Oscillations
13.1 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 non-repetitive. 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, the 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 section.
13.2 PERIODIC AND OSCILLATORY MOTIONS
Fig. 13.1 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 versus time, it would look something like Fig. 13.1 (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. 13.1 (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 in Fig. 13.1 (c). Note that both the curved parts in Fig. 13.1 (c) are sections of a parabola given by the Newton’s equation of motion (see section 2.6),
$h=u t+\frac{1}{2} g t^{2}$ for downward motion, and
$h=u t-\frac{1}{2} g t^{2}$ 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.
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 it 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 the 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.
Any material medium can be pictured as a collection of a large number of coupled oscillators. The collective oscillations of the constituents of a medium manifest themselves as waves. Examples of waves include water waves, seismic waves, electromagnetic waves. We shall study the wave phenomenon in the next chapter.
13.2.1 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 $\left(10^{-6} \mathrm{~s}\right)$ abbreviated as $\mu \mathrm{s}$. 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 $v$. The relation between $v$ and $T$ is
$$ \begin{equation*} v=1 / T \tag{13.1} \end{equation*} $$
The unit of $v$ is thus $\mathrm{s}^{-1}$. After the discoverer of radio waves, Heinrich Rudolph Hertz (1857-1894), a special name has been given to the unit of frequency. It is called hertz (abbreviated as $\mathrm{Hz}$ ). Thus,
1 hertz $=1 \mathrm{~Hz}=1$ oscillation per second $=1 \mathrm{~s}^{-1}$
Note, that the frequency, $v$, is not necessarily an integer.
Example 13.1 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 $=75 /(1 \mathrm{~min})$
$$ \begin{aligned} & =75 /(60 \mathrm{~s}) \\ & =1.25 \mathrm{~s}^{-1} \\ & =1.25 \mathrm{~Hz} \\ \text { The time period } T \quad & =1 /\left(1.25 \mathrm{~s}^{-1}\right) \\ & =0.8 \mathrm{~s} \end{aligned} $$
14.2.2 Displacement
In section 3.2, we defined displacement of a 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 steel 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. 13.2(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. 13.2(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. The voltage across a capacitor, changing with time in an $\mathrm{AC}$ 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
$$ \begin{equation*} f(t)=A \cos \omega t \tag{13.3a} \end{equation*} $$
If the argument of this function, $\omega t$, is increased by an integral multiple of $2 \pi$ radians, the value of the function remains the same. The function $f(t)$ is then periodic and its period, $T$, is given by
$$ \begin{equation*} T=\frac{2 \pi}{\omega} \tag{13.3b} \end{equation*} $$
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 function, $f(t)=A \sin \omega t$. Further, a linear combination of sine and cosine functions like,
$$ \begin{equation*} f(t)=A \sin \omega t+B \cos \omega t \tag{13.3c} \end{equation*} $$
is also a periodic function with the same period $T$. Taking,
$$ A=D \cos \phi \text { and } B=D \sin \phi $$
Eq. (13.3c) can be written as,
$$ \begin{equation*} f(t)=D \sin (\omega t+\phi), \tag{13.3d} \end{equation*} $$
Here $D$ and $\phi$ are constant given by
$$ D=\sqrt{A^{2}+B^{2}} \text { and } \varphi=\tan ^{-1} \frac{B}{A} $$
The great importance of periodic sine and cosine functions is due to a remarkable result proved by the French mathematician, Jean Baptiste Joseph Fourier (1768-1830): Any periodic function can be expressed as a superposition of sine and cosine functions of different time periods with suitable coefficients.
Example 13.2 Which of the following functions of time represent (a) periodic and (b) non-periodic motion? Give the period for each case of periodic motion [$\omega$ is any positive constant]
(i) $\sin \omega t+\cos \omega t$
(ii) $\sin \omega t+\cos 2 \omega t+\sin 4 \omega t$
(iii) $\mathrm{e}^{-\omega t}$
(iv) $\log (\omega t)$
Answer
(i) $\sin \omega t+\cos \omega t$ is a periodic function, it can also be written as $\sqrt{2} \sin (\omega t+\pi / 4)$.
Now $\sqrt{2} \sin (\omega t+\pi / 4)=\sqrt{2} \sin (\omega t+\pi / 4+2 \pi)$
$=\sqrt{2} \sin [\omega(\mathrm{t}+2 \pi / \omega)+\pi / 4]$
The periodic time of the function is $2 \pi / \omega$. (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, $\sin \omega t$ has a period $T_{0}=2 \pi / \omega ; \cos 2 \omega t$ has a period $\pi / \omega=T_{0} / 2$; and $\sin 4 \omega t$ has a period $2 \pi / 4 \omega=T_{o} / 4$. The period of the first term is a multiple of the periods of the last two terms. Therefore, the smallest interval of time after which the sum of the three terms repeats is $T_{0}$, and thus, the sum is a periodic function with a period $2 \pi / \omega$.
(iii) The function $e^{-\omega t}$ is not periodic, it decreases monotonically with increasing time and tends to zero as $t \rightarrow \infty$ and thus, never repeats its value.
(iv) The function $\log (\omega t)$ increases monotonically with time $t$. It, therefore, never repeats its value and is a nonperiodic function. It may be noted that as $t \rightarrow \infty, \log (\omega t)$ diverges to $\infty$. It, therefore, cannot represent any kind of physical displacement.
13.3 SIMPLE HARMONIC MOTION
Consider a particle oscillating back and forth about the origin of an $x$-axis between the limits $+A$ and $-A$ as shown in Fig. 13.3. This oscillatory motion is said to be simple harmonic if the displacement $x$ of the particle from the origin varies with time as :
$$ \begin{equation*} x(t)=A \cos (\omega t+\phi) \tag{13.4} \end{equation*} $$
where $A, \omega$ and $\phi$ are constants.
Thus, simple harmonic motion (SHM) is not any periodic motion but one in which displacement is a sinusoidal function of time. Fig. 13.4 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. 13.5 plots the graph of $x$ versus $t$, which gives the values of displacement as a continuous function of time. The quantities $A$,
$\omega$ and $\phi$ which characterize a given SHM have standard names, as summarised in Fig. 13.6. Let us understand these quantities.
The amplitutde $A$ 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 $\omega$ and $\phi$ but different amplitudes $A$ and $B$, as shown in Fig. 13.7 (a).
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 $(\omega t+\phi)$ in the cosine function. This time-dependent quantity, $(\omega t+\phi)$ is called the phase of the motion. The value of plase at $t=0$ is $\phi$ and is called the phase constant (or phase angle). If the amplitude is known, $\phi$ can be determined from the displacement at $t=0$. Two simple harmonic motions may have the same $A$ and $\omega$ but different phase angle $\phi$, as shown in Fig. 13.7 (b).
