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. Let’s discuss SHM executed by some such systems.

Oscillations due to a Spring

The simplest observable example of the 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. 14.17.

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 maximum displacements to the left and the right of the mean position. According to the English physicist Robert Hooke, 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. 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,
\(
F(x)=-k x \dots(14.19)
\)

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 (14.19) is the same as the force law for SHM and therefore the system executes a simple harmonic motion. From Eq. \((14.14)\) we have,
\(
\omega=\sqrt{\frac{k}{m}} \dots(14.20)
\)
and the period, T, of the oscillator is given by,
\(
T=2 \pi \sqrt{\frac{m}{k}} \dots(14.21)
\)
Stiff springs have a high value of \(k\) (spring constant). A block of small mass \(m\) attached to a stiff spring will have, according to Eq. (14.20), a large oscillation frequency, as expected physically.

Example 1: A \(5 \mathrm{~kg}\) collar is attached to a spring of spring constant \(500 \mathrm{~N} \mathrm{~m}^{-1}\). It slides without friction over a horizontal rod. The collar is displaced from its equilibrium position by \(10.0 \mathrm{~cm}\) and released. Calculate
(a) the period of oscillation.
(b) the maximum speed and
(c) the maximum acceleration of the collar.

Solution:

(a) The period of oscillation as given by Eq. (14.21) is,
\(
\begin{aligned}
T & =2 \pi \sqrt{\frac{m}{k}}=2 \pi \sqrt{\frac{5.0 \mathrm{~kg}}{500 \mathrm{Nm}^{-1}}} \\
& =(2 \pi / 10) \mathrm{s} \\
& =0.63 \mathrm{~s}
\end{aligned}
\)
(b) The velocity of the collar executing SHM is given by,
\(
v(t)=-A \omega \sin (\omega t+\phi)
\)
The maximum speed is given by,
\(
\begin{aligned}
& v_m=A \omega \\
& =0.1 \times \sqrt{\frac{k}{m}} \\
& =0.1 \times \sqrt{\frac{500 \mathrm{~N} \mathrm{~m}^{-1}}{5 \mathrm{~kg}}} \\
& =1 \mathrm{~m} \mathrm{~s}^{-1}
\end{aligned}
\)
and it occurs at \(x=0\)
(c) The acceleration of the collar at the displacement \(x(t)\) from the equilibrium is given by,
\(
\begin{aligned}
a(t) & =-\omega^2 x(t) \\
& =-\frac{k}{m} x(t)
\end{aligned}
\)
Therefore, the maximum acceleration is,
\(
\begin{aligned}
& a_{\text {max }}=\omega^2 A \\
& =\frac{500 \mathrm{~N} \mathrm{~m}^{-1}}{5 \mathrm{~kg}} \times 0.1 \mathrm{~m} \\
& =10 \mathrm{~m} \mathrm{~s}^{-2}
\end{aligned}
\)
and it occurs at the extremities.

The Simple Pendulum

Let us make a pendulum by tying a piece of stone to a long unstretchable thread, approximately 100 cm long. Suspend your pendulum from a 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 displacements from the mean position.

Consider a simple pendulum – a small bob of mass \(m\) tied to an inextensible 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. 14.18(a) shows this system. Fig. 14.18(b) is a kind of ‘free-body’ diagram of the simple pendulum showing the forces acting on the bob.

Let \(\theta\) be the angle made by the string with the vertical. When the bob is at the mean position, \(\theta=0\). 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 \(m g\) can be resolved into the component \(m g \cos \theta\) along the string and \(m g \sin \theta\) perpendicular to it.

