Course Content

Waves Sound and Light: Lesson

Waves  Sound and Light: Quiz 1
Oscillations
Introduction
In our daily life we come across various kinds of motions. You have already learnt about some of them, e.g., rectilinear motion and motion of a projectile. Both these motions are nonrepetitive. We have also learnt about uniform circular motion and orbital motion of planets in the solar system. In these cases, the motion is repeated after a certain interval of time, that is, it is periodic. In your childhood, you must have enjoyed rocking in a cradle or swinging on a swing. Both these motions are repetitive in nature but different from the periodic motion of a planet. Here, the object moves to and fro about a mean position. The pendulum of a wall clock executes a similar motion. Examples of such periodic to and fro motion abound: a boat tossing up and down in a river, the piston in a steam engine going back and forth,etc. Such a motion is termed as oscillatory motion. In this chapter we study this motion.
The study of oscillatory motion is basic to physics; its concepts are required for the understanding of many physical phenomena. In musical instruments, like the sitar, guitar or the violin, we come across vibrating strings that produce pleasing sounds. The membranes in drums and diaphragms in telephone and speaker systems vibrate to and fro about their mean positions. The vibrations of air molecules make the propagation of sound possible. In a solid, the atoms vibrate about their equilibrium positions, the average energy of vibrations being proportional to temperature. AC power supply give voltage that oscillates alternately going positive and negative about the mean value (zero).
The description of a periodic motion, in general, and oscillatory motion, in particular, requires some fundamental concepts, like period, frequency, displacement, amplitude and phase. These concepts are developed in the next Chapter ‘Waves’.
Periodic and Oscillatory Motions
Fig. 1 shows some periodic motions. Suppose an insect climbs up a ramp and falls down, it comes back to the initial point and repeats the process identically. If you draw a graph of its height above the ground verses time, it would look something like Fig. 1(a). If a child climbs up a step, comes down, and repeats the process identically, its height above the ground would look like that in Fig. 1(b). When you play the game of bouncing a ball off the ground, between your palm and the ground, its height versus time graph would look like the one Fig. (c). Note that both the curved parts in Fig. 1(c) are sections of a parabola given by the Newton’s equation of motion.
h = ut + 1/2 gt² for downward motion, and
h = ut – 1/2 gt² for upward motion,
with different values of u in each case. These are examples of periodic motion. Thus, a motion that repeats itself at regular intervals of time is called periodic motion.
Fig. 1 – Examples of periodic motion. The period T is shown in each case.
Very often, the body undergoing periodic motion has an equilibrium position somewhere inside its path. When the body is at this position no net external force acts on it. Therefore, if it is left there at rest, it remains there forever. If the body is given a small displacement from the position, a force comes into play which tries to bring the body back to the equilibrium point, giving rise to oscillations or vibrations. For example, a ball placed in a bowl will be in equilibrium at the bottom. If displaced a little from the point, it will perform oscillations in the bowl. Every oscillatory motion is periodic, but every periodic motion need not be oscillatory. Circular motion is a periodic motion, but is not oscillatory.
There is no significant difference between oscillations and vibrations. It seems that when the frequency is small, we call it oscillation (like, the oscillation of a branch of a tree), while when the frequency is high, we call it vibration (like the vibration of a string of a musical instrument).
Simple harmonic motion is simplest form of oscillatory motion. This motion arises when the force on the oscillating body is directly proportional to its displacement from the mean position, which is also the equilibrium position. Further, at any point in its oscillation, this force is directed towards the mean position.
In practice, oscillating bodies eventually come to rest at their equilibrium positions because of the damping due to friction and other dissipative causes. However, they can be forced to remain oscillating by means of some external periodic agency. We discuss the phenomena of damped and forced oscillations later in the chapter.
Period and frequency
We have seen that any motion that repeats itself at regular intervals of time is called periodic motion. The smallest interval of time after which the motion is repeated is called its period. Let us denote the period by the symbol T. Its SI unit is second. For periodic motions, which are either too fast or too slow on the scale of seconds, other convenient units of time are used. The period of vibrations of a quartz crystal is expressed in units of microseconds (10^{6} s) abbreviated as µs. On the other hand, the orbital period of the planet Mercury is 88 earth days. The Halley’s comet appears after every 76 years.
The reciprocal of T gives the number of repetitions that occur per unit time. This quantity is called the frequency of the periodic motion. It is represented by the symbol υ. The relation between υ and T is
υ = 1/T ……………………………………….(1)
The unit of υ is thus s^{1}. After the discoverer of radio waves, Heinrich Rudolph Hertz(18571894), a special name has been given to the unit of frequency. It is called hertz (abbreviated as Hz). Thus,
1 hertz = 1 Hz = 1 oscillation per second = 1s^{1 }…………………….(2)
Note, that the frequency, υ, is not necessarily an integer.
Example 1
On an average, a human heart is found to beat 75 times in a minute. Calculate its frequency and period.
Answer
The beat frequency of heart = 75/(1 min)
= 75/(60 s)
= 1.25 s^{1}
= 1.25 Hz
The time period T = 1/(1.25 s^{1})
= 0.8 s
Displacement
We defined displacement of particle as the change in its position vector. In this chapter, we use the term displacement in a more general sense. It refers to change with time of any physical property under consideration. For example, in case of rectilinear motion of a set ball on a surface, the distance from the starting point as a function of time is its position displacement. The choice of origin is a matter of convenience. Consider a block attached to a spring, the other end of the spring is fixed to a rigid wall [see Fig. 2(a)]. Generally, it is convenient to measure displacement of the body from its equilibrium position. For an oscillating simple pendulum, the angle from the vertical as a function of time may be regarded as a displacement variable [see Fig. 2(b)]. The term displacement is not always to be referred in the context of position only. There can be many other kinds of displacement variables.
