposed on a particle. Determine the amplitude and epoch of the resultant S.H.M.
A 20 cm wide thin circular disc of mass 200 g is suspended to a rigid support from a thin metallic string. By holding the rim of the disc, the string is twisted through 60° and released. It now performs angular oscillations of period 1 second. Calculate the maximum restoring torque generated in the string under undamped conditions. (\(\pi^{3} \approx 31\))
Find the number of oscillations performed per minute by a magnet vibrating in the plane of a uniform field of \(1.6 \times 10^{-5}\) Wb/m². The magnet has moment of inertia \(3 \times 10^{-6}\) kg m² and magnetic moment 3 A m².
A wooden block of mass \(m\) is kept on a piston that can perform vertical vibrations of adjustable frequency and amplitude. During vibrations, we don’t want the block to leave the contact with the piston. How much maximum frequency is possible if the amplitude of vibrations is restricted to 25 cm? In this case, how much is the energy per unit mass of the block? (\(g \approx \pi^{2} \approx 10\) m s⁻²)