Since the motion of the bob is along a circle of length \(L\) and centre at the support point, the bob has a radial acceleration \(\left(\omega^2 L\right)\) 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 \(\mathrm{T}-\mathrm{mg} \cos \theta\), while the tangential acceleration is provided by \(m g \sin \theta\). It is more convenient to work with torque about the support since the radial force gives zero torque. Torque \(\tau\) about the support is entirely provided by the tangential component of force

\(\tau=-L(m g \sin \theta) \dots(14.22)\)

This is the restoring torque that tends to reduce angular displacement, hence the negative sign. By Newton’s law of rotational motion,

\(\tau=I \alpha \dots(14.23)\)
where \(I(=m L^2)\) is the moment of inertia of the system about the support and \(\alpha\) is the angular acceleration. Thus,

\(I \alpha=-m g \sin \theta \quad L \dots(10.24)\)

Or,
\(
\alpha=-\frac{m g L}{I} \sin \theta \dots(14.25)
\)
We can simplify Eq. (14.25) if we assume that the displacement \(\theta\) is small. We know that \(\sin \theta\) can be expressed as,
\(
\sin \theta=\theta-\frac{\theta^3}{3 !}+\frac{\theta^5}{5 !} \pm \ldots \dots(14.26)
\)
where \(\theta\) is in radians.
Now if \(\theta\) is small, \(\sin \theta\) can be approximated by \(\theta\) and Eq. (14.25) can then be written as,
\(
\alpha=-\frac{m g L}{I} \theta \dots(14.27)
\)

In Table 14.1, we have listed the angle \(\theta\) in degrees, it’s equivalent in radians, and the value of the function \(\sin \theta\). From this table, it can be seen that for \(\theta\) as large as 20 degrees, \(\sin \theta\) is nearly the same as \(\theta\) expressed in radians.

Equation (14.27) is mathematically, identical to Eq. (14.11) except that the variable is angular displacement. Hence we have proved that for small \(\theta\), the motion of the bob is simple harmonic. From Eqs. (14.27) and (14.11),
\(
\omega=\sqrt{\frac{m g L}{I}}
\)
and
\(
T=2 \pi \sqrt{\frac{I}{m g L}} \dots(14.28)
\)
Now since the string of the simple pendulum is massless, the moment of inertia \(I\) is simply \(\mathrm{mL}^2\). Eq. (14.28) then gives the well-known formula for the time period of a simple pendulum.
\(
T=2 \pi \sqrt{\frac{L}{g}} \dots(14.29)
\)

Example 1: What is the length of a simple pendulum, which ticks seconds?

Solution:

From Eq. (14.29), the time period of a simple pendulum is given by,
\(
T=2 \pi \sqrt{\frac{L}{g}}
\)
From this relation one gets,
\(
L=\frac{g T^2}{4 \pi^2}
\)
The time period of a simple pendulum, which ticks seconds, is \(2 \mathrm{~s}\). Therefore, for \(g=9.8 \mathrm{~m} \mathrm{~s}^{-2}\) and \(T=2 \mathrm{~s}, L\) is
\(
\begin{aligned}
& =\frac{9.8\left(\mathrm{~m} \mathrm{~s}^{-2}\right) \times 4\left(\mathrm{~s}^2\right)}{4 \pi^2} \\
& =1 \mathrm{~m}
\end{aligned}
\)

Example 2: Calculate the time period of a simple pendulum of length one meter. The acceleration due to gravity at the place is \(\pi^2 \mathrm{~m} \mathrm{~s}^{-2}\)

Solution:

The time period is
\(
\begin{aligned}
& T=2 \pi \sqrt{L g^{-1}} \\
= & 2 \pi \sqrt{\frac{1.00 \mathrm{~m}}{\pi^2 \mathrm{~m} \mathrm{~s}^{-2}}}=2.0 \mathrm{~s} .
\end{aligned}
\)

Example 3: In a laboratory experiment with simple pendulum it was found that it took \(36 \mathrm{~s}\) to complete 20 oscillations when the effective length was kept at \(80 \mathrm{~cm}\). Calculate the acceleration due to gravity from these data.

Solution:

The time period of a simple pendulum is given by
or,
\(
\begin{aligned}
& T=2 \pi \sqrt{L g^{-1}} \\
& g=\frac{4 \pi^2 L}{T^2} \dots(i)
\end{aligned}
\)
In the experiment described in the question, the time period is
\(
T=\frac{36 \mathrm{~s}}{20}=1 \cdot 8 \mathrm{~s}
\)
Thus, by (i),
\(
g=\frac{4 \pi^2 \times 0.80 \mathrm{~m}}{(1.8 \mathrm{~s})^2}=9.75 \mathrm{~m} \mathrm{~s}^{-2} .
\)

Example 4: A uniform rod of length \(1.00 \mathrm{~m}\) is suspended through an end and is set into oscillation with small amplitude under gravity. Find the time period of oscillation.

