NCERT Exemplar Class 11 Physics Solutions Chapter 14 Oscillations

NCERT Exemplar Class 11 Physics Solutions Chapter 14: MCQI

Question:14.1

The displacement of a particle is represented by the equation $y = 3 \cos\left (\frac{\pi}{4} - 2 \omega t \right )$. The motion of the particle is
(a) simple harmonic with period $\frac{2p}{w}$
(b) simple harmonic with period $\frac{\pi}{\omega}$
(c) periodic but not simple harmonic
(d) non-periodic

Answer:

The answer is option (b), Simple harmonic with period $\frac{\pi}{\omega}$

Question:14.2

The displacement of a particle is represented by the equation $y= \sin^{3} \omega t$. The motion is
(a) non-periodic
(b) periodic but not simple harmonic
(c) simple harmonic with period $\frac{2\pi }{\omega}$
(d) simple harmonic with period $\frac{\pi }{\omega}$

Answer:

The answer is option (b), periodic but not simple harmonic

Question:14.3

The relation between acceleration and displacement of four particles are given below:
(a) $a_{x} = +2x$
(b) $a_{x} = +2x^{2}$
(c) $a_{x} = -2x^{2}$
(d) $a_x = -2x$
Which one of the particle is exempting simple harmonic motion?

Answer:

The answer is option (d) $a_x = -2x$
Explanation: In simple harmonic motion,
Acceleration proportional (as well as opposite) to displacement.
Thus, opt (d)

Question:14.4

The motion of an oscillating liquid column in a U-tube is
(a) periodic but not simple harmonic
(b) non-periodic
(c) simple harmonic and time period is independent of the density of the liquid
(d) simple harmonic and time period is directly proportional to the density of the liquid

Answer:

The answer is the option (c) Simple harmonic and time period is independent of the density of the liquid.
Explanation: Let us take a test tube viz., filled with a liquid of density $\rho$ up to height ‘h’.
When the liquid is lifted in arm Q to a height ‘y’ from A to B, the liquid in arm P drops by the height ‘y’ from A’ to C’. The height difference between the two arms is 2y.
Here, the hydrostatic pressure provides the restoring force, thus,
$F = -V. \rho.g$
$= -A.2y\rho g$ , where,
A = Area of a cross-section of tube, & $F \propto -y,$
Thus, it is simple harmonic motion.
$T = \sqrt{\frac{2\pi m (inertia)}{k (spring)}}= \sqrt{\frac{2\pi A (2h)}{\rho 2 A \rho g}}$
Thus, $T = 2\pi \sqrt{\frac{h}{g}}$
Thus, the motion is harmonic as the time period is independent of density.

Question:14.5

A particle is acted simultaneously by mutually perpendicular simple harmonic motion $x = a \cos \omega t and y = a \sin \omega t$. The trajectory of motion of the particle will be
(a) an ellipse
(b) a parabola
(c) a circle
(d) a straight line

Answer:

The answer is the option (c) a circle.
Explanation: We know that,
Resultant displacement = x + y
$= a \cos \omega t + a \sin \omega t$
Thus, $y^{'} = a \left (\cos \omega t + a \sin \omega t \right )$
$=a \sqrt{2}\left [ \frac{\cos \omega t}{\sqrt{2}}+ \frac {\sin \omega t}{\sqrt{2}} \right ]$
$=a \sqrt{2} \left [ \cos \omega t \cos 45^{\circ} +\sin \omega t \sin 45^{\circ} \right ]$
……. (particle is acted simultaneously by mutually perpendicular direction)
$y^{'}= a\sqrt{2}\cos s\left ( \omega t - 45^{\circ} \right )$ ……. Thus, the displacement can neither be a straight line nor a parabola
Now, let us square and add x & y,
$x^{2} + y^{2} = a^{2} \cos^{2} \omega t + a^{2} \sin^{2} \omega t$
Thus, $x^{2} + y^{2} = a^{2}$
This is the equation of a circle, hence, opt (c)

Question:14.6

The displacement of a particle varies with time according to the relation
$y = a \sin t + b \cos t$
(a) The motion is oscillatory but not SHM
(b) The motion is SHM with amplitude a + b
(c) The motion is SHM with amplitude $a^{2} + b^{2}$
(d) The motion is SHM with amplitude $\sqrt{{a}^{2} + b^{2}}$

Answer:

The answer is the option (d) The motion is SHM with amplitude $\sqrt{a^{2}+b^{2}}$
Explanation: Given: $y = a \sin \omega t + b \cos \omega t$
Now, we know that,
The amplitude of motion, $A = \sqrt{a^{2}+b^{2}}$
Thus, opt (d).