Finally, the quantity $\omega$ can be seen to be related to the period of motion $T$. Taking, for simplicity, $\phi=0$ in Eq. (13.4), we have
$$ \begin{equation*} x(t)=A \cos \omega t \tag{13.5} \end{equation*} $$
Since the motion has a period $T, x(t)$ is equal to $x(t+T)$. That is,
$$ \begin{equation*} A \cos \omega t=A \cos \omega(t+T) \tag{13.6} \end{equation*} $$
Now the cosine function is periodic with period $2 \pi$, i.e., it first repeats itself when the argument changes by $2 \pi$. Therefore,
$$ \omega(t+T)=\omega t+2 \pi $$
$$ \text{that is } \quad \omega=2 \pi / T \tag{13.7}$$
$\omega$ is called the angular frequency of SHM. Its S.I. unit is radians per second. Since the frequency of oscillations is simply $1 / \mathrm{T}, \omega$ is $2 \pi$ times the frequency of oscillation. Two simple harmonic motions may have the same $\mathrm{A}$ and $\phi$, but different $\omega$, as seen in Fig. 13.8. In this plot the curve (b) has half the period and twice the frequency of the curve (a)
Example 13.3 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) $\sin \omega t-\cos \omega t$
(2) $\sin ^{2} \omega t$
Answer
(a) $\sin \omega t-\cos \omega t$
$$ \begin{aligned} &= \sin \omega t-\sin (\pi / 2-\omega t) \\ &= 2 \cos (\pi / 4) \sin (\omega t-\pi / 4) \\ &=\sqrt{ } 2 \sin (\omega t-\pi / 4) \end{aligned} $$
This function represents a simple harmonic motion having a period $T=2 \pi / \omega$ and a phase angle $(-\pi / 4)$ or $(7 \pi / 4)$
(b) $\sin ^{2} \omega t$
$$ =1 / 2-1 / 2 \cos 2 \omega t $$
The function is periodic having a period $T=\pi / \omega$. It also represents a harmonic motion with the point of equilibrium occurring at $1 / 2$ instead of zero.
13.4 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. 13.9) helps us visualise 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 midpoint. 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. 13.10 describes the same situation mathematically. Suppose a particle $\mathrm{P}$ is moving uniformly on a circle of radius $A$ with angular speed $\omega$. The sense of rotation is anticlockwise. The initial position vector of the particle, i.e., the vector $\overline{\mathbf{O P}}$ at $t=0$ makes an angle of $\phi$ with the positive direction of $x$-axis. In time $t$, it will cover a further angle $\omega t$ and its position vector
will make an angle of $\omega t+\phi$ with the + ve $x$-axis. Next, consider the projection of the position vector OP on the $x$-axis. This will be $\mathrm{OP}^{\prime}$. The position of $\mathrm{P}^{\prime}$ on the $x$-axis, as the particle $\mathrm{P}$ moves on the circle, is given by
$$ x(t)=A \cos (\omega t+\phi) $$
which is the defining equation of SHM. This shows that if $\mathrm{P}$ moves uniformly on a circle, its projection $\mathrm{P}^{\prime}$ 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 $\mathrm{P}$ on any diameter, say the $y$-axis. In that case, the displacement $y(t)$ of $\mathrm{P}^{\prime}$ on the $y$-axis is given by
$$ y=A \sin (\omega t+\phi) $$
which is also an SHM of the same amplitude as that of the projection on $x$-axis, but differing by a phase of $\pi / 2$.
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 13.4 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 $x$-projection of the radius vector of the rotating particle $\mathrm{P}$ in each case.
Answer
(a) At $t=0$, OP makes an angle of $45^{\circ}=\pi / 4 \mathrm{rad}$ with the (positive direction of) $x$-axis. After
time $t$, it covers an angle $\frac{2 \pi}{T} t$ in the anticlockwise sense, and makes an angle of $\frac{2 \pi}{T} t+\frac{\pi}{4}$ with the $x$-axis.
The projection of OP on the $\mathrm{x}$-axis at time $t$ is given by,
$$ x(t)=A \cos \frac{2 \pi}{T} t+\frac{\pi}{4} $$
For $T=4 \mathrm{~s}$
$$ x(t)=A \cos \frac{2 \pi}{4} t+\frac{\pi}{4} $$
which is a SHM of amplitude $A$, period $4 \mathrm{~s}$, and an initial phase $*=\frac{\pi}{4}$.
(b) In this case at $t=0$, OP makes an angle of $90^{\circ}=\frac{\pi}{2}$ with the $x$-axis. After a time $t$, it covers an angle of $\frac{2 \pi}{T} t$ in the clockwise sense and makes an angle of $\frac{\pi}{2}-\frac{2 \pi}{T} t$ with the $x$-axis. The projection of OP on the $x$-axis at time $t$ is given by
$$ \begin{array}{r} x(t)=B \cos \frac{\pi}{2}-\frac{2 \pi}{T} t \\ \\ =B \sin \frac{2 \pi}{T} t \end{array} $$
For $T=30 \mathrm{~s}$,
$$ x(t)=B \sin \frac{\pi}{15} t $$
Writing this as $x(t)=B \cos \frac{\pi}{15} t-\frac{\pi}{2}$, and comparing with Eq. (13.4). We find that this represents a SHM of amplitude $B$, period $30 \mathrm{~s}$, and an initial phase of $-\frac{\pi}{2}$.
13.5 VELOCITY AND ACCELERATION IN SIMPLE HARMONIC MOTION
The speed of a particle $v$ in uniform circular motion is its angular speed $\omega$ times the radius of the circle $A$.
$$ \begin{equation*} V=\omega A \tag{13.8} \end{equation*} $$
The direction of velocity $\overline{\mathbf{v}}$ 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. 13.11, it is clear that the velocity of the projection particle $\mathrm{P}^{\prime}$ at time $t$ is
$$ \begin{equation*} v(t)=-\omega A \sin (\omega t+\phi) \tag{13.9} \end{equation*} $$
where the negative sign shows that $v(\mathrm{t})$ has a direction opposite to the positive direction of $x$-axis. Eq. (13.9) gives the instantaneous velocity of a particle executing SHM, where displacement is given by Eq. (13.4). We can, of course, obtain this equation without using geometrical argument, directly by differentiating (Eq. 13.4) with respect of $t$ :
$$ \begin{equation*} v(t)=\frac{\mathrm{d}}{\mathrm{d} t} x(t) \tag{13.10} \end{equation*} $$
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 $\mathrm{P}$ in uniform circular motion has a magnitude $v^{2} / \mathrm{A}$ or $\omega^{2} \mathrm{~A}$, and it is directed towards the centre i.e., the direction is along PO. The instantaneous acceleration of the projection particle $\mathrm{P}^{\prime}$ is then (See Fig. 13.12)
$$ \begin{align*} a(t) & =-\omega^{2} A \cos (\omega t+\phi) \\ & =-\omega^{2} x(t) \tag{13.11} \end{align*} $$
Eq. (13.11) gives the acceleration of a particle in SHM. The same equation can again be obtained directly by differentiating velocity $v(t)$ given by Eq. (13.9) with respect to time:
$$ \begin{equation*} a(t)=\frac{\mathrm{d}}{\mathrm{d} t} v(t) \tag{13.12} \end{equation*} $$
We note from Eq. (13.11) the important property that acceleration of a particle in SHM is proportional to displacement. For $\mathrm{x}(t)>0$, $a(t)<0$ and for $x(t)<0, a(t)>0$. 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 $\phi=0$ and write the expression for $x(t), v(t)$ and $a(t)$
$x(t)=A \cos \omega t, v(t)=-\omega A \sin \omega t, a(t)=-\omega^{2} A \cos \omega t$ The corresponding plots are shown in Fig. 13.13. 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 $-\omega A$ to $\omega A$ and $a(t)$ from $-\omega^{2} A$ to $\omega^{2}$. With respect to displacement plot, velocity plot has a phase difference of $\pi / 2$ and acceleration plot has a phase difference of $\pi$.