A block attached to a spring, the other end of which is fixed to a rigid wall. The block moves on a frictionless surface. The motion of the block can be described in terms of its distance or displacement x from the equilibrium position.
The voltage across a capacitor, changing with time in an AC circuit, is also a displacement variable. In the same way, pressure variations in time in the propagation of sound wave, the changing electric and magnetic fields in a light wave are examples of displacement in different contexts. The displacement variable may take both positive and negative values. In experiments on oscillations, the displacement is measured for different times.
The displacement can be represented by a mathematical function of time. In case of periodic motion, this function is periodic in time. One of the simplest periodic functions is given by
f(t) = A cos ωt …………………………………………………………….(3a)
If the argument of this function, ωt, is increased by an integral multiple of 2π radians, the value of the function remains the same. The function f(t) is then periodic and its period, T, is given by,
T = 2π/ω …………………………………………………….(3b)
Thus, the function f(t) is periodic with period T,
f(t) = f(t+T)
The same result is obviously correct, if we consider a sine and cosine function, f(t) = A sin ωt. Further, a linear combination of sine and cosine functions like,
f(t) = A sin ωt + B cos ωt ……………………………………..(3c)
is also a periodic function with the same period T. Taking,
A = D cos Φ and B = D sin Φ
Eq. (3c) can be written as,
f(t) = D sin (ωt + Φ),
Here D and Φ are constant given by
D = √(A² + B²) and Φ = tan^{1} (B/A)
The great importance of periodic sine and cosine functions is due to remarkable result proved by the French mathematician, Kean Baptiste Joseph Fourier (17681880): Any periodic function can be expressed as a superposition of sine and cosine functions of different time periods with suitable coefficients.
Example 2
Which of the following functions of time represent (a) periodic and (b) nonperiodic motion? Give the period for each case of periodic motion [ω is any positive constant].
(i) sin ωt + cosωt
(ii) sin ωt + cos 2ωt + sin 4ωt
(iii) e^{–ωt}
(iv) log ωt
Answer
(i)sin ωt + cosωt is a periodic function, it can also be written as √[2 sin (ωt + π/4)].
Now √[2 sin (ωt + π/4)] = √[2 sin (ωt + π/4 +2 π)]
= √(2 sin [ω (t + 2 π/ω) +π/4)]
The periodic time of the function is 2 π/ω.
(ii) This is an example of a periodic motion. It can be noted that each term represents a periodic function with a different angular frequency. Since period is the least interval of time after which a function repeats its value, since sin ωt has a period T_{0} = 2 π/ω; cos 2ωt has a period 2π/4ω = T_{0}/4. The period of the first term is a multiple of the periods of the last two terms. Therefore, the smallest interval of time repeats is T_{0}, and thus, the sum is a periodic function with a period 2π/ω.
(iii) The function e^{–ωt} is not periodic, it decreases monotonically with increasing time and tends to zero as t →∞, and thus, never repeats its value.
(iv) The function log(ωt) increases monotonically with time t. It, therefore, never repeats its value and is a non periodic function. It may be noted that as t →∞, log(ωt) diverges to ∞. It, therefore, cannot represent any kind of physical displacement.
Simple Harmonic Motion
Consider a particle oscillating back and forth about the origin of an xaxis between the limits +A and –A as shown in Fig. 3. This oscillatory motion is said to be simple harmonic if the
displacement x of the particle from the origin varies with time as:
x(t) = A cos (ωt + Φ) …………………………………………………(4)
where A, ω and Φ are constants.
Thus, simple harmonic motion (SHM) is not any periodic motion but one in which displacement is a sinusoidal function of time. Fig. 4 shows the positions of a particle executing SHM at discrete value of time, each interval of time being T/4, where T is the period of motion.
Fig 4 – The location of the particle SHM at the discrete values t = 0, T/4, T/2. 3T/4, T, 5T/4. The time after which motion repeats itself is T. T will remain fixed, no matter what location you choose as the initial (t = 0) location. The speed is maximum for zero displacement (at x = 0) and zero at the extremes of motion.
Fig. 5 plots the graph of x versus t, which gives the values of displacement as a continuous function of time. The quantities A, ω and Φ which characterize a given SHM have standard names, as summarized in Fig. 6. Let us understand these quantities.
x(t) : displacement x as a function of time t
A : amplitude
ω : angular frequency
ωt + Φ : phase (timedependent)
Φ : phase constant
The meaning of standard symbols in Eq. (4)
The amplitude of A of SHM is the magnitude of maximum displacement of the particle. [Note, A can be taken to be positive without any loss of generality]. As the cosine function of time varies from +1 to 1, the displacement varies between the extremes A and A. Two simple harmonic motions may have same ω and Φ but different amplitudes A and B, as shown in Fig. 7(a).
Fig 7(a) – A plot of displacement as a function of time as obtained from Eq. (4) with Φ = 0. The curves 1 and 2 are for two different amplitudes A and B.
While the amplitude A is fixed for a given SHM, the state of motion (position and velocity) of the particle at any time t is determined by the argument [ωt + Φ] in the cosine function. This timedependent quantity, [ωt + Φ] is called the phase of the motion. The value of phase at t =0 is Φ and is called phase constant (or phase angle). If the amplitude is known, Φ can be determined from the displacement at t = 0. Two simple harmonic motions may have the same A and ω but different phase angle Φ, as shown in Fig. 7(b).
Finally, the quantity ω can be seen to be related to the period of motion T. Taking, for simplicity, Φ = 0 in Eq. (4), we have
x(t) = A cos ωt …………………………………………………………….(5)
Since the motion has a period T, x(t) is equal to x(t + T). That is,
A cos ωt = A cos ω(t + T) ……………………………….(6)
Now the cosine function is periodic with period 2π, i.e., it first repeats itself when the argument changes by 2π. Therefore,
ω(t + T) = ωt + 2π
that is is ω = 2π/T ………………………………….(7)
ω is called the angular frequency of SHM. Its S.I unit is radians per second. Since the frequency of oscillations is simply 1/T, ω is 2π times the frequency of oscillation. Two simple harmonic motions may have the same A and Φ, but different ω, as seen in Fig. 8. In this plot the curve (b) has half the period and twice the frequency of the curve (a).