Question:14.7

Four pendulums A, B, C and D are suspended from the same elastic support as shown in the figure. A and C are of the same length, while B is smaller than A and D is larger than A. If A is given a transverse displacement,
(a) D will vibrate with maximum amplitude
(b) C will vibrate with maximum amplitude
(c) B will vibrate with maximum amplitude
(d) All the four will oscillate with equal amplitude.

Answer:

The answer is the option (b) C will vibrate with maximum amplitude.
Explanation: If the pendulum vibrates with transverse vibration,
Time period, $T=2\pi \sqrt{\frac{l}{g}}$, where,
l = length of pendulum A & C.
Now, the disturbance produced is transmitted to all pendulums, i.e., B, C & D, where the time period (T) of C & A is the same. C will vibrate with a maximum in resonance, as the periodic force of period T produces resonance in C.
Hence, opt (b).

Question:14.8

Figure shows the circular motion of a particle. The radius of the circle, the period, sense of revolution and the initial position are indicated on the figure. The simple harmonic motion of the x-projection of the radius vector of the rotating particle P is


$(a)x(t)= B\sin \left ( \frac{2\pi t}{30} \right )$
$(b)x(t)= B\sin \left ( \frac{\pi t}{15} \right )$
$(c)x(t)= B\sin \left ( \frac{\pi t}{15}+\frac{\pi}{2} \right )$
$(d)x(t)= B\cos \left ( \frac{\pi t}{15}+\frac{\pi}{2} \right )$

Answer:

The answer is the option $(a)x(t)= B\sin \left ( \frac{2\pi t}{30} \right )$
Explanation: Circular motion is executed by P with radius B.
Here, $x = OQ \cos (90^{\circ} - \theta)$
=$OQ \sin \theta$
= $B \sin \omega$
Thus, $\theta = \omega t$
$x = B \sin \frac{2 \pi }{T} t$
$x = B \sin \frac{2 \pi }{30}t$
Thus, opt (a)

Question:14.9

The equation of motion of a particle is $x = a \cos (\alpha t)^{2}.$
The motion is
(a) periodic but not oscillatory.
(b) periodic and oscillatory.
(c) oscillatory but not periodic.
(d) neither periodic nor oscillatory.

Answer:

The answer is the option (c) Oscillatory but not periodic.

Question:14.10

A particle executing S.H.M. has a maximum speed of $30 \frac{cm}{s}$ and a maximum acceleration of $60 \frac{cm}{s^{2}}$. The period of oscillation is
$(a) \pi s$.
$(b) \left ( \frac{\pi}{2} \right ) s$.
$(c) 2\pi s$.
$(d) \left (\frac{\pi}{t} \right ) s.$

Answer:

Answer: The answer is the option $(a) \pi s$
Explanation: Let us consider the equation of a SHM
Thus,$y = a \sin \omega t$
$v = \frac{dy}{dt}$
$= a\omega \cos \omega t$ ……. (i)
$a = \frac{dv}{dt}$
$= -a\omega^ 2 \sin \omega t$ ….. (ii)
$V_{max} = 30 \frac{cm}{s}$ …. (given)
From (i),
$V_{max} = a \omega$
Thus, $a \omega=30$ ….. (iii)
From (ii),
Now, $a_{max} = a\omega ^{2} = 60$ …….. (iv)
Thus, $60 = \omega \times 30$
Therefore, $\omega = 2 \frac{rad}{s}$
Now, $\frac{2 \pi }{T} = 2$
Thus, $T = \pi$.
Hence opt (a).