Example 13.5 A body oscillates with SHM according to the equation (in SI units),
$$ x=5 \cos [2 \pi t+\pi / 4] . $$
At $t=1.5 \mathrm{~s}$, calculate the (a) displacement, (b) speed and (c) acceleration of the body.
Answer The angular frequency $\omega$ of the body $=2 \pi \mathrm{s}^{-1}$ and its time period $T=1 \mathrm{~s}$.
At $t=1.5 \mathrm{~s}$
(a) displacement $=(5.0 \mathrm{~m}) \cos \left[\left(2 \pi \mathrm{s}^{-1}\right) \times\right.$ $1.5 \mathrm{~s}+\pi / 4]$
$$ \begin{aligned} & =(5.0 \mathrm{~m}) \cos [(3 \pi+\pi / 4)] \\ & =-5.0 \times 0.707 \mathrm{~m} \\ & =-3.535 \mathrm{~m} \end{aligned} $$
(b) Using Eq. (13.9), the speed of the body
$=-(5.0 \mathrm{~m})\left(2 \pi \mathrm{s}^{-1}\right) \sin \left[\left(2 \pi \mathrm{s}^{-1}\right) \times 1.5 \mathrm{~s}\right.$ $+\pi / 4]$
$=-(5.0 \mathrm{~m})\left(2 \pi \mathrm{s}^{-1}\right) \sin [(3 \pi+\pi / 4)]$
$=10 \pi \times 0.707 \mathrm{~m} \mathrm{~s}^{-1}$
$=22 \mathrm{~m} \mathrm{~s}^{-1}$
(c) Using Eq.(13.10), the acceleration of the body
$$ =-\left(2 \pi \mathrm{s}^{-1}\right)^{2} \times \text { displacement } $$
$$ =-\left(2 \pi \mathrm{s}^{-1}\right)^{2} \times(-3.535 \mathrm{~m}) $$
$$=140 \mathrm{~m} \mathrm{~s}^{-2}$$
13.6 FORCE LAW FOR SIMPLE HARMONIC MOTION
Using Newton’s second law of motion, and the expression for acceleration of a particle undergoing SHM (Eq. 13.11), the force acting on a particle of mass $m$ in SHM is
$$ \begin{align*} F(t) & =m a \\ & =-m \omega^{2} x(t) \end{align*} $$
$$ \text{i.e.,} \quad F(t)=-k x(t) \tag{13.13}$$
$$ \text{where} \quad k=m \omega^{2} \tag{13.14a}$$
$$ \text{or} \quad \omega=\sqrt{\frac{k}{m}} \tag{13.14b}$$
Like acceleration, force is always directed towards the mean position-hence 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. (13.4) for displacement or by Eq. (13.13) that gives its force law. Going from Eq. (13.4) to Eq. (13.13) required us to differentiate two times. Likewise, by integrating the force law Eq. (13.13) two times, we can get back Eq. (13.4).
Note that the force in Eq. (13.13) 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 $x^{2}, x^{3}$, etc. These then are called non-linear oscillators.
Example 13.6 Two identical springs of spring constant $k$ are attached to a block of mass $m$ and to fixed supports as shown in Fig. 13.14. 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. 13.15. 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,
$$ \begin{array}{ll} F_{1}=-k x &\text { (force exerted by the spring on } \\ &\text { the left side, trying to pull the } \\ &\text { mass towards the mean position } \\ F_{2}=-k x &\text { (force exerted by the spring on } \\ & \text { the right side, trying to push the } \\ & \text { mass towards the mean } \\ & \text { position) } \end{array} $$
The net force, $F$, acting on the mass is then given by,
$$ F=-2 k x $$
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,
$$ T=2 \pi \sqrt{\frac{m}{2 k}} $$
13.7 ENERGY IN SIMPLE HARMONIC MOTION
Both kinetic and potential energies of a particle in SHM vary between zero and their maximum values.
In section 13.5 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
$$ \begin{align*} K & =\frac{1}{2} m v^{2} \\ & =\frac{1}{2} m \omega^{2} A^{2} \sin ^{2}(\omega t+\phi) \\ & =\frac{1}{2} k A^{2} \sin ^{2}(\omega t+\phi) \tag{13.15} \end{align*} $$
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 Chapter 6, we have seen that the concept of potential energy is possible only for conservative forces. The spring force $F=-k x$ is a conservative force, with associated potential energy
$$ \begin{equation*} U=\frac{1}{2} k x^{2} \tag{13.16} \end{equation*} $$
Hence the potential energy of a particle executing simple harmonic motion is,
$$ \begin{align*} & U(x)=\frac{1}{2} k x^{2} \\ & =\frac{1}{2} k A^{2} \cos ^{2}(\omega t+\phi) \tag{13.17} \end{align*} $$
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. (13.15) and (13.17) that the total energy, $E$, of the system is,
$$ \begin{aligned} & E=U+K \\ & =\frac{1}{2} k A^{2} \cos ^{2}(\omega t+\phi)+\frac{1}{2} k A^{2} \sin ^{2}(\omega t+\phi) \\ & =\frac{1}{2} k A^{2}\left[\cos ^{2}(\omega t+\phi)+\sin ^{2}(\omega t+\phi)\right] \end{aligned} $$
Using the familiar trigonometric identity, the value of the expression in the brackets is unity. Thus,
$$ \begin{equation*} E=\frac{1}{2} k A^{2} \tag{13.18} \end{equation*} $$
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. 13.16.
Observe that both kinetic energy and potential energy in SHM are seen to be always positive in Fig. 13.16. 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 $x=0$, the energy is kinetic; at the extremes $x= \pm A$, it is all potential energy. In the course of motion between these limits, kinetic energy increases at the expense of potential energy or vice-versa.
Example 13.7 A block whose mass is $1 \mathrm{~kg}$ is fastened to a spring. The spring has a spring constant of $50 \mathrm{~N} \mathrm{~m}^{-1}$. The block is pulled to a distance $x=10 \mathrm{~cm}$ 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 $5 \mathrm{~cm}$ away from the mean position.