Example 3
Which of the following functions of time represent (a) simple harmonic motion and (b) periodic but not simple harmonic? Give the period for each case.
(1) sin ωt – cos ωt
(2) sin²ωt
Answer
(a) sin ωt – cos ωt = sin ωt – sin (π/2 – ωt)
= 2 cos (π/4) sin (ωt – π/4)
= √2 sin (ωt – π/4)
This function represents a simple harmonic motion having a period T = 2π/ ω and a phase angle(π/4) or (7π/4)
(b) sin²ωt = ½ – ½ cos 2ωt
The function is periodic having a period T = π/ ω. It also represents a harmonic motion with the point of equilibrium occurring at ½ instead of zero.
Simple Harmonic Motion and Uniform Circular Motion
In this section, we show that the projection of uniform circular motion on a diameter of the circle follows simple harmonic motion. A simple experiment [Fig. 9] helps us visualize this connection. Tie a ball to the end of a string and make it move in a horizontal plane about a fixed point with a constant angular speed. The ball would then perform a uniform circular motion in the horizontal plane. Observe the ball sideways or from the front, fixing your attention in the plane of motion. The ball will appear to execute to and fro motion along a horizontal line with the point of rotation as the mid point. You could alternatively observe the shadow of the ball on a wall which is perpendicular to the plane of the circle. In this process what we are observing is the motion of the ball on a diameter of the circle normal to the direction of viewing.
Fig 9 – Circular motion of a ball in a plane viewed edgeon is SHM.
Fig. 10 describes the same situation mathematically. Suppose a particle P is moving uniformly on a circle of radius A with angular speed ω. The sense of rotation is anticlockwise. The initial position vector of the particle i.e.,the vector OP at t = 0 makes an angle Φ with the positive direction of xaxis. In time t, it will cover a further angle ωt and its position vector will make an angle of ωt + Φ with the +ve xaxis. Next, consider the projection of the position vector OP on the xaxis. This will be OP’.
The position P’ on the x axis, as the particle P moves on the circle, is given by
x(t) = A cos (ωt + Φ)
which is the defining equation of SHM. This shows that if P moves uniformly on a circle, its projection P’ on a diameter of the circle is given by x(t) = A cos (ωt + Φ)
which is the defining equation of SHM. This shows that if P moves uniformly on a circle, its projection P’ on a diameter of the circle executes SHM. The particle P and the circle on which it moves are sometimes referred to as the reference particle and the reference circle, respectively.
We can take projection of the motion of P on any diameter, say the y –axis. In that case, the displacement y(t) of P’ on the yaxis is given by
y = A sin (ωt + Φ)
which is also an SHM of the same amplitude as that of the projection on xaxis, but differing by a phase of π/2.
In spite of this connection between circular motion and SHM, the force acting on a particle in linear simple harmonic motion is very different from the centripetal force needed to keep a particle, in uniform circular motion.
Example 4
The figure given below depicts two circular motions. The radius of the circle, the period of revolution, the initial position and the sense of revolution are indicated in the figures. Obtain the simple harmonic motions of the xprojection of the radius vector of the rotating particle P in each case.
Answer
(a) At t = 0, OP makes an angle of 45° = π/4 rad with the (positive direction of ) xaxis. After time t, it covers an angle (2π/T )t in the anticlockwise sense, and makes an angle of (2π/T )t +π/4 with the xaxis.
The projection OP on the xaxis at time t is given by,
x(t) = A cos [(2π/T )t + π/4]
For T = 4 s,
x(t) = A cos [(2π/4 )t + π/4]
which is a SHM of amplitude A, period 4 s,
and an initial phase^{∗ }= π/4.
^{∗ }The natural unit of angle is radian, defined through the ratio of arc to radius. Angle is a dimensionless quantity. Therefore it is not always necessary to mention the unit ‘radian’ when we use π, its multiples or submultiples. The conversion between radian and degree is not similar to that between metre and centimeter or mile. If the argument of a trigonometric function is stated without unit, it is understood that the unit is radian. On the other hand, if degree is to be used as the unit of angle, then it must be shown explicitly. For example, sin(15°) means sine of 15 degree, but in sin(15) means sine of 15 radians. Hereafter, we will often drop ‘rad’ as the unit and it should be understood that whenever angle is mentioned as a numerical value, without units, it is to be taken as radians.
(b) In this case at t = 0, OP makes an angle of 90° = π/2 with the axis. After a time t, it covers an angle of (2π/T )t in the clockwise sense and makes an angle of [(π/2) – (2π/T )t] with the xaxis. The projection OP on the xaxis at time t is given by
x(t) = B cos [(π/2) – (2π/T )t]
= B sin [(2π/T )t]
For T = 30 s,
x(t) = B sin (π/15 )t
Writing this as x(t) = B cos ([(π/15)t – (π/2 )], and comparing with Eq. (4). We find that this represents a SHM of amplitude B, period 30 s, and an initial phase π/2.