Question:14.11

When a mass m is connected individually to two springs $S_{1}$ and $S_{2}$, the oscillation frequencies are $v_{1}$ and $v_{2}$. If the same mass is attached to the two springs, as shown in Figure, the oscillation frequency would be

Answer:

Answer: The answer is the option $(b) \sqrt{v_{1}^{2}+v_{2}^{2}}$
Explanation: If we connect a mass (m) to a spring on a frictionless horizontal surface, then, their frequencies will be –
$V_{1} = \frac{1}{2\pi}\sqrt{\frac{K_{1}}{m}}$ & $V_{2} = \frac{1}{2\pi}\sqrt{\frac{K_{2}}{m}}$
…. (i) …. (ii)
Since the springs are parallel, their equivalent will be-
$K_{p} = K_{1} + K_{2}$
& frequency will be –
$V_{p} = V_{1} =\frac{ 1}{2 \pi}\sqrt{\frac{K_{1} + K_{2}}{m}}$
$=\frac{ 1}{2 \pi}\left [ \frac{K_{1}}{m}+\frac{K_{2}}{m} \right ]^{\frac{1}{2}}$
From (i),
$\frac{ K_{1}}{m} = (2\pi v_{1})^{2} = 4\pi^{ 2}v_{1}^{2}$
& $\frac{ K_{2}}{m} = (2\pi v_{1})^{2} = 4\pi^{ 2}v_{2}^{2}$
Thus, $V_{p} = \frac{1}{2\pi} [4\pi ^{2}v_{1}^{2} + 4\pi ^{2}v_{2}^{2}]$
$V_{p} = \sqrt{v_{1}^{2}+v_{2}^{2}}$
Hence, opt (b)

NCERT Exemplar Class 11 Physics Solutions Chapter 14: MCQII

Question:14.12

The rotation of earth about its axis is
(a) periodic motion
(b) simple harmonic motion
(c) periodic but not simple harmonic motion
(d) non-periodic motion

Answer:

The answer is the option (a) periodic motion and (c) periodic but not SHM
Explanation:
(i) The earth completes one revolution in a regular interval of time.
(ii) The motion of the earth is circular about its own axis.
(iii) But this motion is not SHM as we cannot measure its displacement since it is not about a fixed point. Also, it does not move both sides.

Question:14.13

Motion of a ball bearing inside a smooth curved bowl, when released from a point slightly above the lower point is
(a) simple harmonic motion
(b) non-periodic motion
(c) periodic motion
(d) periodic but not SHM

Answer:

The answer is the option (a) Simple harmonic motion and (b) non - periodic motion
Explanation: Lift the ball from point A to B and smoothly release it to reach C & then return to first A and then B. Thus, it is periodic motion. Here, mg sin ? balances the restoring force (R), also, a restoring force $(mg \sin \theta)$acts on the ball.
Thus, $ma = mg \sin \theta$
$a = g \sin \theta$ or $\frac{d^{2}x}{dt^{2} }= - g \sin \theta$ ….. (when the ball moves upward)
$\frac{d^{2}x}{dt^{2} }= - g \left ( \frac{x}{r} \right )$
Thus, $\frac{d^{2}x}{dt^{2} }\propto (-x)$
Thus, it is a simple harmonic motion.
$\Omega =\sqrt{\frac{ g}{r}}$ or $T = \frac{2\pi }{\omega} = 2\pi \sqrt{\frac{r}{g}}$
Thus, this motion is periodic as well as simple harmonic.

Question:14.14

Displacement versus time curve for a particle executing SHM is shown in figure. Choose the correct statements.
(a) Phase of the oscillator is same at t = 0s and t = 2 s
(b) Phase of the oscillator is same at t = 2s and t = 6 s
(c) Phase of the oscillator is same at t = 1s and t = 7 s
(d) Phase of the oscillator is same at t = 1s and t = 5 s

Answer:

The answer is the option (b) Phase of the oscillator is same at t = 2s & t = 6s and (d) The phase of the oscillator is the same at t = 1s & t = 5s
Explanation: If the mode of vibration of two particles is the same, then they are said to be in the same phase. That means, their distance will be $n\lambda$, where, n = 1, 2, 3,4, …..
(a) Here the particles are not in the same phase as the distance between them is $\frac{\lambda }{2}$.
(b) The particles are at a distance $\lambda$. Thus, they are in the same phase.
(c) The distance between particles is $\lambda =\frac{\lambda }{2}=\frac{3\lambda }{2}$. Thus, they are not in the same phase.
(d) The distance between the particles is $\lambda$. Thus, they are in the same phase.