Answer The block executes SHM, its angular frequency, as given by Eq. (13.14b), is
$$ \omega=\sqrt{\frac{k}{m}} $$
$$ \begin{aligned} & =\sqrt{\frac{50 \mathrm{~N} \mathrm{~m}^{-1}}{1 \mathrm{~kg}}} \\ \\ & =7.07 \mathrm{rad} \mathrm{s}^{-1} \end{aligned} $$
Its displacement at any time $t$ is then given by,
$$ x(t)=0.1 \cos (7.07 t) $$
Therefore, when the particle is $5 \mathrm{~cm}$ away from the mean position, we have
$$ 0.05=0.1 \cos (7.07 t) $$
Or $\cos (7.07 t)=0.5$ and hence
$\sin (7.07 t)=\frac{\sqrt{3}}{2}=0.866$
Then, the velocity of the block at $x=5 \mathrm{~cm}$ is
$$ \begin{aligned} & =0.1 \times 7.070 .866 \mathrm{~m} \mathrm{~s}^{-1} \\ & =0.61 \mathrm{~m} \mathrm{~s}^{-1} \end{aligned} $$
Hence the K.E. of the block,
$$ \begin{aligned} & =\frac{1}{2} m v^{2} \\ & =1 / 2\left[1 \mathrm{~kg} \times\left(0.6123 \mathrm{~m} \mathrm{~s}^{-1}\right)^{2}\right] \\ & =0.19 \mathrm{~J} \end{aligned} $$
The P.E. of the block,
$$ \begin{aligned} & =\frac{1}{2} k x^{2} \\ & =1 / 2\left(50 \mathrm{~N} \mathrm{~m}^{-1} \times 0.05 \mathrm{~m} \times 0.05 \mathrm{~m}\right) \\ & =0.0625 \mathrm{~J} \end{aligned} $$
The total energy of the block at $x=5 \mathrm{~cm}$,
$$ =\text { K.E. + P.E. } $$
$$ =0.25 \mathrm{~J} $$
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,
$$ \begin{aligned} & =1 / 2\left(50 \mathrm{~N} \mathrm{~m}^{-1} \times 0.1 \mathrm{~m} \times 0.1 \mathrm{~m}\right) \\ & =0.25 \mathrm{~J} \end{aligned} $$
which is same as the sum of the two energies at a displacement of $5 \mathrm{~cm}$. This is in conformity with the principle of conservation of energy.
13.8 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 $100 \mathrm{~cm}$ 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 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. 13.17(a) shows this system. Fig. 13.17(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 $\mathrm{T}$ 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 tangental 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 $\mathrm{T}-m g \cos \theta$, 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 tangental component of force
$$ \begin{equation*} \tau=-L(m g \sin \theta) \tag{13.19} \end{equation*} $$
This is the restoring torque that tends to reduce angular displacement - hence the negative sign. By Newton’s law of rotational motion,
$$ \begin{equation*} \tau=I \alpha \tag{13.20} \end{equation*} $$
where $I$ is the moment of inertia of the system about the support and $\alpha$ is the angular acceleration. Thus,
$$ \begin{equation*} I \alpha=-m g \sin \theta \quad L \tag{13.21} \end{equation*} $$
Or,
$$ \begin{equation*} \alpha=-\frac{m g L}{I} \sin \theta \tag{13.22} \end{equation*} $$
We can simplify Eq. (13.22) if we assume that the displacement $\theta$ is small. We know that $\sin \theta$ can be expressed as,
$$ \begin{equation*} \sin \theta=\theta-\frac{\theta^{3}}{3 !}+\frac{\theta^{5}}{5 !} \pm \ldots \tag{13.23} \end{equation*} $$
where $\theta$ is in radians.
Now if $\theta$ is small, $\sin \theta$ can be approximated by $\theta$ and Eq. (13.22) can then be written as,
$$ \begin{equation*} \alpha=-\frac{m g L}{I} \theta \tag{13.24} \end{equation*} $$
In Table 13.1, we have listed the angle $\theta$ in degrees, its 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. Table $13.1 \sin \theta$ as ma function of angle $\theta$
| $\theta$ (degrees) | $\theta$ (radians) | $\sin \theta$ |
|---|---|---|
| 0 | 0 | 0 |
| 5 | 0.087 | 0.087 |
| 10 | 0.174 | 0.174 |
| 15 | 0.262 | 0.259 |
| 20 | 0.349 | 0.342 |
Equation (13.24) is mathematically, identical to Eq. (13.11) except that the variable is angular displacement. Hence we have proved that for small q, the motion of the bob is simple harmonic. From Eqs. (13.24) and (13.11),
$$ \omega=\sqrt{\frac{m g L}{I}} $$
and
$$ \begin{equation*} T=2 \pi \sqrt{\frac{I}{m g L}} \tag{13.25} \end{equation*} $$
Now since the string of the simple pendulum is massless, the moment of inertia $I$ is simply $\mathrm{mL}^{2}$. Eq. (13.25) then gives the well-known formula for time period of a simple pendulum.
$$ \begin{equation*} T=2 \pi \sqrt{\frac{L}{g}} \tag{13.26} \end{equation*} $$
Example 13.8 What is the length of a simple pendulum, which ticks seconds?
Answer From Eq. (13.26), 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
$$ \begin{aligned} & =\frac{9.8\left(\mathrm{~m} \mathrm{~s}^{-2}\right) \times 4\left(\mathrm{~s}^{2}\right)}{4 \pi^{2}} \ & =1 \mathrm{~m} \end{aligned} $$
SUMMARY
1. The motion that repeats itself is called periodlic motion.
2. The period $T$ is the time reequired for one complete oscillation, or cycle. It is related to the frequency $v$ by,
$$ T=\frac{1}{v} $$
The frequency $v$ of periodic or oscillatory motion is the number of oscillations per unit time. In the SI, it is measured in hertz :
$$ 1 \text { hertz }=1 \mathrm{~Hz}=1 \text { oscillation per second }=1 \mathrm{~s}^{-1} $$
3. In simple harmonic motion (SHM), the displacement $x(t)$ of a particle from its equilibrium position is given by,
$$ x(t)=A \cos (\omega t+\phi) \quad \text { (displacement) } $$
in which $A$ is the amplitude of the displacement, the quantity $(\omega t+\phi)$ is the phase of the motion, and $\phi$ is the phase constant. The angular frequency $\omega$ is related to the period and frequency of the motion by,
$$ \omega=\frac{2 \pi}{T}=2 \pi \nu \quad \text { (angular frequency). } $$
4. Simple harmonic motion can also be viewed as the projection of uniform circular motion on the diameter of the circle in which the latter motion occurs.
5. The particle velocity and acceleration during SHM as functions of time are given by,
$$ \begin{array}{rlr} v(t) & =-\omega A \sin (\omega t+\phi) & \text { (velocity), } \\ \\ a(t) & =-\omega^{2} A \cos (\omega t+\phi) & \\ \\ & =-\omega^{2} x(t) & \text { (acceleration), } \end{array} $$
Thus we see that both velocity and acceleration of a body executing simple harmonic motion are periodic functions, having the velocity amplitude $v_{m}=\omega A$ and acceleration amplitude $a_{m}=\omega^{2} A$, respectively.
6. The force acting in a simple harmonic motion is proportional to the displacement and is always directed towards the centre of motion.
7. A particle executing simple harmonic motion has, at any time, kinetic energy $K=1 / 2 m v^{2}$ and potential energy $U=1 / 2 \mathrm{kx}^{2}$. If no friction is present the mechanical energy of the system, $E=K+U$ always remains constant even though $K$ and $U$ change with time.