Velocity and Acceleraton in Simple Harmonic Motion
The speed of a particle v in uniform circular motion is its angular speed ω times the radius of the circle A.
v = ωA ………………………………………………..(8)
The direction of velocity v at a time t is along the tangent to the circle at the point where the particle is located at that instant. From the geometry of Fig 11, it is clear that the velocity of the projection particle P’ at time t is
v(t) = ωA sin (ωt + Φ) ………………………………….(9)
where the negative sign shows that v(t) has a direction opposite to the positive direction of xaxis. Eq. (9) gives the instantaneous velocity of a particle executing SHM, where displacement is given by Eq. (4). We can, of course, obtain this equation without using geometrical argument, directly by differentiating Eq. (4) with respect to t:
v(t) = (d/dt)x(t) ……………………………………(10)
The method of reference circle can be similarly used for obtaining instantaneous acceleration of a particle undergoing SHM. We know that the centripetal acceleration of a particle P in uniform circular motion has a magnitude v²/A or ω²A, and is directed towards the centre i.e., the direction is along PO. The instantaneous acceleration of the projection particle P’ is then (see Fig 12)
α(t) = – ω²A cos (ωt +Φ)
= – ω²x(t) …………………………………..(11)
Eq. (11) gives the acceleration of a particle in SHM. The same eqauation can again be obtained directly by differentiating velocity v(t) given by Eq. (9) with respect to time:
a(t) = (d/dt)v(t) ………………………………………..(12)
We note from Eq. (11) the important property that acceleration of a particle in SHM is proportional to displacement. For x(t) > 0, a(t) <0 and x(t) < 0, a(t)>o. Thus, whatever the value of x between –A and A, the acceleration a(t) is always directed towards the centre.
For simplicity, let us put Φ = 0 and write the expression for x(t), v(t) and a(t).
x(t) = A cos ωt, v(t) = ωA sin ωt, a(t)= ω²A cos ωt. The corresponding plots are shown in Fig. 13. All quantities vary sinusoidally with time; only their maxima differ and the different plots differ in phase, x varies between –A to A; v(t) varies from ωA to ωA and a(t) from –ω²A to ω²A. With respect to displacement plot, velocity plot has a phase difference of π/2 and acceleration plot has phase difference of π.
Example 5
A body oscillates with SHM according to the equation (in S I units).
x = 5 cos [2πt + π/4].
At t = 1.5 s, calculate the (a) displacement, (b) speed (c) acceleration of the body.
Answer
The angular frequency ω of the body = 2πs^{–}^{1 }and its time period T = 1 s.
At t = 1.5 s
(a) displacement = (5.0 m) cos [(2πs^{–}^{1}) ×1.5 s + π/4]
= (5.0 m) cos [(3 π + π/4)]
= 5.0 ×0.707 m
= 3.535 m
(b) Using Eq. (9), the speed of the body = [5.0 m][2πs^{–}^{1}]sin [(2πs^{–}^{1}) × 1.5 s + π/4]
= [5.0 m][2πs^{–}^{1}] sin [(3π +π/4)]
= 10π × 0.707 m s^{–}^{1}
= 22 m s^{–}^{1}
(c) Using Eq. (10), the acceleration of the body = [2πs^{–}^{1}]² × displacement
= [2πs^{–}^{1}]² × [3.535 m]
= 140 m s^{–}^{2}
Force Law for Simple Harmonic Motion
Using Newton’s second law of motion, and the expression for acceleration of a particle undergoing SHM (Eq. 11), the force acting on a particle of mass m in SHM is
F(t) = ma
= mω²x(t)
i.e., F(t) = kx(t) ……………………………..(13)
where k = mω² ……………………………(14a)
or ω = √(k/m) ………………………….(14b)
Like acceleration, force is always directed towards the mean positionhence it is sometimes called the restoring force in SHM. To summarise the discussion so far, simple harmonic motion can be defined in two equivalent ways, either by Eq. (4) to Eq. (13) required us to differentiate two times. Likewise, by integrating the force law Eq. (13) two times, we can get back Eq. (4).
Note that the force in Eq. (13) is linearly proportional to x(t). A particle oscillating under such a force is, therefore, calling a linear harmonic oscillator. In the real world, the force may contain small additional terms proportional to x², x³, etc. These then are called nonlinear oscillators.
Example 6
Two identical springs of spring constant k are attached to a block of mass m and to fixed supports as shown in Fig. 4. Show that when the mass is displaced from its equilibrium position on either side, it executes a simple harmonic motion. Find the period of oscillations.
Answer
Let the mass be displaced by a small distance x to the right side of the equilibrium position, as shown in Fig. 15. Under this situation the spring
on the left side gets elongated by a length equal to x and that on the right side gets compressed by the same length. The forces acting on the mass are then,
F_{1} = kx (force exerted by the spring on the left side, trying to pull the mass towards the mean position)
F_{2} = kx (force exerted by the spring on the right side, trying to push the mass towards the mean position)
The net force, F, acting on the mass is then given by,
F = 2kx
Hence the force acting on the mass is proportional to the displacement and is directed towards the mean position; therefore, the motion executed by the mass is simple harmonic. The time period of oscillations is,
T = 2π√(m/2k)
Energy in Simple Harmonic Motion
Both kinetic and potential energies of a particle in SHM vary between zero and their maximum values.
In section 5, we have seen that the velocity of a particle executing SHM, is a periodic function of time. It is zero at the extreme positions of displacement. Therefore, the kinetic energy (K) of such a particle, which is defined as
K = 1/2 mv²
= 1/2 mω²A² sin²[ωt + Φ]
=1/2 kA² sin²[ωt + Φ]
is also a periodic function of time, being zero when the displacement is maximum and maximum when the particle is at the mean position. Note, since the sign of v is immaterial in K, the period of K is T/2.
What is the potential energy (U) of a particle executing simple harmonic motion? In the Chapter ‘Work, Power and Energy’ we have seen that the concept of potential energy is possible only for conservative forces. The spring force F=kx is a conservative force, with associated potential energy
U = 1/2 kx² ………………………………………..(16)
Hence the potential energy of a particle executing simple harmonic motion is,
U(x) = 1/2 kx²
= 1/2 kA²cos²[ωt + Φ] ………………………………………..(17)
Thus, the potential energy of a particle executing simple harmonic motion is also periodic, with period T/2, being zero at the mean position and maximum at the extreme displacements.
