OSCILLATIONS-Execise
Physics - Exercise
Q1. 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 N-S direction and released.
(c) A hydrogen molecule rotating about its centre of mass.
(d) An arrow released from a bow.
Solution
A motion is said to be periodic if it repeats itself at regular intervals of time, returning to the same state of position and motion after each interval. Each given example is examined on this basis.
In the case of a swimmer completing one return trip from one bank of a river to the other and back, the motion does not repeat itself in a regular and identical manner. Factors such as varying speed, fatigue, water currents, and non-uniform timing prevent the motion from being exactly repetitive. Hence, this motion cannot be regarded as periodic.
A freely suspended bar magnet, when displaced from its north–south direction and released, oscillates about its equilibrium position under the action of restoring torque due to Earth’s magnetic field. The motion repeats itself at equal intervals of time as long as damping is negligible. Therefore, this motion is periodic in nature.
A hydrogen molecule rotating about its centre of mass executes rotational motion in which its orientation repeats after every complete rotation. Since the same physical state recurs after equal intervals of time, this motion is periodic.
An arrow released from a bow moves forward under the action of the initial force and gravity, following a projectile path. It does not return to its initial position or repeat its motion at regular time intervals. Hence, this motion is non-periodic.
Thus, among the given examples, the motion of the freely suspended bar magnet and the rotation of a hydrogen molecule represent periodic motion, while the swimmer’s motion and the motion of the arrow do not.
Q2. Which of the following examples represent (nearly) simple harmonic motion and
which represent periodic but not simple harmonic motion?
(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.
Solution
Simple harmonic motion is a special type of periodic motion in which the restoring force or acceleration is directly proportional to the displacement from the equilibrium position and is always directed towards it. Each of the given motions can be analysed by examining whether this condition is satisfied.
The rotation of the Earth about its axis is a periodic motion because the Earth returns to the same orientation after equal intervals of time. However, there is no restoring force acting to bring the Earth back when it is angularly displaced, as the motion continues due to angular momentum rather than a restoring torque proportional to displacement. Hence, this motion is periodic but not simple harmonic.
In a U-tube filled with mercury, when the liquid column is slightly displaced from its equilibrium position, a difference in liquid levels is produced in the two arms. This difference creates a restoring force due to gravity that is proportional to the displacement for small oscillations. As a result, the mercury column executes nearly simple harmonic motion about its mean position.
When a ball bearing is placed in a smooth curved bowl and released from a point slightly above the lowest point, it rolls back and forth under the action of gravity. For small displacements from the lowest point, the component of gravitational force along the surface is proportional to the displacement, giving rise to nearly simple harmonic motion. For larger displacements, this proportionality does not hold, but for small oscillations the motion is approximately SHM.
The general vibrations of a polyatomic molecule about its equilibrium configuration involve several modes of vibration with complex force–displacement relations. Although the motion is periodic, the restoring forces are not, in general, directly proportional to the displacement in a single coordinate. Therefore, such vibrations are periodic but not simple harmonic.
Thus, the oscillations of the mercury column in a U-tube and the motion of the ball bearing in a curved bowl for small displacements represent nearly simple harmonic motion, while the rotation of the Earth and the general vibrations of a polyatomic molecule are periodic but not simple harmonic.
Q3. Fig. 13.18 depicts four x-t plots for linear motion of a particle. Which of the plots represent periodic motion? What is the period of motion (in case of periodic motion) ?
Solution
A motion is periodic if the displacement–time graph repeats itself after equal intervals of time, such that the particle returns to the same position with the same direction of motion. Each of the given \(x\)–\(t\) plots is examined with this criterion.
In plot (a), the displacement increases continuously with time and the graph does not repeat its shape at any later instant. Since there is no definite interval after which the motion repeats, this motion is non-periodic.
In plot (b), the displacement pattern repeats itself regularly. The graph between \(t=-3\ \text{s}\) and \(t=-1\ \text{s}\) is identical to that between \(t=-1\ \text{s}\) and \(t=1\ \text{s}\), and again between \(t=1\ \text{s}\) and \(t=3\ \text{s}\). Hence, the motion is periodic. The time taken for one complete repetition is \[ \begin{aligned} T &= (-1)-(-3) \\ &= 2\ \text{s} \end{aligned} \]
In plot (c), the displacement varies in an irregular manner and the shapes of successive portions of the graph are different. Since no identical pattern repeats after a fixed interval of time, this motion is non-periodic.
In plot (d), the displacement varies smoothly in a sinusoidal manner. The portion of the graph from \(t=-3\ \text{s}\) to \(t=-1\ \text{s}\) is repeated from \(t=-1\ \text{s}\) to \(t=1\ \text{s}\), and again from \(t=1\ \text{s}\) to \(t=3\ \text{s}\). Thus, the motion is periodic. The period is \[ \begin{aligned} T &= (-1)-(-3) \\ &= 2\ \text{s} \end{aligned} \]
Therefore, plots (b) and (d) represent periodic motion, each with a period of \(2\ \text{s}\), while plots (a) and (c) represent non-periodic motion.