Solution:

For small amplitude, the angular motion is nearly simple harmonic and the time period is given by
\(
\begin{aligned}
T & =2 \pi \sqrt{\frac{I}{m g l}}=2 \pi \sqrt{\frac{\left(m L^2 / 3\right)}{m g L / 2}} \\
& =2 \pi \sqrt{\frac{2 L}{3 g}}=2 \pi \sqrt{\frac{2 \times 1.00 \mathrm{~m}}{3 \times 9.80 \mathrm{~m} \mathrm{~s}^{-2}}}=1.64 \mathrm{~s} .
\end{aligned}
\)

Torsional  Pendulum

In a torsional pendulum, an extended body is suspended by a light thread or a wire. The body is rotated through an angle about the wire as the axis of rotation (figure below).

The wire remains vertical during this motion but a twist is produced in the wire. The lower end of the wire is rotated through an angle with the body but the upper end remains fixed with the support. Thus, a twist \(\theta\) is produced. The twisted wire exerts a restoring torque on the body to bring it back to its original position in which the twist \(\theta\) in the wire is zero. This torque has a magnitude proportional to the angle of twist which is equal to the angle rotated by the body. The proportionality constant is called the torsional constant of the wire. Thus, if the torsional constant of the wire is \(k\) and the body is rotated through an angle \(\theta\), the torque produced is \(\tau=-k \theta\).

If \(I\) be the moment of inertia of the body about the vertical axis, the angular acceleration is
\(
\alpha=\frac{\tau}{I}=-\frac{k}{I} \theta
\)
\(
\begin{aligned}
& =-\omega^2 \theta \\
\end{aligned}
\)

\(\text { where } \quad \omega=\sqrt{\frac{k}{I}}\)
Thus, the motion of the body is simple harmonic and the time period is
\(
T=\frac{2 \pi}{\omega}=2 \pi \sqrt{\frac{I}{k}}
\)

Example 5: A uniform disc of radius \(5.0 \mathrm{~cm}\) and mass \(200 \mathrm{~g}\) is fixed at its centre to a metal wire, the other end of which is fixed with a clamp. The hanging disc is rotated about the wire through an angle and is released. If the disc makes torsional oscillations with time period \(0.20 \mathrm{~s}\), find the torsional constant of the wire.

Solution:

The situation is shown in the figure below. The moment of inertia of the disc about the wire is

\(
\begin{aligned}
I & =\frac{m r^2}{2}=\frac{(0 \cdot 200 \mathrm{~kg})\left(5 \cdot 0 \times 10^{-2} \mathrm{~m}\right)^2}{2} \\
& =2 \cdot 5 \times 10^{-4} \mathrm{~kg}-\mathrm{m}^2 .
\end{aligned}
\)

The time period is given by
\(
T=2 \pi \sqrt{\frac{I}{k}}
\)
or,
\(
k=\frac{4 \pi^2 I}{T^2}
\)
\(
\begin{aligned}
& =\frac{4 \pi^2\left(2 \cdot 5 \times 10^{-4} \mathrm{~kg}-\mathrm{m}^2\right)}{(0 \cdot 20 \mathrm{~s})^2} \\
& =0.25 \frac{\mathrm{kg}-\mathrm{m}^2}{\mathrm{~s}^2}
\end{aligned}
\)

Example 6: Describe the motion of the mass \(m\) shown in the figure below. The walls and the block are elastic.

Solution:

The block reaches the spring with a speed \(v\). It now compresses the spring. The block is decelerated due to the spring force, comes to rest when \(\frac{1}{2} m v^2=\frac{1}{2} k x^2\) and returns back. It is accelerated due to the spring force till the spring acquires its natural length. The contact of the block with the spring is now broken. At this instant, it has regained its speed \(v\) (towards left) as the spring is unstretched and no potential energy is stored. This process takes half the period of oscillation, i.e., \(\pi \sqrt{m / k}\). The block strikes the left wall after a time \(L / v\) and as the collision is elastic, it rebounds with the same speed \(v\). After a time \(L / v\), it again reaches the spring and the process is repeated. The block thus undergoes periodic motion with time period \(\pi \sqrt{m / k}+\frac{2 L}{v}\).