It follows from Eqs. (15) and (17) that the total energy, E, of the system is,
E = U + K
= 1/2 kA²cos²(ωt + Φ) + 1/2 kA²sin²(ωt + Φ)
= 1/2 kA²[cos²(ωt + Φ) + sin²(ωt + Φ)]
Using the familiar trigonometric identity, the value of the expression in the brackets is unity. Thus,
E = 1/2 kA² ………………………………………(18)
The total mechanical energy of a harmonic oscillator is thus independent of time as expected for motion under any conservative force. The time and displacement dependence of the potential and kinetic energies of a linear simple harmonic oscillator are shown in Fig. 16.
Kinetic energy, potential energy and total energy as a function of time (shown in (a)) and displacement (shown in (b)) of a particle in SHM. The kinetic energy and potential energy both repeat after a period T/2. The total energy remains constant at all or x.
Observe that kinetic energy and potential energy in SHM are not seen to be always positive in Fig. 16. Kinetic energy can, of course, be never negative, since it is proportional to the square of speed. Potential energy is positive by choice of the undermined constant in potential energy. Both kinetic energy and potential energy peak twice during each period of SHM. For x = o, the energy is kinetic; at the extremes x = + A, it is all potential energy. In the course of motion between these limits, kinetic energy increases at the expense of potential or viceversa.
Example 7
A block whose mass is 1 kg is fastened to a spring. The spring has a spring constant of 50 N m^{–}^{1}. The block is pulled to a distance x = 10 cm from its equilibrium position at x = 0 on a frictionless surface from rest at t = 0. Calculate the kinetic, potential and total energies of the block when it is 5 cm away from the mean position.
Answer
The block executes SHM, its angular frequency, as given by Eq. (14b) is
ω =√(k/m)
= √(50 N m^{–}^{1}/1 kg)
= 7.07 rad s^{–}^{1}
Its displacement at any time t is then given by,
x(t) = 0.1 cos (7.07 t)
Therefore, when the particle is 5 cm away from the mean position, we have
0.05 = 0.1 cos (7.07 t)
Or cos(7.07 t) = 0.5 and hence
sin (7.07 t) = √3/2 = 0.866
Then, the velocity of the block at x = 5 cm is
= 0.1 × 7.07 × 0.866 m s^{–}^{1}
= 0.61 m s^{–}^{1}
Hence K.E of the block,
= ½m v²
= ½ [1 kg × (0.6123 m s^{–}^{1})²]
= 0.19 J
The P.E. of the block,
= ½kx²
= ½[50 N m^{–}^{1}× 0.05 m×0.05 m)
= 0.0625 J
The total energy of the block at x = 5 cm,
= K.E + P.E
= 0.25 J
We also know that at maximum displacement, K.E is zero and hence the total energy of the system is equal to the P.E. Therefore, the total energy of the system,
= ½[50 N m^{–}^{1}× 0.1 m×0.1 m)
= 0.25 J
which is same as the sum of the two energies at a displacement of 5 cm. This is in conformity with the principle of conservation of energy.
Some Systems Executing Simple Harmonic Motion
There are no physical examples of absolutely pure simple harmonic motion. In practice we come across systems that execute simple harmonic motion approximately under certain conditions. In the subsequent part of this section, we discuss the motion executed by some such systems.
Oscillations due to a spring
The simplest observable example of simple harmonic motion is the small oscillations of a block of mass m fixed to a spring, which in turn is fixed to a rigid wall as shown in Fig. 17. The block is placed on a frictionless horizontal surface. If the block is pulled on one side and is released, it then executes a to and fro motion about the mean position. Let x =0, indicate the position of the centre of the block when the spring is in equilibrium. The positions marked as –A and +A indicate the
Fig. 17 –
A linear simple harmonic oscillator consisting of a block of mass m attached to a spring. The block moves over a frictionless surface. The box, when pulled or pushed and released, executes simple harmonic motion.
maximum displacements to the left and the right of the mean position. We have already learnt that springs have special properties, which were first discovered by the English physicist Robert Hooke. He had shown that such a system when deformed, is subject to a restoring force, the magnitude of which is proportional to the deformation or the displacement and acts in opposite direction. This is known as Hooke’s law (See Chapter ‘Mechanical Properties of Solids’ ). It holds good for displacements small in comparison to the length of the spring. At any time t, if the displacement of the block from its mean position is x, the restoring force F acting on the block is,
F(x) = kx …………………………………………………(19)
The constant of proportionality, k, is called the spring constant, its value is governed by the elastic properties of the spring. A stiff spring has large k and a soft spring has small k. Equation (19) is same as the force law of SHM and therefore the system executes a simple harmonic motion. From Eq. (14) we have,
ω = √(k/m) ……………………………………….(20)
and the period, T, of the oscillator is given by,
T = 2π√(m/k) …………………………………….(21)
Stiff springs have high value of k (spring constant). A block of small mass m attached to a stiff spring will have, according to Eq. (20), large oscillation frequency, as expected physically.
Example 8
A 5 kg collar is attached to a spring of spring constant 500 N m^{–}^{1} . It slides without friction over a horizontal rod. The collar is displaced from its equilibrium position by 10.0 cm and released. Calculate(a) the period of oscillation, (b) the maximum speed and (c) the maximum acceleration of the collar.
Answer
(a) The period of oscillation as given by Eq. (21) is,
T =2π√(m/k) = 2π√(5.0 kg/500 Nm^{–}^{1})
= (2π/10)s
= 0.63 s
(b) The velocity of the collar executing SHM is given by
v(t) = ωA sin (wt + Φ)
The maximum speed is given by,
v_{m} = Aω
= 0.1 × √(k/m)
= 0.1 × √(500 Nm^{–}^{1}/5 kg)
= 1 m s^{–}^{1}
and it occurs at x = 0
(c) The acceleration of the collar at the displacement x(t) from the equilibrium is given by,
a(t) = ω²x(t)
= (k/m)x(t)
Therefore, the maximum acceleration is,
a_{max} = ω² A
= (500 Nm^{–}^{1}/5 kg) × 0.1 m
= 10 ms^{–}^{2}
and it occurs at extremities.