8. A particle of mass $m$ oscillating under the influence of Hooke’s law restoring force given by $F=-k x$ exhibits simple harmonic motion with
$$ \begin{array}{ll} \omega=\sqrt{\frac{k}{m}} & \text { (angular frequency) } \\ \\ T=2 \pi \sqrt{\frac{m}{k}} & \text { (period) } \end{array} $$
Such a system is also called a linear oscillator.
9. The motion of a simple pendulum swinging through small angles is approximately simple harmonic. The period of oscillation is given by,
$$ T=2 \pi \sqrt{\frac{L}{g}} $$
| Physical quantity | Symbol | Dimensions | Unit | Remarks |
|---|---|---|---|---|
| Period | $T$ | $[\mathrm{~T}]$ | $\mathrm{s}$ | The least time for motion to repeat itself |
| Frequency | $v(o r f)$ | $\left[\mathrm{T}^{-1}\right]$ | $\mathrm{s}^{-1}$ | $v=\frac{1}{T}$ |
| Angular frequency | $\omega$ | $\left[\mathrm{T}^{-1}\right]$ | $\mathrm{s}^{-1}$ | $\omega=2 \pi v$ |
| Phase constant | $\phi$ | Dimensionless | rad | Initial value of phase of displacement in SHM |
| Force constant | $k$ | $\left[\mathrm{MT}^{-2}\right]$ | $\mathrm{N} \mathrm{m}^{-1}$ | Simple harmonic motion $F=-k x$ |
POINTS TO PONDER
1. The period $T$ is the least time after which motion repeats itself. Thus, motion repeats itself after $n T$ where $n$ is an integer.
2. Every periodic motion is not simple harmonic motion. Only that periodic motion governed by the force law $F=-k x$ is simple harmonic.
3. Circular motion can arise due to an inverse-square law force (as in planetary motion) as well as due to simple harmonic force in two dimensions equal to: $-m \omega^{2} r$. In the latter case, the phases of motion, in two perpendicular directions ( $x$ and $y$ ) must differ by $\pi / 2$. Thus, for example, a particle subject to a force $-m \omega^{2} r$ with initial position $(0$, $A)$ and velocity $(\omega A, 0)$ will move uniformly in a circle of radius $A$.
4. For linear simple harmonic motion with a given $\omega$, two initial conditions are necessary and sufficient to determine the motion completely. The initial conditions may be (i) initial position and initial velocity or (ii) amplitude and phase or (iii) energy and phase.
5. From point 4 above, given amplitude or energy, phase of motion is determined by the initial position or initial velocity.
6. A combination of two simple harmonic motions with arbitrary amplitudes and phases is not necessarily periodic. It is periodic only if frequency of one motion is an integral multiple of the other’s frequency. However, a periodic motion can always be expressed as a sum of infinite number of harmonic motions with appropriate amplitudes.
7. The period of SHM does not depend on amplitude or energy or the phase constant. Contrast this with the periods of planetary orbits under gravitation (Kepler’s third law).
8. The motion of a simple pendulum is simple harmonic for small angular displacement.
9. For motion of a particle to be simple harmonic, its displacement $x$ must be expressible in either of the following forms :
$$ \begin{aligned} & x=A \cos \omega t+B \sin \omega t \\ & x=A \cos (\omega t+\alpha), x=B \sin (\omega t+\beta) \end{aligned} $$
The three forms are completely equivalent (any one can be expressed in terms of any other two forms).
Thus, damped simple harmonic motion is not strictly simple harmonic. It is approximately so only for time intervals much less than $2 \mathrm{~m} / \mathrm{b}$ where $b$ is the damping constant.
EXERCISES
13.1 Which of the following examples represent periodic motion?
(a) A swimmer completing one (return) trip from one bank of a river to the other and back.
(b) A freely suspended bar magnet displaced from its $\mathrm{N}-\mathrm{S}$ direction and released.
(c) A hydrogen molecule rotating about its centre of mass.
(d) An arrow released from a bow.
Answer (b) and (c) The swimmer’s motion is not periodic. The motion of the swimmer between the banks of a river is back and forth. However, it does not have a definite period. This is because the time taken by the swimmer during his back and forth journey may not be the same. The motion of a freely-suspended magnet, if displaced from its N-S direction and released, is periodic. This is because the magnet oscillates about its position with a definite period of time. When a hydrogen molecule rotates about its centre of mass, it comes to the same position again and again after an equal interval of time. Such motion is periodic. An $\to$ released from a bow moves only in the forward direction. It does not come backward. Hence, this motion is not a periodic.Show Answer
(a) the rotation of earth about its axis.
(b) motion of an oscillating mercury column in a U-tube.
(c) motion of a ball bearing inside a smooth curved bowl, when released from a point slightly above the lower most point.
(d) general vibrations of a polyatomic molecule about its equilibrium position.
Answer (b) and (c) are SHMs and (d) are periodic, but not SHMs During its rotation about its axis, earth comes to the same position again and again in equal intervals of time. Hence, it is a periodic motion. However, this motion is not simple harmonic. This is because earth does not have a to and fro motion about its axis. An oscillating mercury column in a U-tube is simple harmonic. This is because the mercury moves to and fro on the same path, about the fixed position, with a certain period of time. The ball moves to and fro about the lowermost point of the bowl when released. Also, the ball comes back to its initial position in the same period of time, again and again. Hence, its motion is periodic as well as simple harmonic. A polyatomic molecule has many natural frequencies of oscillation. Its vibration is the superposition of individual simple harmonic motions of a number of different molecules. Hence, it is not simple harmonic, but periodic.Show Answer
Answer (b) and (d) are periodic It is not a periodic motion. This represents a unidirectional, linear uniform motion. There is no repetition of motion in this case. In this case, the motion of the particle repeats itself after $2 s$. Hence, it is a periodic motion, having a period of $2 s$. It is not a periodic motion. This is because the particle repeats the motion in one position only. For a periodic motion, the entire motion of the particle must be repeated in equal intervals of time. In this case, the motion of the particle repeats itself after $2 s$. Hence, it is a periodic motion, having a period of $2 s$.Show Answer
(a) $\sin \omega t-\cos \omega t$
(b) $\sin ^{3} \omega t$
(c) $3 \cos (\pi / 4-2 \omega t)$
(d) $\cos \omega t+\cos 3 \omega t+\cos 5 \omega t$
(e) $\exp \left(-\omega^{2} t^{2}\right)$
(f) $1+\omega t+\omega^{2} t^{2}$
Answer (a) SHM The given function is: $
\begin{aligned}
& \sin \omega t-\cos \omega t \\
& =\sqrt{2}[\frac{1}{\sqrt{2}} \sin \omega t-\frac{1}{\sqrt{2}} \cos \omega t] \\
& =\sqrt{2}[\sin \omega t \times \cos \frac{\pi}{4}-\cos \omega t \times \sin \frac{\pi}{4}] \\
& =\sqrt{2} \sin (\omega t-\frac{\pi}{4})
\end{aligned}
$ This function represents SHM as it can be written in the form: ${ }^{a} \sin (\omega t+\phi)$ Its period is: $\frac{2 \pi}{\omega}$ (b) Periodic, but not SHM The given function is: $\sin ^{3} \omega t$ $=\frac{1}{2}[3 \sin \omega t-\sin 3 \omega t]$ The terms $\sin \omega t$ and $\sin \omega t$ individually represent simple harmonic motion (SHM). However, the superposition of two SHM is periodic and not simple harmonic. (c) SHM The given function is: $3 \cos [\frac{\pi}{4}-2 \omega t]$ $=3 \cos [2 \omega t-\frac{\pi}{4}]$ This function represents simple harmonic motion because it can be written in the form: $a \cos (\omega t+\phi)$ Its period is: $\frac{2 \pi}{2 \omega}=\frac{\pi}{\omega}$ (d) Periodic, but not SHM The given function is $\cos \omega t+\cos 3 \omega t+\cos 5 \omega t$. Each individual cosine function represents SHM. However, the superposition of three simple harmonic motions is periodic, but not simple harmonic. (e) Non-periodic motion The given function $\exp (-\omega^{2} t^{2})$ is an exponential function. Exponential functions do not repeat themselves. Therefore, it is a non-periodic motion. (f) The given function $1+\omega t+\omega^{2} t^{2}$ is non-periodic.Show Answer
(a) at the end $\mathrm{A}$,
(b) at the end $\mathrm{B}$,
(c) at the mid-point of $\mathrm{AB}$ going towards $\mathrm{A}$,
(d) at $2 \mathrm{~cm}$ away from $\mathrm{B}$ going towards $\mathrm{A}$,
(e) at $3 \mathrm{~cm}$ away from $A$ going towards $B$, and
(f) at $4 \mathrm{~cm}$ away from $\mathrm{B}$ going towards $\mathrm{A}$.