Q4. Which of the following functions of time represent (a) simple harmonic, (b) periodic
but not simple harmonic, and (c) non-periodic motion? Give period for each case of
periodic motion (ω is any positive constant):
(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) \(exp (–\omega^2t^2)\)
(f) \(1 + \omega t + \omega^2t^2\)
Solution
A motion is said to be simple harmonic if it can be represented by a single sine or cosine function of time with constant angular frequency. A motion is periodic if it repeats itself after a fixed interval of time, even if it is not a pure sine or cosine. Each given function is analysed accordingly.
For the function \(x(t)=\sin \omega t-\cos \omega t\), the two terms have the same angular frequency \(\omega\). They can be combined into a single sinusoidal form using trigonometric identities: \[ \begin{aligned} x(t) &= \sin \omega t-\cos \omega t \\ &= \sqrt{2}\,\sin\!\left(\omega t-\frac{\pi}{4}\right) \end{aligned} \] Since the motion is described by a single sine function, it represents simple harmonic motion with period \[ T=\frac{2\pi}{\omega} \]
For the function \(x(t)=\sin^{3}\omega t\), using the identity \[ \sin^{3}\theta=\frac{3\sin\theta-\sin3\theta}{4} \] the displacement can be written as \[ x(t)=\frac{3\sin\omega t-\sin3\omega t}{4} \] This expression contains more than one angular frequency, although the motion repeats itself after a definite time. Hence, the motion is periodic but not simple harmonic. The fundamental period is \[ T=\frac{2\pi}{\omega} \]
For the function \(x(t)=3\cos\!\left(\frac{\pi}{4}-2\omega t\right)\), this is a cosine function with angular frequency \(2\omega\) and a constant phase. It represents simple harmonic motion. The period of oscillation is \[ T=\frac{2\pi}{2\omega}=\frac{\pi}{\omega} \]
For the function \(x(t)=\cos\omega t+\cos3\omega t+\cos5\omega t\), the displacement is a superposition of cosine terms with different angular frequencies. Although the resulting motion is not simple harmonic, all frequencies are integral multiples of \(\omega\), so the motion repeats after a fixed time. Therefore, the motion is periodic but not simple harmonic, with period \[ T=\frac{2\pi}{\omega} \]
For the function \(x(t)=\exp(-\omega^{2}t^{2})\), the displacement continuously decreases with time and does not repeat itself. There is no fixed interval after which the motion recurs. Hence, this motion is non-periodic.
For the function \(x(t)=1+\omega t+\omega^{2}t^{2}\), the displacement increases monotonically with time and does not repeat its values at regular intervals. Therefore, this motion is also non-periodic.
Thus, functions (a) and (c) represent simple harmonic motion, functions (b) and (d) represent periodic but not simple harmonic motion, and functions (e) and (f) represent non-periodic motion.
Q5. A particle is in linear simple harmonic motion between two points, A and B, 10 cm
apart. Take the direction from A to B as the positive direction and give the signs of
velocity, acceleration and force on the particle when it is
(a) at the end A,
(b) at the end B,
(c) at the mid-point of AB going towards A,
(d) at 2 cm away from B going towards A,
(e) at 3 cm away from A going towards B, and
(f) at 4 cm away from B going towards A.
Solution
The particle executes linear simple harmonic motion between two extreme points A and B which are 10 cm apart. The mean position lies at the midpoint of AB, that is 5 cm from each end. The direction from A to B is taken as positive. In SHM, the acceleration and hence the restoring force are always directed towards the mean position and are proportional to the displacement from it.
At the extreme position A, the particle is momentarily at rest before moving towards B. Therefore, the velocity is zero. Since A lies to the left of the mean position, the displacement is negative, so the restoring acceleration and force act towards the mean position, that is towards B, which is the positive direction. Hence, the velocity is zero, while acceleration and force are positive.
At the extreme position B, the particle is again momentarily at rest. Here the displacement from the mean position is positive. The restoring acceleration and force act towards the mean position, that is towards A, which is the negative direction. Thus, the velocity is zero, while acceleration and force are negative.
At the midpoint of AB, the displacement is zero. When the particle is passing through this point going towards A, its velocity is directed towards A and is therefore negative. Since the displacement is zero at the mean position, both acceleration and restoring force are zero at that instant.
At a point 2 cm away from B going towards A, the particle is still on the right side of the mean position because its distance from the mean position is \(5-2=3\) cm towards B. Hence the displacement is positive. The velocity is towards A and therefore negative. The restoring acceleration and force act towards the mean position, which is towards A, so both acceleration and force are negative.