The Simple Pendulum
It is said that Galileo measured the periods of a swinging chandelier in a church by his pulse beats. He observed that the motion of the chandelier was periodic. The system is a kind of pendulum. You can also make your own pendulum by tying a piece of stone to a long unstretchable thread, approximately 100 cm long. Suspend your pendulum from suitable support so that it is free to oscillate. Displace the stone to one side by a small distance and let it go. The stone executes a to and fro motion, it is periodic with a period of about two seconds.
We shall show that this periodic motion is simple harmonic for small harmonic for small displacements from the mean position. Consider simple pendulum a small bob of massless string of length L. The other end of the string is fixed to a rigid support. The bob oscillates in a plane about the vertical line through the support. Fig. 18(a) shows this system. Fig. 18(b) is a kind of ‘freebody’ diagram of the simple pendulum showing the forces acting on the bob.
Fig 18 – (a) A bob oscillating about its mean position. (b) The radial force Tmg cosθ provides centripetal force but no torque about the support. The tangential force mg sinθ provides the restoring torque.
Let θ be the angle made by the string with the vertical. When the bob is at the mean position, θ = 0.
There are only two forces acting on the bob; the tension T along the string and the vertical force due to gravity (=mg). The force mg can be resolved in to the component mg cosθ along the mg sin θ perpendicular to it. Since the motion of the bob is along a circle of length L and centre of support point, the bob has a radial acceleration (ω²L) and also a tangential acceleration; the latter arises since motion along the arc of the circle is not uniform. The radial acceleration is provided by the net radial force Tmg cosθ, while the tangential acceleration is provided by mg sinθ. It is more convenient to work with torque about the support since the radial force gives zero torque. Torque τ about the support is entirely provided by the tangential component of force
τ = L (mg sinθ) …………………………………………….(22)
This is the restoring torque that tends to reduce angular displacement – hence the negative sign. By Newton’s law of rotational motion,
τ = I α ………………………………………………….(23)
where I is the moment of inertia of the system about the support and α is the angular acceleration. Thus,
I α = mg sinθ L ………………………………………..(24)
Or,
α = –(mgL/I)sinθ ……………………………………(25)
We can simplify Eq. (25) if we assume that the displacement θ is small. We know that sin θ can be expressed as,
sinθ = θ –(θ³/3!) + (θ^{5}/5!) + …………………….(26)
where θ is in radians.
Now if θ is small, sin θ can be approximated by θ and Eq. (25) can then be written as,
α = –(mgL/I) θ ………………………….(27)
In Table 1, we have listed the angle θ in degrees, its equivalent in radians, and the value of the function sin θ.
Table 1 sin θ as a function of angle θ
θ (degrees)  θ (radians)  sin θ 

0  0  0 
5  0.087  0.087 
10  0.174  0.174 
15  0.262  0.259 
20  0.349  0.342 
From this table it can be seen that for θ as large as 20 degrees, sin θ is nearly the same as θ expressed in radians.
SHM – how small should the amplitude be?
When you perform the experiment to determine the time period of a simple pendulum, your teacher tells you to keep the amplitude small. But have you ever asked how small is small? Should the amplitude to 5°, 2°, 1°, or 0.5°? Or could it be 10°, 20°or 30°?
To appreciate this, it would be better to measure the time period for different, amplitudes, up to large amplitudes. Of course, for large oscillations, you will have to take care that the pendulum oscillates in a vertical plane. Let us denote the time period for smallamplitude oscillations as T(O) and write the time period for amplitude θ_{0 }as T(θ_{0}) =c T(O), where c is the multiplying factor. If you plot a graph of c versus θ_{0 }, you will get values somewhat like this:
θ_{0 }: 20° 45° 50° 70° 90°
c : 1.02 1.04 1.05 1.10 1.18
This means that the error in the time period is about 2% at an amplitude of 20°, 5% at an amplitude of 50° and 10% at an amplitude of 70° and 18% an amplitude of 90°.
In the experiment, you will never be able to measure T(O) because this means there are no oscillations. Even theoretically, sin θ is exactly equal to θ only θ = 0. There will be some inaccuracy for all other values of θ. The difference increases with increasing θ. Therefore we have to decide how much error we can tolerate. No measurements is ever perfectly accurate. You must consider questions like these: What is the accuracy of the stopwatch? What is your own accuracy in starting and stopping the stopwatch? You will realise that the accuracy in your measurements at this level is never better than 5% or 10%. Since the above table shows that the time period of the pendulum increases hardly by 5% at an amplitude of 50° over its low amplitude value, you could very well keep the amplitude to be 50° in your experiments.
Equation (27) is mathematically, identical to Eq. (11) except that the variable is angular displacement. Hence we have proved that for small θ, the motion of the bob is simple harmonic. From Eqs. (27) and (11),
ω = √(mgL/I)
and
T = 2π √(I/mgL) …………………………………………………….(28)
Now since the string of the simple pendulum is massless, the moment of inertia I is simply mL². Eq. (28) then gives the well known formula for the time period of a simple pendulum.
T = 2π √(L/g) ………………………………………………..(29)
Example 9
What is the length of a simple pendulum, which ticks seconds?