Answer (a) Zero, Positive, Positive (b) Zero, Negative, Negative (c) Negative, Zero, Zero (d) Negative, Negative, Negative (e) Zero, Positive, Positive (f) Negative, Negative, Negative Explanation: (a) The given situation is shown in the following figure. Points $A$ and $B$ are the two end points, with $AB=10 cm$. $O$ is the midpoint of the path. A particle is in linear simple harmonic motion between the end points At the extreme point A, the particle is at rest momentarily. Hence, its velocity is zero at
this point. Its acceleration is positive as it is directed along AO. Force is also positive in this case as the particle is directed rightward. (b) At the extreme point B, the particle is at rest momentarily. Hence, its velocity is zero at this point. Its acceleration is negative as it is directed along B. Force is also negative in this case as the particle is directed leftward. (c) The particle is executing a simple harmonic motion. $O$ is the mean position of the particle. Its velocity at the mean position $O$ is the maximum. The value for velocity is negative as the particle is directed leftward. The acceleration and force of a particle executing SHM is zero at the mean position. (d) The particle is moving toward point $O$ from the end $B$. This direction of motion is opposite to the conventional positive direction, which is from A to B. Hence, the particle’s velocity and acceleration, and the force on it are all negative. (e) The particle is moving toward point $O$ from the end $A$. This direction of motion is from $A$ to B, which is the conventional positive direction. Hence, the values for velocity, acceleration, and force are all positive. (f) This case is similar to the one given in (d).Show Answer
(a) $a=0.7 x$
(b) $\quad a=-200 x^{2}$
(c) $a=-10 x$
(d) $\quad a=100 x^{3}$
Answer A motion represents simple harmonic motion if it is governed by the force law: $F=-k x$ $m a=-k$ $\therefore a=-\frac{k}{m} x$ Where, $F$ is the force $m$ is the mass (a constant for a body) $x$ is the displacement $a$ is the acceleration $k$ is a constant Among the given equations, only equation $a=-10 x$ is written in the above form with $\frac{k}{m}=10$. Hence, this relation represents SHM.Show Answer
$$ x(t)=A \cos (\omega t+\phi) $$
If the initial $(t=0)$ position of the particle is $1 \mathrm{~cm}$ and its initial velocity is $\omega \mathrm{cm} / \mathrm{s}$, what are its amplitude and initial phase angle? The angular frequency of the particle is $\pi \mathrm{s}^{-1}$. If instead of the cosine function, we choose the sine function to describe the SHM : $x=B \sin (\omega t+\alpha)$, what are the amplitude and initial phase of the particle with the above initial conditions.
Answer Initially, at $t=0$ : Displacement, $x=1 cm$ Initial velocity, $v=\omega cm / sec$. Angular frequency, $\omega=\pi rad / s^{-1}$ It is given that: $
\begin{aligned}
& x(t)=A \cos (\omega t+\phi) \\
& 1=A \cos (\omega \times 0+\phi)=A \cos \phi
\end{aligned}
$ $A \cos \phi=1$ Velocity, $v=\frac{d x}{d t}$ $$
\begin{align*}
& \omega=-A \omega \sin (\omega t+\phi) \\
& 1=-A \sin (\omega \times 0+\phi)=-A \sin \phi \\
& A \sin \phi=-1 \tag{ii}
\end{align*}
$$ Squaring and adding equations ( $i$ ) and (ii), we get: $
\begin{aligned}
& A^{2}(\sin ^{2} \phi+\cos ^{2} \phi)=1+1 \\
& A^{2}=2 \\
& \therefore A=\sqrt{2} cm
\end{aligned}
$ Dividing equation (ii) by equation $(i)$, we get: $
\begin{aligned}
& \tan \phi=-1 \\
& \therefore \phi=\frac{3 \pi}{4}, \frac{7 \pi}{4}, \ldots \ldots
\end{aligned}
$ SHM is given as: $
x=B \sin (\omega t+\alpha)
$ Putting the given values in this equation, we get: $$
\begin{equation*}
1=B \sin [\omega \times 0+\alpha] \tag{iii}
\end{equation*}
$$ $B \sin \alpha=1$ Velocity, $v=\omega B \cos (\omega t+\alpha)$ Substituting the given values, we get: $$
\begin{align*}
& \pi=\pi B \sin \alpha \\
& B \sin \alpha=1 \tag{iv}
\end{align*}
$$ Squaring and adding equations (iii) and (iv), we get: $
\begin{aligned}
& B^{2}[\sin ^{2} \alpha+\cos ^{2} \alpha]=1+1 \\
& B^{2}=2 \\
& \therefore B=\sqrt{2} cm
\end{aligned}
$ Dividing equation (iii) by equation (iv), we get:
$\frac{B \sin \alpha}{B \cos \alpha}=\frac{1}{1}$ $\tan \alpha=1=\tan \frac{\pi}{4}$ $\therefore \alpha=\frac{\pi}{4}, \frac{5 \pi}{4}, \ldots \ldots$.Show Answer
Answer Maximum mass that the scale can read, $M=50 kg$ Maximum displacement of the spring $=$ Length of the scale, $l=20 cm=0.2 m$ Time period, $T=0.6 s$ Maximum force exerted on the spring, $F=M g$ Where, $g=$ acceleration due to gravity $=9.8 m / s^{2}$ $F=50 \times 9.8=490$ $\therefore$ Spring constant, $
k=\frac{F}{l}=\frac{490}{0.2}=2450 Nm^{-1}
$ Mass $m$, is suspended from the balance. Time period, $T=2 \pi \sqrt{\frac{m}{k}}$
$\therefore m=(\frac{T}{2 \pi})^{2} \times k=(\frac{0.6}{2 \times 3.14})^{2} \times 2450=22.36 kg$ $\therefore$ Weight of the body $=m g=22.36 \times 9.8=219.167 N$ Hence, the weight of the body is about $219 N$.Show Answer
Determine (i) the frequency of oscillations, (ii) maximum acceleration of the mass, and (iii) the maximum speed of the mass.