At a point 3 cm away from A going towards B, the particle lies on the left side of the mean position since its distance from the mean position is \(5-3=2\) cm towards A. The displacement is therefore negative. The velocity is towards B and hence positive. The restoring acceleration and force act towards the mean position, that is towards B, so both acceleration and force are positive.
At a point 4 cm away from B going towards A, the particle is 1 cm to the right of the mean position because its distance from the mean position is \(5-4=1\) cm towards B. Thus the displacement is positive. The velocity is towards A and is therefore negative. The restoring acceleration and force act towards the mean position, that is towards A, so both acceleration and force are negative.
Thus, the signs of velocity depend on the direction of motion, while the signs of acceleration and force are always opposite to the sign of displacement, as required for simple harmonic motion.
Q6.Which of the following relationships between the acceleration a and the displacement
x of a particle involve simple harmonic motion?
(a) \(a = 0.7x\)
(b) \(a = –200x^2\)
(c) \(a = –10x\)
(d) \(a = 100x^3\)
Solution
For a particle to execute simple harmonic motion, its acceleration must be directly proportional to its displacement from the equilibrium position and must always be directed towards that position. Mathematically, this condition is expressed as \[ a=-\omega^{2}x, \] where \(\omega\) is a positive constant. Each given relation between acceleration and displacement is examined using this criterion.
In the relation \(a=0.7x\), the acceleration is proportional to the displacement, but it is in the same direction as the displacement. This implies that the acceleration pushes the particle away from the equilibrium position rather than towards it. Hence, this relation does not represent simple harmonic motion.
For the relation \(a=-200x^{2}\), the acceleration depends on the square of the displacement. Since it is not linearly proportional to \(x\), the restoring force condition required for simple harmonic motion is not satisfied. Therefore, this motion is not simple harmonic.
In the relation \(a=-10x\), the acceleration is directly proportional to the displacement and is opposite in direction to it. This exactly matches the standard form of acceleration in simple harmonic motion, with \(\omega^{2}=10\). Hence, this relation represents simple harmonic motion.
For the relation \(a=100x^{3}\), the acceleration varies as the cube of the displacement and is not proportional to \(x\). As a result, the motion does not satisfy the fundamental condition of simple harmonic motion and is therefore not SHM.
Thus, among the given relations, only the relation \(a=-10x\) corresponds to simple harmonic motion, while the others do not.
Q7. The motion of a particle executing simple harmonic motion is described by the displacement function, \[x(t) = A \cos (\omega t + \phi )\] If the initial (t = 0) position of the particle is 1 cm and its initial velocity is \(\omega\ cm/s\), what are its amplitude and initial phase angle ? The angular frequency of the particle is \(\pi s^{–1}\). If instead of the cosine function, we choose the sine function to describe the SHM : \(x = B \sin (\omega t + α)\), what are the amplitude and initial phase of the particle with the above initial conditions.
Solution
The displacement of a particle executing simple harmonic motion is given by \[ x(t)=A\cos(\omega t+\phi) \] where the angular frequency is \(\omega=\pi\ \text{s}^{-1}\)
Using the initial position condition at \(t=0\), \[ x(0)=A\cos\phi=1\ \text{cm}\tag{1} \]
The velocity of the particle is obtained by differentiating the displacement: \[ v(t)=\frac{dx}{dt}=-A\omega\sin(\omega t+\phi) \] At \(t=0\), \[ v(0)=-A\omega\sin\phi \] Given that the initial velocity is \(\omega\ \text{cm s}^{-1}\), \[ -A\omega\sin\phi=\omega \] or \[ A\sin\phi=-1\tag{2} \]
Squaring and adding equations (1) and (2), \[ (A\cos\phi)^2+(A\sin\phi)^2=1^2+(-1)^2 \] \[ A^2=2 \] \[ A=\sqrt{2}\ \text{cm} \]
From equations (1) and (2), \[ \cos\phi=\frac{1}{\sqrt{2}}, \qquad \sin\phi=-\frac{1}{\sqrt{2}} \] which gives the initial phase angle \[ \phi=-\frac{\pi}{4} \]
Hence, for the cosine form of SHM, the amplitude is \(\sqrt{2}\ \text{cm}\) and the initial phase angle is \(-\dfrac{\pi}{4}\).
If the same motion is described using the sine function, \[ x(t)=B\sin(\omega t+\alpha) \]
Applying the initial position condition, \[ B\sin\alpha=1\tag{3} \]
The velocity is now \[ v(t)=B\omega\cos(\omega t+\alpha) \] At \(t=0\), \[ v(0)=B\omega\cos\alpha=\omega \] which gives \[ B\cos\alpha=1\tag{4} \]
Squaring and adding equations (3) and (4), \[ (B\sin\alpha)^2+(B\cos\alpha)^2=1^2+1^2 \] \[ B^2=2 \] \[ B=\sqrt{2}\ \text{cm} \]
From equations (3) and (4), \[ \sin\alpha=\frac{1}{\sqrt{2}}, \qquad \cos\alpha=\frac{1}{\sqrt{2}} \] which gives \[ \alpha=\frac{\pi}{4} \]
Thus, when the SHM is written in sine form, the amplitude is \(\sqrt{2}\ \text{cm}\) and the initial phase angle is \(\dfrac{\pi}{4}\).