Answer
From Eq. (29), the time period of a simple pendulum is given by,
T = 2π √(L/g)
From this relation one gets,
L = gT²/4π²
The time period of a simple pendulum, which ticks seconds, is 2s. Therefore, for g = 9.8 m s^{2} and T = 2 s. L is
= 9.8 (m s^{2})× 4(s²)/4π²
= 1 m
Damped Simple Harmonic Motion
We know that the motion of a simple pendulum, swinging in air, dies out eventually. Why does it happen? This is because the air drag and the friction at the support oppose the motion of the pendulum and dissipate its energy gradually. The pendulum is said to execute damped oscillations. In damped oscillations, the energy of the system is dissipated continuously; but for small damping, the oscillations remain approximately periodic. The dissipating forces are generally the friction forces. To understand the effect of such external forces on the motion of an oscillator, let us consider a system as shown in in Fig. 19. Here a block of mass m connected to an elastic spring of spring constant k oscillates vertically. If the block is pushed down a little and released, its angular frequency of oscillation is ω = √(k/m), as seen in Eq. (20).
However, in practice, the surrounding medium (air) will exert a damping force on the motion of the block and the mechanical energy of the blockspring system will decrease. The energy loss will appear as heat of the surrounding medium (and the block also) [Fig. 19].
The damping force depends on the nature of the surrounding medium. If the block is immersed in a liquid, the magnitude of damping will be much greater and the dissipation of energy much faster. The damping force is generally proportional to velocity of the bob, [Remember Stokes law, Eq. (19) given in Chapter, ‘Mechanical Properties of Fluids’] and acts opposite to the direction of velocity. If the damping force is denoted by F_{d, }we have
F_{d }= b v ……………………………………….(30)
where the positive constant b depends on characteristics of the medium (viscosity, for example) and the size and shape of the block. etc. Eq. (30) is usually valid only for small velocity.
When the mass m is attached to the spring (hung vertically as shown in Fig. 19) and released, the spring will elongate a little and the mass will settle at some height. This position shown by O in Fig. 19, is the equilibrium position of the mass. If the mass is pulled down or pushed up a little, the restoring force on the block due to the spring is F_{s }= kx where x is the displacement^{* }of the mass from its equilibrium position. Thus, the total force acting on the mass at any time t, is F = –kx –bv.
If a(t) is the acceleration of mass at time t, then by Newton’s Law of Motion applied along the direction of motion, we have
m a(t) = k x(t) b v(t) ……………………………………….(31)
Here we have dropped the vector notation because we are discussing onedimensional motion.
Using the first and second derivatives of x(t) for v(t) and a(t), respectively, we have
m(d²x/dt²) + b(dx/dt) + kx = 0 …………………………………..(32)
The solution of Eq. (32) describes the motion of the block under the influence of a damping force which is proportional to velocity. The solution is found to be of the form
x(t) = A e^{bt/2m} cos(ω’t + Φ) …………………………………..(33)
where A is the amplitude and ω’ is the angular frequency of the dampeed oscillator given by,
* Under gravity, the block will be at a certain equilibrium position O on the spring; x here represents the displacement from that position.
ω’ = √[(k/m) – (b²/4m²)] ………………………………….(34)
In this function, the cosine function has a period 2π/ω’ but the function x(t) is not strictly periodic because of the factor e^{bt/2m} which decreases continuously with time. However, if the decrease is small in one time period T, the motion represented by Eq. (33) is approximately periodic.
The solution, Eq. (33), can be graphically represented as shown in Fig. 20. We can regard it as a cosine function whose amplitude, which is Ae^{bt/2m} , gradually decreases with time.
Now the mechanical energy of the undamped oscillator, is ½kA². For a damped oscillator, the amplitude is not constant but depends on time. For small damping, we may use the same expression but regard the amplitude Ae^{bt/2m}
E(t) = ½kA²e^{bt/2m }……………………………………..(35)
Equation (35) shows that the total energy of the system decreases exponentially with time. Note that small damping means that the dimensionless ratio [b/√(km)] is much less than 1.
Of course, as expected, if we put b=0, all equations of a damped oscillator in this section reduce to the corresponding equations of an undamped oscillator.
Example 10
For the damped oscillator shown in Fig. 19, the mass m of the block is 200 g, k = 90 N m^{1} and the damping constant b is 40 g s^{1}. Calculate (a) the period of oscillation (b) time taken for its amplitude of vibrations to drop to half of its initial value, and (c) the time taken for its mechanical energy to drop to half its initial value.
Answer
(a) We see that km = 90 ×0.2 = 18 kg Nm^{1}= 18 kg² s^{2}; therefore √km = 4.243 kg s^{1}, and b = 0.04 kg s^{1}. Therefore, b is much less than √km. Hence, the time period T from Eq. (34) is given by
T = 2π√(m/k)
= 2π√(0.2 kg/90 N m^{1})
= 0.3 s
(b) Now, from Eq. (33), the time, T_{1/2 }for the amplitude to drop to half of its initial value is given by,
T_{1/2 }= ln(1/2)/(b/2m)
= (0.693/40)×2×200 s
= 6.93 s
(c) For calculating the time, t_{1/2 }, for its mechanical energy to drop to half its initial value we make use of Eq. (35). From this equation we have,
E(t_{1/2 })/E(0) = exp (bt_{1/2}/m)
Or ½ = exp (bt_{1/2}/m)
ln(½) = (bt_{1/2}/m)
Or t_{1/2}= (0.693/40 g s^{1})×200 g
= 3.46 s
This is just half of the decay period for amplitude. This is not surprising, because, according to Eqs. (33) and (35), energy depends on the square of the amplitude. Notice that there is a factor of 2 in the exponents of the two exponentials.