Answer Spring constant, $k=1200 N m^{-1}$ Mass, $m=3 kg$ Displacement, $A=2.0 cm=0.02 cm$ Frequency of oscillation $v$, is given by the relation: $v=\frac{1}{T}=\frac{1}{2 \pi} \sqrt{\frac{k}{m}}$ Where, $T$ is the time period $
\therefore v=\frac{1}{2 \times 3.14} \sqrt{\frac{1200}{3}}=3.18 m / s
$ Hence, the frequency of oscillations is $3.18 m / s$. Maximum acceleration $(a)$ is given by the relation: $a=\omega^{2} A$ Where, $\omega=$ Angular frequency $=\sqrt{\frac{k}{m}}$ $A=$ Maximum displacement $\therefore a=\frac{k}{m} A=\frac{1200 \times 0.02}{3}=8 m s^{-2}$ Hence, the maximum acceleration of the mass is $8.0 m / s^{2}$. Maximum velocity, $v _{\max }=A \omega$ $
=A \sqrt{\frac{k}{m}}=0.02 \times \sqrt{\frac{1200}{3}}=0.4 m / s
$ Hence, the maximum velocity of the mass is $0.4 m / s$.Show Answer
(a) at the mean position,
(b) at the maximum stretched position, and
(c) at the maximum compressed position.
In what way do these functions for SHM differ from each other, in frequency, in amplitude or the initial phase?
Answer (a) $x=2 \sin 20 t$ (b) $x=2 \cos 20 t$ (c) $x=-2 \cos 20 t$ The functions have the same frequency and amplitude, but different initial phases. Distance travelled by the mass sideways, $A=2.0 cm$ Force constant of the spring, $k=1200 N m^{-1}$ Mass, $m=3 kg$ Angular frequency of oscillation: $
\begin{aligned}
& \omega=\sqrt{\frac{k}{m}} \\
& =\sqrt{\frac{1200}{3}}=\sqrt{400}=20 rad s^{-1}
\end{aligned}
$ When the mass is at the mean position, initial phase is 0 . Displacement, $x=A \sin \omega t$ $=2 \sin 20 t$ At the maximum stretched position, the mass is toward the extreme right. Hence, the initial phase is $\frac{\pi}{2}$. Displacement, $\quad x=A \sin (\omega t+\frac{\pi}{2})$ $
=2 \sin (20 t+\frac{\pi}{2})
$ $=2 \cos 20 t$ At the maximum compressed position, the mass is toward the extreme left. Hence, the
initial phase is $\frac{3 \pi}{2}$. Displacement, $x=A \sin (\omega t+\frac{3 \pi}{2})$ $=2 \sin (20 t+\frac{3 \pi}{2})=-2 \cos 20 t$ The functions have the same frequency $(\frac{20}{2 \pi} Hz)$ and amplitude $(2 cm)$, but different initial phases $(0, \frac{\pi}{2}, \frac{3 \pi}{2})$.Show Answer
Obtain the corresponding simple harmonic motions of the $x$-projection of the radius vector of the revolving particle $\mathrm{P}$, in each case.
Answer Time period, $T=2 s$ Amplitude, $A=3 cm$ At time, $t=0$, the radius vector OP makes an angle $\frac{\pi}{2}$ with the positive $x$-axis, i.e., phase angle $\phi=+\frac{\pi}{2}$ Therefore, the equation of simple harmonic motion for the $x$-projection of $OP$, at time $t$, is given by the displacement equation: $
\begin{aligned}
& x=A \cos [\frac{2 \pi t}{T}+\phi] \\
& =3 \cos (\frac{2 \pi t}{2}+\frac{\pi}{2})=-3 \sin (\frac{2 \pi t}{2})
\end{aligned}
$ $\therefore x=-3 \sin \pi t cm$ Time period, $T=4 s$ Amplitude, $a=2 m$ At time $t=0$, OP makes an angle $\pi$ with the $x$-axis, in the anticlockwise direction. Hence, phase angle, $\Phi=+\pi$ Therefore, the equation of simple harmonic motion for the $x$-projection of $OP$, at time $t$, is given as: $
\begin{aligned}
& x=a \cos (\frac{2 \pi t}{T}+\phi)=2 \cos (\frac{2 \pi t}{4}+\pi) \\
& \therefore x=-2 \cos (\frac{\pi}{2} t) m
\end{aligned}
$Show Answer
(a) $x=-2 \sin (3 t+\pi / 3)$
(b) $x=\cos (\pi / 6-t)$
(c) $\quad x=3 \sin (2 \pi t+\pi / 4)$
(d) $x=2 \cos \pi t$
Answer $
\begin{aligned}
& x=-2 \sin (3 t+\frac{\pi}{3})=+2 \cos (3 t+\frac{\pi}{3}+\frac{\pi}{2}) \\
& =2 \cos (3 t+\frac{5 \pi}{6})
\end{aligned}
$ If this equation is compared with the standard SHM equation $x=A \cos (\frac{2 \pi}{T} t+\phi)$, then we get: Amplitude, $A=2 cm$ Phase angle, $\phi=\frac{5 \pi}{6}=150^{\circ}$ Angular velocity, $\omega=\frac{2 \pi}{T}=3 rad / sec$. The motion of the particle can be plotted as shown in the following figure. $
x=\cos (\frac{\pi}{6}-t)=\cos (t-\frac{\pi}{6})
$ If this equation is compared with the standard SHM equation $x=A \cos (\frac{2 \pi}{T} t+\phi)$, then we get: Amplitude, $A=1$ Phase angle, $\phi=-\frac{\pi}{6}=-30^{\circ}$ Angular velocity, $\omega=\frac{2 \pi}{T}=1 rad / s$ The motion of the particle can be plotted as shown in the following figure. $
\begin{aligned}
& x=3 \sin (2 \pi t+\frac{\pi}{4}) \\
& =-3 \cos [(2 \pi t+\frac{\pi}{4})+\frac{\pi}{2}]=-3 \cos (2 \pi t+\frac{3 \pi}{4})
\end{aligned}
$ If this equation is compared with the standard SHM equation $x=A \cos (\frac{2 \pi}{T} t+\phi)$, then we get: Amplitude, $A=3 cm$ Phase angle, $\phi=\frac{3 \pi}{4}=135^{\circ}$ Angular velocity, $\omega=\frac{2 \pi}{T}=2 \pi rad / s$ The motion of the particle can be plotted as shown in the following figure. $x=2 \cos \pi t$ If this equation is compared with the standard SHM equation $x=A \cos (\frac{2 \pi}{T} t+\phi)$, then we get: Amplitude, $A=2 cm$ Phase angle, $\Phi=0$ Angular velocity, $\omega=\pi rad / s$ The motion of the particle can be plotted as shown in the following figure.Show Answer
(a) What is the maximum extension of the spring in the two cases ?