Q8. A spring balance has a scale that reads from 0 to 50 kg. The length of the scale is 20 cm. A body suspended from this balance, when displaced and released, oscillates with a period of 0.6 s. What is the weight of the body ?
Solution
The spring balance has a scale reading from 0 to 50 kg over a length of 20 cm. Hence, the extension produced per kilogram load is \[ \frac{20\ \text{cm}}{50\ \text{kg}}=0.4\ \text{cm kg}^{-1}=4\times10^{-3}\ \text{m kg}^{-1} \]
For a spring balance, the extension produced by a load \(m\) is given by \[ x=\frac{mg}{k} \] where \(k\) is the spring constant. Thus, the extension per unit mass is \[ \frac{x}{m}=\frac{g}{k} \] Using the scale calibration, \[ \frac{g}{k}=4\times10^{-3} \] Hence, \[ \begin{aligned} k=\frac{g}{4\times10^{-3}}&=\frac{9.8}{4\times10^{-3}}\\\\&=2.45\times10^{3}\ \text{N m}^{-1} \end{aligned} \]
When a body of mass \(m\) is suspended from the spring and oscillates, the time period of simple harmonic motion is \[ T=2\pi\sqrt{\frac{m}{k}} \] Given \(T=0.6\ \text{s}\), \[ 0.6=2\pi\sqrt{\frac{m}{2.45\times10^{3}}} \]
Rearranging, \[ \sqrt{\frac{m}{2.45\times10^{3}}}=\frac{0.6}{2\pi} \] \[ \frac{m}{2.45\times10^{3}}=\left(\frac{0.6}{2\pi}\right)^2 \] \[ \begin{aligned} m&=2.45\times10^{3}\left(\frac{0.6}{2\pi}\right)^2\\\\&\approx 22.3\ \text{kg} \end{aligned} \]
The weight of the body is therefore \[ W=mg=22.3\times9.8\approx2.18\times10^{2}\ \text{N} \]
Hence, the body suspended from the spring balance has a weight of approximately \(2.2\times10^{2}\ \text{N}\), corresponding to a mass of about \(22\ \text{kg}\).
Q9. A spring having with a spring constant 1200 N m–1 is mounted on a horizontal table
as shown in Fig. 13.19. A mass of 3 kg is attached to the free end of the spring. The
mass is then pulled sideways to a distance of 2.0 cm and released.
Determine
(i) the frequency of oscillations,
(ii) maximum acceleration of the mass,
and
(iii) the maximum speed of the mass.
Solution
A mass–spring system executes simple harmonic motion on a horizontal table. The given data are: \[ k=1200\ \text{N m}^{-1}, \quad m=3\ \text{kg}, \quad A=2.0\ \text{cm}=0.020\ \text{m} \]
For a mass attached to a spring, the angular frequency of SHM is \[ \omega=\sqrt{\frac{k}{m}} \] Substituting the given values, \[ \begin{aligned} \omega&=\sqrt{\frac{1200}{3}}\\\\&=\sqrt{400}\\\\&=20\ \text{rad s}^{-1} \end{aligned} \]
The frequency of oscillation is related to angular frequency by \[ f=\frac{\omega}{2\pi} \] Hence, \[ \begin{aligned} f&=\frac{20}{2\pi}\\\\&=\frac{10}{\pi}\\\\&\approx3.18\ \text{Hz} \end{aligned} \]
The maximum acceleration in SHM occurs at the extreme position and is given by \[ a_{\max}=\omega^{2}A \] Using the known values, \[ \begin{aligned} a_{\max}&=(20)^2\times0.020\\&=400\times0.020\\&=8.0\ \text{m s}^{-2} \end{aligned} \]
The maximum speed occurs at the mean position and is given by \[ v_{\max}=\omega A \] Thus, \[ \begin{aligned} v_{\max}&=20\times0.020\\&=0.40\ \text{m s}^{-1} \end{aligned} \]
Therefore, the frequency of oscillation of the mass is \(3.18\ \text{Hz}\), the maximum acceleration is \(8.0\ \text{m s}^{-2}\), and the maximum speed of the mass is \(0.40\ \text{m s}^{-1}\).
Q10. In Exercise 13.9, let us take the position of mass when the spring is unstreched as
x = 0, and the direction from left to right as the positive direction of
x-axis. Give x as a function of time t for the oscillating mass if at the moment we
start the stopwatch (t = 0), the mass is
(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?