Forced Oscillations and Resonance
When a system (such as simple pendulum or a block attached to a spring) is displaced from its equilibrium position and released, it oscillates with its natural frequency ω, and the oscillations are called free oscillations. All free oscillations eventually die out because of the ever present damping forces. However, an external agency can maintain these oscillations. These are called forced or driven oscillations. We consider the case when the external force is itself periodic, with a frequency ω_{d }called the driven frequency. The most important fact of forced periodic oscillations is that the system oscillates not with its natural frequency ω, but at the frequency ω_{d} of the external agency; the free oscillations die out due to damping. The most familiar example of forced example is when a child in a garden swing periodically presses his feet against the ground (or someone else periodically gives the child a push) to maintain the oscillations.
Suppose an external F(t) of amplitude F_{0 }that varies periodically with time is applied to a damped oscillator. Such a force can be represented as,
F(t) = F_{0}_{ }cosω_{d}t ……………………………………………………(36)
The motion of a particle under the combined action of a linear restoring force, damping force and a time dependent driving force represented by Eq. (36) is given by,
ma(t) = kx(t) bv(t) + F_{0}cosω_{d}t …………………………………………(37a)
Substituting d²x/dt² for acceleration in Eq. (37a) and rearranging it, we get
m(d²x/dt²) + b(dx/dt) + kx = F_{0}cosω_{d}t ……………………………….(37b)
This is the equation of an oscillator of mass m on which a periodic force of (angular) frequency ω_{d }is applied. The oscillator, initially, oscillates with its natural frequency ω. When we apply the external periodic force, the oscillations with the natural frequency die out, and then the body oscillates with the (angular) frequency of the external periodic force. Its displacement, after, the natural oscillations die out is given by,
x(t) = A cos (cosω_{d}t + Φ) ………………………………………….(38)
where t is the time measured from the moment when we apply the periodic force.
The amplitude A is a function of the forced frequency ω_{d }and the natural frequency ω. Analysis shows that it is given by
A = F_{0}/{m²(ω² – ω²_{d})² + ω²_{d}b²}½ ………………………..(39a)
and tan Φ = –v_{0}/ω_{d}x_{0 …………………………………………..}_{(39b)}
where m is the mass of the particle and v_{0 }and x_{0 }are the velocity and the displacement of the particle at time t = 0, which is the moment when we apply the periodic force. Equation (39) shows that the amplitude of the forced oscillator depends on the (angular) frequency of the driving force. We can see a different behaviour of the oscillator when ω_{d }it is close to ω. We consider these two cases.
(a) Small Damping, Driving Frequency far from Natural Frequency: In this case, ω_{d}b_{ }will be much smaller than m(w² – ω²_{d}), and we can neglect them. Then Eq. (39) reduces to dited to John Dalton. He proposed the atomic theory to
A = F_{0}/{m²(ω² – ω²_{d}) ………………………..(40)
Fig. (21) shows the dependence of the displacement amplitude of an oscillator on the angular frequency of the driving force for different amounts of damping present in the system. It may be noted that in all cases the amplitude is the greatest when ω_{d}/ω = 1. The curves in this figure show that smaller the damping, the taller and narrower is the resonance peak.
If we go on changing the driving frequency, the amplitude tends to infinity when it equals the natural frequency. But this is the ideal case of zero damping, a case which never arises in a real system as the damping is never perfectly zero. You must have experienced in a swing that when the timing of your push exactly matches with the time period of the swing, your swing gets the maximum amplitude. This amplitude is large, but not infinity, because there is always some damping in your swing. This will become clear in the (b).
(b) Driving Frequency Close to Natural Frequency: If ω_{d }is very close to ω, m[ω – (ω_{d})²] would bee much less than ω_{d}b, for any reasonable value of b, then Eq. (39) reduces to
A = F_{0}/ω_{d}b …………………………………………………………………(41)
This makes it clear that the maximum possible amplitude for a given driving frequency is governed by the driving frequency and the damping, and is never infinity. The phenomenon of increase in amplitude when the driving force is close to the natural frequency of the oscillator is called resonance.
In our daily life, we encounter phenomena which involve resonance. Your experience with swings is a good example of resonance. You might have realised that the skill in swinging to greater heights lies in the synchronization of the rhythm of pushing against the ground with the natural frequency of the swing.
To illustrate this point further, let us consider a set of five simple pendulums of assorted lengths suspended from a common rope as shown in Fig. 22. The pendulums 1 and 4 have the same lengths and the others have different lengths. Now, let us set pendulum 1 in to motion. The energy from this pendulum gets transferred to other pendulums through the connecting rope and they start oscillating. The driving force is provided through the connecting rope. The frequency of this force is the frequency with which pendulum 1 oscillates. If we observe the response of pendulums 2, 3 and 5, they first start oscillating with their natural frequencies of oscillations and different amplitudes, but this motion is gradually damped and not sustained. Their frequencies of
oscillation gradually change, and ultimately, they oscillate with the frequency of pendulum 1, i.e., the frequency of the driving force but with different amplitudes. They oscillate with small amplitudes. The response of pendulum 4 is in contrast to this set pendulums. It oscillates with the same frequency as that of pendulum 1 and its amplitude gradually picks up and becomes very large. A resonancelike response is seen. This happens because in this condition for resonance is satisfied, i.e., the natural frequency of the system coincides with that of the driving force.
We have so far considered oscillating systems which have just one natural frequency. In general, a system may have several natural frequencies. You will see examples of such system (vibrating strings, air columns, etc.) in the Chapter ‘Waves’. Any mechanical structure, like a building, a bridge, or an aircraft may have several possible natural frequencies. An external periodic force or disturbance will set the system in forced oscillation. If accidentally, the forced frequency ω_{d } happens to be close to one of the natural frequencies of the system, the amplitude of oscillation will shoot up (resonance), resulting in possible damage. This is why, soldiers go out of step while crossing a bridge. For the same reason, an earthquake will not cause uniform damage to all buildings in an affected area, even if they are built with the same strength and material. The natural frequencies of a building depend its height, other size parameters, and the nature of building material. The one with its natural frequency close to the frequency of seismic wave is likely to be damaged more.