Example 7: A block of mass \(m\) is suspended from the ceiling of a stationary standing elevator through a spring of spring constant \(k\). Suddenly, the cable breaks and the elevator starts falling freely. Show that the block now executes a simple harmonic motion of amplitude \(m g / k\) in the elevator.

Solution:

When the elevator is stationary, the spring is stretched to support the block. If the extension is \(x\), the tension is \(k x\) which should balance the weight of the block.

Thus, \(x=m g / k\). As the cable breaks, the elevator starts falling with acceleration ‘ \(g\) ‘. We shall work in the frame of reference of the elevator. Then we have to use a pseudo force \(m g\) upward on the block. This force will ‘balance’ the weight. Thus, the block is subjected to a net force \(k x\) by the spring when it is at a distance \(x\) from the position of unstretched spring. Hence, its motion in the elevator is simple harmonic with its mean position corresponding to the unstretched spring. Initially, the spring is stretched by \(x=m g / k\), where the velocity of the block (with respect to the elevator) is zero. Thus, the amplitude of the resulting simple harmonic motion is \(\mathrm{mg} / \mathrm{k}\).

Example 8: A simple pendulum is suspended from the ceiling of a car accelerating uniformly on a horizontal road. If the acceleration is \(a_0\) and the length of the pendulum is \(l\), find the time period of small oscillations about the mean position.

Solution:

We shall work in the car frame. As it is accelerated with respect to the road, we shall have to apply a pseudo force \(m a_0\) on the bob of mass \(\mathrm{m}\). For mean position, the acceleration of the bob with respect to the car should be zero. If \(\theta\) be the angle made by the string with the vertical, the tension, weight and the pseudo force will add to zero in this position.

Suppose, at some instant during oscillation, the string is further deflected by an angle \(\alpha\) so that the displacement of the bob is \(x\). Taking the components perpendicular to the string,
component of \(T=0\),
component of \(m g=m g \sin (\alpha+\theta)\) and
component of \(m a_0=-m a_0 \cos (\alpha+\theta)\).
Thus, the resultant component \(F\)
\(
=m\left[g \sin (\alpha+\theta)-a_0 \cos (\alpha+\theta)\right] .
\)
Expanding the sine and cosine and putting \(\cos \alpha \approx 1\), \(\sin \alpha \approx \alpha=x / l\), we get
\(
F=m\left[g \sin \theta-a_0 \cos \theta+\left(g \cos \theta+a_0 \sin \theta\right) \frac{x}{l}\right] \dots(i)
\)
At \(x=0\), the force \(F\) on the bob should be zero, as this is the mean position. Thus by (i),
\(
0=m\left[g \sin \theta-a_0 \cos \theta\right] \dots(ii)
\)
\(
\text { giving } \quad \tan \theta=\frac{a_0}{g}
\)
\(
\text { Thus, } \quad \sin \theta=\frac{a_0}{\sqrt{a_0^2+g^2}} \dots(iii)
\)
\(
\cos \theta=\frac{g}{\sqrt{a_0^2+g^2}} \dots(iv)
\)
Putting (ii), (iii) and (iv) in (i), \(F=m \sqrt{g^2+a_0^2} \frac{x}{l}\) or, \(\quad F=m \omega^2 x\), where \(\omega^2=\frac{\sqrt{g^2+a_0^2}}{l}\).
This is an equation of simple harmonic motion with time period
\(
t=\frac{2 \pi}{\omega}=2 \pi \frac{\sqrt{ } l}{\left(g^2+a_0^2\right)^{1 / 4}}
\)
An easy working rule may be found out as follows. In the mean position, the tension, the weight and the pseudo force balance.
From the figure, the tension is
\(
\begin{array}{rlrl}
& T=\sqrt{\left(m a_0\right)^2+(m g)^2} \\
\text { or, } & \frac{T}{m} =\sqrt{a_0^2+g^2}
\end{array}
\)

This plays the role of effective ‘ \(g\) ‘. Thus the time period is
\(
t=2 \pi \sqrt{\frac{l}{T / m}}=2 \pi \frac{\sqrt{l}}{\left[g^2+a_0^2\right]^{1 / 4}} .
\)

QUESTIONS FOR SHORT ANSWER

Q1. A person goes to bed at sharp \(10. 00 \mathrm{pm}\) every day. Is it an example of periodic motion? If yes, what is the time period? If no, why?