Solution
The mass attached to the spring executes simple harmonic motion. The position of the mass when the spring is unstretched is taken as the origin, \(x=0\), and motion from left to right is taken as the positive direction of the \(x\)-axis. For SHM, the displacement can be written in the general form \[ x(t)=A\cos(\omega t+\phi), \] where \(A\) is the amplitude, \(\omega\) is the angular frequency, and \(\phi\) is the initial phase determined by the initial condition at \(t=0\).
If at \(t=0\) the mass is at the mean position, then its displacement from the origin is zero. This condition is satisfied by choosing the sine form of SHM. Hence, the displacement as a function of time is \[ x(t)=A\sin(\omega t). \] Here, the initial phase is zero and the particle passes through the mean position at the start.
If at \(t=0\) the mass is at the maximum stretched position, the displacement is equal to the positive amplitude \(A\). This condition is satisfied by the cosine function with zero phase. Therefore, the displacement is given by \[ x(t)=A\cos(\omega t). \] In this case, the initial phase is zero and the mass starts from the extreme right position.
If at \(t=0\) the mass is at the maximum compressed position, the displacement is equal to \(-A\). This situation corresponds to a phase difference of \(\pi\) from the maximum stretched position. Hence, the displacement can be written as \[ \begin{aligned} x(t) &= A\cos(\omega t+\pi) \\ &= -A\cos(\omega t). \end{aligned} \] Here, the mass starts from the extreme left position.
In all three cases, the angular frequency \(\omega\) and the amplitude \(A\) remain the same because they depend only on the physical parameters of the spring–mass system. The only difference between these expressions lies in the initial phase, which changes according to the initial position of the mass at \(t=0\).
Thus, the different initial conditions of the same SHM lead to different mathematical forms of \(x(t)\) that differ only in their initial phase, while having the same frequency and amplitude.
Q11. Figures 13.20 correspond to two circular motions. The radius of the circle, the
period of revolution, the initial position, and the sense of revolution (i.e. clockwise
or anti-clockwise) are indicated on each figure.
Obtain the corresponding simple harmonic motions of the x-projection of the radius
vector of the revolving particle P, in each case.
Solution
In uniform circular motion, the projection of the radius vector of a particle on any fixed diameter executes simple harmonic motion. The displacement along the \(x\)-axis is given by \[ x(t)=A\cos(\omega t+\phi), \] where \(A\) is the radius of the circle, \(\omega=2\pi/T\) is the angular speed, and \(\phi\) is the initial phase determined by the initial position and the sense of revolution.
For figure (a), the radius of the circular path is \(A=3\ \text{cm}\) and the period of revolution is \(T=2\ \text{s}\). Hence, \[ \omega=\frac{2\pi}{T}=\frac{2\pi}{2}=\pi\ \text{rad s}^{-1}. \] At \(t=0\), the particle is at the lowest point of the circle, so its projection on the \(x\)-axis is zero. From the indicated direction of motion, the particle moves such that the \(x\)-projection initially increases in the positive direction. Therefore, the appropriate SHM is described by a sine function: \[ x(t)=3\sin(\pi t)\ \text{cm}. \]
For figure (b), the radius of the circular path is \(A=2\ \text{m}\) and the period of revolution is \(T=4\ \text{s}\). Thus, \[ \omega=\frac{2\pi}{T}=\frac{2\pi}{4}=\frac{\pi}{2}\ \text{rad s}^{-1}. \] At \(t=0\), the particle is at the leftmost point of the circle, so its \(x\)-projection is at the negative extreme, \(x=-A\). The sense of revolution shown indicates that the particle initially moves upward, so the \(x\)-projection starts from the negative maximum value. This corresponds to a cosine function with a phase of \(\pi\): \[ \begin{aligned} x(t) &= A\cos\!\left(\omega t+\pi\right) \\ &= -A\cos(\omega t). \end{aligned} \] Substituting the given values, \[ x(t)=-2\cos\!\left(\frac{\pi}{2}t\right)\ \text{m}. \]
Thus, the simple harmonic motion corresponding to figure (a) is \(x(t)=3\sin(\pi t)\ \text{cm}\), while that corresponding to figure (b) is \(x(t)=-2\cos\!\left(\dfrac{\pi}{2}t\right)\ \text{m}\). In both cases, the SHM arises as the projection of uniform circular motion, with the amplitude equal to the radius of the circle and the angular frequency determined by the period of revolution.
Q12. Plot the corresponding reference circle for each of the following simple harmonic
motions. Indicate the initial (t =0) position of the particle, the radius of the circle,
and the angular speed of the rotating particle. For simplicity, the sense of rotation
may be fixed to be anticlockwise in every case: (x is in cm and t is in s).