(b) If the mass in Fig. (a) and the two masses in Fig. (b) are released, what is the period of oscillation in each case?
Answer For the one block system: When a force $F$, is applied to the free end of the spring, an extension $l$, is produced. For the maximum extension, it can be written as: $F=k l$ Where, $k$ is the spring constant Hence, the maximum extension produced in the spring, $l=\frac{F}{k}$ For the two block system: The displacement $(x)$ produced in this case is: $x=\frac{l}{2}$ Net force, $F=+2 k x=2 k \frac{l}{2}$ $\therefore l=\frac{F}{k}$ For the one block system: For mass $(m)$ of the block, force is written as: $F=m a=m \frac{d^{2} x}{d t^{2}}$ Where, $x$ is the displacement of the block in time $t$ $\therefore m \frac{d^{2} x}{d t^{2}}=-k x$ It is negative because the direction of elastic force is opposite to the direction of displacement. $\frac{d^{2} x}{d t^{2}}=-(\frac{k}{m}) x=-\omega^{2} x$ Where, $\omega^{2}=\frac{k}{m}$ $\omega=\sqrt{\frac{k}{m}}$ Where, $\omega$ is angular frequency of the oscillation $\therefore$ Time period of the oscillation, $T=\frac{2 \pi}{\omega}$ $=\frac{2 \pi}{\sqrt{\frac{k}{m}}}=2 \pi \sqrt{\frac{m}{k}}$ For the two block system: $F=m \frac{d^{2} x}{d t^{2}}$ $m \frac{d^{2} x}{d t^{2}}=-2 k x$ It is negative because the direction of elastic force is opposite to the direction of displacement. $
\frac{d^{2} x}{d t^{2}}=-[\frac{2 k}{m}] x=-\omega^{2} x
$ Where, Angular frequency, $\omega=\sqrt{\frac{2 k}{m}}$ $\therefore$ Time period, $T=\frac{2 \pi}{\omega}=2 \pi \sqrt{\frac{m}{2 k}}$Show Answer
Answer Angular frequency of the piston, $\omega=200 rad / min$. Stroke $=1.0 m$ Amplitude, $A=\frac{1.0}{2}=0.5 m$ The maximum speed $(v _{\max })$ of the piston is give by the relation: $
\begin{aligned}
v_{\max } & =A \omega \\
& =200 \times 0.5=100 m / min
\end{aligned}
$Show Answer
Answer Acceleration due to gravity on the surface of moon, $g^{\prime}=1.7 m s^{-2}$ Acceleration due to gravity on the surface of earth, $g=9.8 m s^{-2}$ Time period of a simple pendulum on earth, $T=3.5 s$ $
T=2 \pi \sqrt{\frac{l}{g}}
$ Where, $l$ is the length of the pendulum $
\begin{aligned}
\therefore l & =\frac{T^{2}}{(2 \pi)^{2}} \times g \\
& =\frac{(3.5)^{2}}{4 \times(3.14)^{2}} \times 9.8 m
\end{aligned}
$ The length of the pendulum remains constant. On moon’s surface, time period, $T^{\prime}=2 \pi \sqrt{\frac{l}{g^{\prime}}}$ $
=2 \pi \sqrt{\frac{\frac{(3.5)^{2}}{\frac{4 \times(3.14)^{2}}{1.7}} \times 9.8}{1.7}}=8.4 s
$ Hence, the time period of the simple pendulum on the surface of moon is $8.4 s$.Show Answer
Answer The bob of the simple pendulum will experience the acceleration due to gravity and the centripetal acceleration provided by the circular motion of the car. Acceleration due to gravity $=g$ Centripetal acceleration $=\frac{v^{2}}{R}$ Where, $v$ is the uniform speed of the car $R$ is the radius of the track Effective acceleration $(a _{\text{eff }})$ is given as: $a _{\text{eff }}=\sqrt{g^{2}+(\frac{v^{2}}{R})^{2}}$ Time period, $T=2 \pi \sqrt{\frac{l}{a _{\text{eff }}}}$ Where, $l$ is the length of the pendulum $\therefore$ Time period, $T=2 \pi \sqrt{\frac{l}{g^{2}+\frac{v^{4}}{R^{2}}}}$Show Answer
$$ T=2 \pi \sqrt{\frac{h \rho}{\rho_{1} g}} $$
where $\rho$ is the density of cork. (Ignore damping due to viscosity of the liquid).
Answer Base area of the cork $=A$ Height of the cork $=h$ Density of the liquid $=\rho_l$ Density of the cork $=\rho$ In equilibrium: Weight of the cork = Weight of the liquid displaced by the floating cork Let the cork be depressed slightly by $x$. As a result, some extra water of a certain volume is displaced. Hence, an extra up-thrust acts upward and provides the restoring force to the cork. Up-thrust $=$ Restoring force, $F=$ Weight of the extra water displaced $F=-($ Volume $\times$ Density $\times g)$ Volume $=$ Area $\times$ Distance through which the cork is depressed Volume $=A x$ $\therefore F=-A x^{\rho_l} g \ldots(i)$ According to the force law: $F=k x$ $k=\frac{F}{x}$ Where, $k$ is a constant $$
\begin{equation*}
k=\frac{F}{x}=-A \rho_1 g \tag{ii}
\end{equation*}
$$ The time period of the oscillations of the cork: $$
\begin{equation*}
T=2 \pi \sqrt{\frac{m}{k}} \tag{iii}
\end{equation*}
$$ Where,
$m=$ Mass of the cork $=$ Volume of the cork $\times$ Density $=$ Base area of the cork $\times$ Height of the cork $\times$ Density of the cork $=A h \rho$ Hence, the expression for the time period becomes: $T=2 \pi \sqrt{\frac{A h \rho}{A \rho_l g}}=2 \pi \sqrt{\frac{h \rho}{\rho_l g}}$Show Answer
Show Answer
Answer
Area of cross-section of the U-tube $=A$
Density of the mercury column $=\rho$
Acceleration due to gravity $=g$
Restoring force, $F=$ Weight of the mercury column of a certain height
$F=-($ Volume $\times$ Density $\times g)$
$F=-(A \times 2 h \times \rho \times g)=-2 A \rho g h=-k \times$ Displacement in one of the arms $(h)$
Where,
$2 h$ is the height of the mercury column in the two arms
$k$ is a constant, given by $k=-\frac{F}{h}=2 A \rho g$
Time period, $T=2 \pi \sqrt{\frac{m}{k}}=2 \pi \sqrt{\frac{m}{2 A \rho g}}$
Where,
$m$ is the mass of the mercury column
Let $l$ be the length of the total mercury in the U-tube.
Mass of mercury, $m=$ Volume of mercury $\times$ Density of mercury
$=A l \rho$
$\therefore \quad T=2 \pi \sqrt{\frac{A l \rho}{2 A \rho g}}=2 \pi \sqrt{\frac{l}{2 g}}$
Hence, the mercury column executes simple harmonic motion with time period $2 \pi \sqrt{\frac{l}{2 g}}$.