Answer:

No, Here, the person is going to bed sharp at 10:00.
But what about his motion, if we take his bed as the reference he might take different paths every day. Someday he might follow.
(1) Bed – Bathroom – Bus stop – Office – Bus stop – Bed
(2) Bed – Bathroom – Cab – Office – Restaurant – Bed
So every day the path might be different, even if the path is same the velocity might be different (at the same time on different days)
Example: He might go to his office every day but the speed of a bike and a car and a bus will be different. Clearly, the motion is not periodic.

Q2. A particle executing simple harmonic motion comes to rest at extreme positions. Is the resultant force on the particle zero at these positions according to Newton’s first law?

Answer: No. The resultant force on the particle is maximum at the extreme positions. In a simple harmonic motion, the magnitude of the resultant force is proportional to the displacement from the mean position. At the extreme positions, the displacement is maximum hence the magnitude of the force is also maximum.

From Newton’s First Law we infer that if a body remains at rest then the resultant force on it is zero but in this case, the body comes to rest momentarily and moves back towards the mean position because the force and acceleration are not zero at extreme points.

Q3. Can simple harmonic motion take place in a noninertial frame? If yes, should the ratio of the force applied with the displacement be constant?

Answer: Yes. Simple harmonic motion can take place in a non-inertial frame. However, the ratio of the force applied to the displacement cannot be constant because a non-inertial frame has some acceleration with respect to the inertial frame. Therefore, a fictitious force should be added to explain the motion.

Q4. A particle executes simple harmonic motion. If you are told that its velocity at this instant is zero, can you say what is its displacement? If you are told that its velocity at this instant is maximum, can you say what is its displacement?

Answer: No, we cannot say anything from the given information. To determine the displacement of the particle using its velocity at any instant, its mean position has to be known.

Q5. A small creature moves with constant speed in a vertical circle on a bright day. Does its shadow formed by the sun on a horizontal plane move in a simple harmonic motion?

Answer: Yes, its shadow on a horizontal plane moves in simple harmonic motion. The projection of a uniform circular motion executes simple harmonic motion along its diameter (which is the shadow on the horizontal plane), with the mean position lying at the centre of the circle.

Q6. A particle executes simple harmonic motion. Let \(P\) be a point near the mean position and \(Q\) be a point near an extreme. The speed of the particle at \(P\) is larger than the speed at \(Q\). Still the particle crosses \(P\) and \(Q\) equal number of times in a given time interval. Does it make you unhappy?

Answer: No. It does not make me unhappy because the number of times a particle crosses the mean and extreme positions does not depend on the speed of the particle.

Q7. In measuring time period of a pendulum, it is advised to measure the time between consecutive passage through the mean position in the same direction. This is said to result in better accuracy than measuring time between consecutive passage through an extreme position. Explain.

Answer: The mean position of a particle executing simple harmonic motion is fixed, whereas its extreme position keeps on changing. Therefore, when we use stopwatch to measure the time between consecutive passage, we are certain about the mean position.

Q8. It is proposed to move a particle in simple harmonic motion on a rough horizontal surface by applying an external force along the line of motion. Sketch the graph of the applied force against the position of the particle. Note that the applied force has two values for a given position depending on whether the particle is moving in positive or negative direction.

Answer: Figure (a) shows the graph of the applied force against the position of the particle. 

Q9. Can the potential energy in a simple harmonic motion be negative? Will it be so if we choose zero potential energy at some point other than the mean position?

Answer: No. It cannot be negative because the minimum potential energy of a particle executing simple harmonic motion at the mean position is zero. The potential energy increases in the positive direction at the extreme position.
However, if we choose zero potential energy at some other point, say extreme position, the potential energy can be negative at the mean position.