(a) \(x = –2 sin (3t + \pi/3)\)
(b) \(x = cos (\pi/6 – t)\)
(c) \(x = 3 sin (2\pi t + \pi/4)\)
(d) \(x = 2 cos \pi t\)
Solution
In simple harmonic motion, the displacement can always be interpreted as the projection of a uniform circular motion on a diameter of a reference circle. If the motion is written in the form \[ x(t)=A\sin(\omega t+\phi)\quad \text{or}\quad x(t)=A\cos(\omega t+\phi), \] then the radius of the reference circle is \(A\), the angular speed of the rotating particle is \(\omega\), and the initial position on the circle is fixed by the phase angle \(\phi\). The sense of rotation is taken to be anticlockwise in all cases.
For the motion \(x=-2\sin(3t+\pi/3)\), the negative sign may be absorbed into the phase: \[ \begin{aligned} x &= -2\sin(3t+\pi/3) \\ &= 2\sin\!\left(3t+\pi/3+\pi\right). \end{aligned} \] Thus, the radius of the reference circle is \(2\ \text{cm}\) and the angular speed is \(\omega=3\ \text{rad s}^{-1}\). At \(t=0\), the particle on the circle makes an angle \(\pi/3+\pi=4\pi/3\) with the positive \(x\)-axis.
For the motion \(x=\cos(\pi/6-t)\), this may be rewritten as \[ x=\cos\!\left(t-\frac{\pi}{6}\right), \] which represents SHM of amplitude \(1\ \text{cm}\) and angular speed \(\omega=1\ \text{rad s}^{-1}\). At \(t=0\), the rotating particle on the reference circle is at an angle \(-\pi/6\) measured from the positive \(x\)-axis, moving anticlockwise.
For the motion \(x=3\sin(2\pi t+\pi/4)\), the amplitude is \(3\ \text{cm}\) and the angular speed is \(\omega=2\pi\ \text{rad s}^{-1}\). At \(t=0\), the phase angle is \(\pi/4\), so the particle on the reference circle is initially located at an angle \(\pi/4\) from the positive \(x\)-axis.
For the motion \(x=2\cos(\pi t)\), the amplitude is \(2\ \text{cm}\) and the angular speed is \(\omega=\pi\ \text{rad s}^{-1}\). Since there is no additional phase constant, the particle starts at \(t=0\) from the extreme right position on the reference circle, that is, at zero phase angle.
In each case, the corresponding reference circle has a radius equal to the amplitude of SHM, the particle rotates anticlockwise with angular speed equal to the coefficient of \(t\), and the initial angular position on the circle is fixed by the phase constant of the given displacement equation.
Q13. Figure 13.21(a) shows a spring of force constant k clamped rigidly at one end and a
mass m attached to its free end. A force \(\vec{F}\) applied at the free end stretches the
spring. Figure 13.21 (b) shows the same spring with both ends free and attached to
a mass m at either end. Each end of the spring in Fig. 13.21(b) is stretched by the
same force \(vec{F}\).
(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 ?
Solution
In both arrangements, the spring has force constant \(k\). The extension produced in a spring depends only on the force applied and the spring constant, according to Hooke’s law, \[ F = kx. \] This relation is used to determine the maximum extension in each case.
In Fig. 13.21(a), one end of the spring is rigidly clamped and a force \(F\) is applied to the free end. If the extension of the spring is \(x_1\), then \[ \begin{aligned} F &= kx_1 \\ x_1 &= \frac{F}{k}. \end{aligned} \] Thus, the maximum extension of the spring in case (a) is \(F/k\).
In Fig. 13.21(b), the same spring has both ends free and each end is pulled by a force \(F\) in opposite directions. The tension throughout the spring is still equal to \(F\). If the total extension of the spring is \(x_2\), then \[ \begin{aligned} F &= kx_2 \\ x_2 &= \frac{F}{k}. \end{aligned} \] Hence, the maximum extension of the spring in case (b) is also \(F/k\). Therefore, the extension of the spring is the same in both cases.
Next, the periods of oscillation are determined after releasing the masses. In Fig. 13.21(a), a single mass \(m\) is attached to a spring of force constant \(k\). The time period of simple harmonic motion is \[ \begin{aligned} T_1 &= 2\pi\sqrt{\frac{m}{k}}. \end{aligned} \]
In Fig. 13.21(b), two equal masses \(m\) are attached at the two ends of the spring. When the system oscillates, the motion is symmetric about the midpoint of the spring. Each mass effectively behaves as if it is attached to a spring of force constant \(2k\), because a displacement of one mass by a distance \(x\) produces a restoring force twice as large as that produced by the same displacement in case (a). Hence, for each mass, \[ \begin{aligned} T_2 &= 2\pi\sqrt{\frac{m}{2k}}. \end{aligned} \]
Thus, the maximum extension of the spring is the same, \(F/k\), in both arrangements. The period of oscillation is \(2\pi\sqrt{m/k}\) for the single-mass system in Fig. 13.21(a), while for the two-mass system in Fig. 13.21(b) it is \(2\pi\sqrt{m/(2k)}\).
Q14. The piston in the cylinder head of a locomotive has a stroke (twice the amplitude) of 1.0 m. If the piston moves with simple harmonic motion with an angular frequency of 200 rad/min, what is its maximum speed ?
Solution
The piston executes simple harmonic motion. The stroke of the piston is given as 1.0 m, which is equal to twice the amplitude. Hence, the amplitude of oscillation is \[ \begin{aligned} A &= \frac{1.0}{2} \\\\&= 0.50\ \text{m} \end{aligned} \]
The angular frequency of the motion is given as \(200\ \text{rad min}^{-1}\). To use standard SHM relations, this must be converted into radians per second: \[ \begin{aligned} \omega &= \frac{200}{60} \\\\&= \frac{10}{3}\ \text{rad s}^{-1} \end{aligned} \]
In simple harmonic motion, the maximum speed occurs at the mean position and is given by \[ v_{\max} = \omega A \] Substituting the known values, \[ \begin{aligned} v_{\max} &= \frac{10}{3} \times 0.50 \\ &= \frac{5}{3}\ \text{m s}^{-1}. \end{aligned} \]
Therefore, the maximum speed of the piston is \(\dfrac{5}{3}\ \text{m s}^{-1}\), which is approximately \(1.67\ \text{m s}^{-1}\).
Q15. The acceleration due to gravity on the surface of moon is \(1.7\ m\ s^{–2}\). What is the time period of a simple pendulum on the surface of moon if its time period on the surface of earth is \(3.5\ s\) ? (g on the surface of earth is \(9.8\ m\ s^{–2}\))
Solution
For a simple pendulum executing small oscillations, the time period is given by \[ T = 2\pi \sqrt{\frac{l}{g}} \] where \(l\) is the length of the pendulum and \(g\) is the acceleration due to gravity at the place of observation.
For the same pendulum, the length \(l\) remains unchanged whether it is on the Earth or on the Moon. Hence, the time period depends only on the value of \(g\). If \(T_E\) is the time period on Earth and \(T_M\) is the time period on the Moon, then \[ \begin{aligned} \frac{T_M}{T_E} &= \sqrt{\frac{g_E}{g_M}} \end{aligned} \]
Substituting the given values, \[ \begin{aligned} T_M &= 3.5 \sqrt{\frac{9.8}{1.7}} \end{aligned} \]
Evaluating the expression, \[ \begin{aligned} \frac{9.8}{1.7} &\approx 5.76, \\ \sqrt{5.76} &= 2.4 \end{aligned} \] Thus, \[ \begin{aligned} T_M &= 3.5 \times 2.4 \\&= 8.4\ \text{s} \end{aligned} \]
Therefore, the time period of the simple pendulum on the surface of the Moon is \(8.4\ \text{s}\).
Q16. A simple pendulum of length l and having a bob of mass M is suspended in a car. The car is moving on a circular track of radius R with a uniform speed v. If the pendulum makes small oscillations in a radial direction about its equilibrium position, what will be its time period ?
Solution
When the car moves uniformly on a circular track of radius \(R\) with speed \(v\), it has a centripetal acceleration directed towards the centre of the circular path. The magnitude of this acceleration is \[ a_c=\frac{v^{2}}{R} \] In the non-inertial frame of the car, this acceleration gives rise to a pseudo force acting on the pendulum bob in the radially outward direction.
Thus, the bob experiences two constant accelerations: the gravitational acceleration \(g\) acting vertically downward and the pseudo-acceleration \(v^{2}/R\) acting horizontally outward. The effective acceleration \(\vec{g}_{\text{eff}}\) is the vector sum of these two accelerations. Its magnitude is \[ \begin{aligned} g_{\text{eff}} &= \sqrt{g^{2}+\left(\frac{v^{2}}{R}\right)^{2}} \end{aligned} \] The pendulum oscillates about a new equilibrium position along the direction of this effective acceleration.
For small oscillations about the equilibrium position, the time period of a simple pendulum depends only on the effective acceleration and is given by \[ T=2\pi\sqrt{\frac{l}{g_{\text{eff}}}} \] Substituting the expression for \(g_{\text{eff}}\), \[ \begin{aligned} T &= 2\pi\sqrt{\frac{l}{\sqrt{g^{2}+\left(\dfrac{v^{2}}{R}\right)^{2}}}} \end{aligned} \]
Hence, the time period of the pendulum executing small radial oscillations inside the car moving on a circular track is \[ \boxed{T = \displaystyle{2\pi\sqrt{\dfrac{l}{\sqrt{g^{2}+\left(\dfrac{v^{2}}{R}\right)^{2}}}}}} \]
Q17. A cylindrical piece of cork of density of base area A and height h floats in a liquid of
density \(\rho_1\).