Q10. The energy of a system in simple harmonic motion is given by \(E=\frac{1}{2} m \omega^2 A^2\). Which of the following two statements is more appropriate?
(A) The energy is increased because the amplitude is increased.
(B) The amplitude is increased because the energy is increased.

Answer: Statement A is more appropriate because the energy of a system in simple harmonic motion is given by \(E=\frac{1}{2} m \omega^2 A^2\). If the mass \((m)\) and angular frequency \((\omega)\) are made constant, Energy \((E)\) becomes proportional to the square of amplitude \(\left(A^2\right)\)
i.e. \(E \propto A^2\)
Therefore, according to the relation, energy increases as the amplitude increases.

Q11. A pendulum clock gives the correct time at the equator. Will it gain time or lose time as it is taken to the poles?

Answer: According to the relation : \(T=2 \pi \sqrt{\frac{l}{g}}\) The time period \((T)\) of the pendulum becomes proportional to the square root of the inverse of \(g\) if the length of the pendulum is kept constant.
i.e. \(T \propto \sqrt{\frac{1}{g}}\)
Also, acceleration due to gravity \((g)\) at the poles is more than that at the equator. Therefore, the time period decreases and the clock gains time.

Q12. Can a pendulum clock be used in an earth-satellite?

Answer: No. According to the relation:
\(
T=2 \pi \sqrt{\frac{l}{g}}
\)
The time period of the pendulum clock depends upon the acceleration due to gravity. As the earth-satellite is a free-falling body and its \(g_{\text {effective }}\) (effective acceleration due to gravity ) is zero at the satellite, the time period of the clock is infinite.

Q13. A hollow sphere filled with water is used as the bob of a pendulum. Assume that the equation for simple pendulum is valid with the distance between the point of suspension and centre of mass of the bob acting as the effective length of the pendulum. If water slowly leaks out of the bob, how will the time period vary?

Answer: The time period of a pendulum depends on the length and is given by the relation, \(T=2 \pi \sqrt{\frac{l}{g}}\). As the effective length of the pendulum first increases and then decreases, the time period first increases and then decreases.

Q14. A block of known mass is suspended from a fixed support through a light spring. Can you find the time period of vertical oscillation only by measuring the extension of the spring when the block is in equilibrium?

Answer: Yes.
The time period of a spring mass system is given by, \(T=2 \pi \sqrt{\frac{m}{k}} \ldots(1) \quad\) where \(m\) is mass of the block, and \(k\) is the spring constant
The time period is also given by the relation,
\(
T=2 \pi \sqrt{\frac{x_0}{g}} \dots(2)
\)
where, \(x_0\) is extension of the spring, and
\(g\) is acceleration due to gravity
From equations (1) and (2), we have :
\(
\begin{aligned}
& m g=k x_0 \\
& \Rightarrow k=\frac{m g}{x_0}
\end{aligned}
\)
Substituting the value of \(k\) in the above equation, we get:
\(
T=2 \pi \sqrt{\frac{m}{\frac{m g}{x_0}}}=2 \pi \sqrt{\frac{x_0}{g}}
\)
Thus, we can find the time period if the value of extension \(x_0\) is known.

Q15. A platoon of soldiers marches on a road in steps according to the sound of a marching band. The band is stopped and the soldiers are ordered to break the steps while crossing a bridge. Why?

Answer: When the frequency of soldiers’ feet movement becomes equal to the natural frequency of the bridge, and resonance occurs between soldiers’ feet movement and movement of the bridge, maximum transfer of energy occurs from soldiers’ feet to the bridge, which increases the amplitude of vibration. A continued increase in the amplitude of vibration, however, may lead to collapsing of the bridge.

Q16. The force acting on a particle moving along \(X\)-axis is \(F=-k\left(x-v_0 t\right)\) where \(k\) is a positive constant. An observer moving at a constant velocity \(v_0\) along the \(X\)-axis looks at the particle. What kind of motion does he find for the particle?

Answer: As the observer moves with a constant velocity along the same axis, he sees the same force on the particle and finds the motion of the particle is not simple harmonic motion.