The cork is depressed slightly and then released. Show that the cork
oscillates up and down simple harmonically with a period
\[T=2\pi\sqrt{\dfrac{h\rho}{\rho_1g}}\]
where ρ is the density of cork. (Ignore damping due to viscosity of the liquid).
Solution
Let the cylindrical cork have uniform cross-sectional area \(A\), height \(h\), and density \(\rho\). It floats vertically in a liquid of density \(\rho_1\). When the cork is in equilibrium, a certain length of it remains submerged such that the buoyant force balances its weight.
The mass of the cork is \[ \begin{aligned} m &= \rho \, A h \end{aligned} \] Hence, its weight is \[ \begin{aligned} W &= mg = \rho A h g \end{aligned} \] At equilibrium, this is balanced by the buoyant force equal to the weight of the displaced liquid.
Suppose the cork is now depressed downward through a small distance \(x\). The additional volume of liquid displaced is \(Ax\), and the corresponding increase in buoyant force is \[ \begin{aligned} \Delta F &= \rho_1 g (Ax) \end{aligned} \] This additional buoyant force acts upward and tends to restore the cork to its equilibrium position. Therefore, it is a restoring force.
Taking downward displacement as positive, the restoring force acting on the cork is \[ \begin{aligned} F &= - \rho_1 g A x \end{aligned} \] By Newton’s second law, \[ \begin{aligned} m\frac{d^2x}{dt^2} &= - \rho_1 g A x \end{aligned} \] Substituting \(m=\rho A h\), \[ \begin{aligned} \rho A h \frac{d^2x}{dt^2} &= - \rho_1 g A x \end{aligned} \]
Dividing both sides by \(\rho A h\), \[ \begin{aligned} \frac{d^2x}{dt^2} &= -\frac{\rho_1 g}{\rho h} x \end{aligned} \] This equation is of the standard form of simple harmonic motion, \[ \frac{d^2x}{dt^2} = -\omega^2 x \] where \[ \begin{aligned} \omega^2 &= \frac{\rho_1 g}{\rho h} \end{aligned} \]
Hence, the angular frequency of oscillation is \[ \begin{aligned} \omega &= \sqrt{\frac{\rho_1 g}{\rho h}} \end{aligned} \] The time period of oscillation is therefore \[ \begin{aligned} T &= \frac{2\pi}{\omega} = 2\pi\sqrt{\frac{h\rho}{\rho_1 g}} \end{aligned} \]
Thus, when slightly depressed and released, the cork oscillates vertically with simple harmonic motion having a time period \[ \boxed{T = 2\pi\sqrt{\dfrac{h\rho}{\rho_1 g}}} \]
Q18. One end of a U-tube containing mercury is connected to a suction pump and the other end to atmosphere. A small pressure difference is maintained between the two columns. Show that, when the suction pump is removed, the column of mercury in the U-tube executes simple harmonic motion.
Solution
Consider a U-tube of uniform cross-sectional area \(A\) containing mercury of density \(\rho\). One limb is connected to a suction pump and the other is open to the atmosphere. Due to the small pressure difference, the mercury levels in the two limbs differ, and the system is initially at rest in a displaced configuration.
Let the mercury column be displaced slightly from its equilibrium position such that the level in one limb rises by a distance \(x\), while the level in the other limb falls by the same amount. The total difference in levels between the two limbs is therefore \(2x\).
This difference in levels produces a pressure difference between the two sides equal to \[ \begin{aligned} \Delta P &= \rho g (2x) \end{aligned} \] This pressure difference acts to restore the mercury column towards its equilibrium position.
The restoring force acting on the mercury column is obtained by multiplying the pressure difference by the cross-sectional area: \[ \begin{aligned} F &= -A \Delta P \\ &= -2\rho g A x \end{aligned} \] The negative sign indicates that the force acts opposite to the displacement.
The mass of the oscillating mercury column is the mass of mercury contained in both limbs that moves during oscillation. If the total length of the oscillating column is \(l\), its mass is \[ \begin{aligned} m &= \rho A l \end{aligned} \] Applying Newton’s second law, \[ \begin{aligned} \rho A l \frac{d^2x}{dt^2} &= -2\rho g A x \end{aligned} \]
Dividing both sides by \(\rho A l\), \[ \begin{aligned} \frac{d^2x}{dt^2} &= -\frac{2g}{l} x \end{aligned} \] This equation is of the standard form of simple harmonic motion, \[ \frac{d^2x}{dt^2} = -\omega^2 x \] where \[ \begin{aligned} \omega^2 &= \frac{2g}{l} \end{aligned} \]
Hence, the angular frequency of oscillation is \[ \begin{aligned} \omega &= \sqrt{\frac{2g}{l}} \end{aligned} \] The corresponding time period is \[ \begin{aligned} T &= 2\pi\sqrt{\frac{l}{2g}} \end{aligned} \]
Thus, when the suction pump is removed, the mercury column in the U-tube oscillates about its equilibrium position with simple harmonic motion.