Question:14.15

Which of the following statements is/are true for a simple hannonic oscillator?
(a) Force acting is directly proportional to displacement from the mean position and opposite to it
(b) Motion is periodic
(c) Acceleration of the oscillator is constant
(d) The velocity is periodic

Answer:

The answer is the option (a) Force acting is directly proportional to the displacement from the mean position and opposite to it, (b) Motion is periodic, and (d) The velocity is periodic.
Explanation: Let, $x = a \sin \omega t$, be a SHM …… (i)
$V = \frac{dx}{dt} = \alpha \omega \cos \omega$ …. (ii)
$V = \frac{dv}{dt} = \alpha \omega\left ( - \omega \sin \omega t \right )$
Thus, $A = - \alpha \omega ^{2} \sin \omega t$ or A = -ω2x
$mA = -m\omega ^{2}x$
Hence, $F \propto (-x)$
Now, x(t) = x(t + T)
Thus, the motion is periodic and simple harmonic.
Now, from (ii),
v(t) = v(t + T)
Hence, opt, (a), (b) & (d) are correct options.

Question:14.16

The displacement time graph of a particle executing S.H.M. is shown in Figure. Which of the following statement is/are true?


(a) The force is zero at $t = \left (\frac{3T}{4} \right )$
(b) The acceleration is maximum at $t = \left (\frac{4T}{4} \right )$
(c) The velocity is maximum at $t = \left (\frac{T}{4} \right )$
(d) The P.E. is equal to K.E. of oscillation at $t = \left (\frac{T}{2} \right )$

Answer:

The answer is the option (a) The force is zero at $t = \left (\frac{3T}{4} \right )$, (b) The acceleration is maximum at $t = \left (\frac{4T}{4} \right )$, and (c) The velocity is maximum at $t = \left (\frac{T}{4} \right )$
Explanation: (a) The particle is at its mean position at $t = \left (\frac{3T}{4} \right )$, so the force acting on it zero, but due to inertia of mass the motion continues
a = 0 thus, F = 0.
(b) Particles velocity changes increasing to decrease so maximum at the change in velocity, at $t = \left (\frac{4T}{4} \right )$. Thus, acceleration is maximum here.
(c) The velocity is maximum at its mean position at $t = \left (\frac{T}{4} \right )$ as there is no retarding force on it.
(d) K.E. = 0 at $t = \frac{T}{2} = \frac{2T}{4}.$
Thus, P.E is not equal to kinetic energy.

Question:14.17

A body is performing SHM, then its
(a) The average total energy per cycle is equal to its maximum kinetic energy
(b) The average kinetic energy per cycle is equal to half of its maximum kinetic energy
(c) mean velocity over a complete cycle is equal to $\frac{2}{\pi}$ times of its maximum velocity
(d) root mean square velocity is $\frac{1}{\sqrt{2}}$ times of its maximum velocity

Answer:

The answer is the option (a) Average total energy per cycle is equal to its maximum kinetic energy, (b) Average kinetic energy per cycle is equal to half its maximum kinetic energy, and (d) Root square mean velocity is equal to $\frac{1}{\sqrt{2}}$ times its maximum velocity.
Explanation: (a) let$x = a \sin \omega t$ be a periodic SHM
Let m be the mass executing SHM
$v = \frac{dx}{dt}$
= $a\omega \cos \omega t$
$v_{max} = \alpha \omega ..... (since \cos \omega t = 1)$
Now, $\frac{K.E.}{P.E}$ = Total mechanical energy
Thus, $T.E. = \frac{1}{2} ma^{2} \omega ^{2}$
Thus, $K.E. = \frac{1}{2} ma^{2} \omega ^{2}$
Or we can say that the average total energy is K.E.max
(b) Let, amplitude = a
Angular frequency = $\omega$
Thus, maximum velocity = αω, which varies according to the sine law.
Thus, the rms value of a complete cycle = $-\frac{1}{\sqrt{2}} \alpha \omega .$
Thus, the average
$K.E. = \frac{1}{2} mv_{rms}^{2}$
= $\frac{1}{2}m \frac{1}{2} a^{2}\omega ^{2}$
= $\frac{1}{2}\left [m \frac{1}{2} a^{2}\omega ^{2} \right ]$
= $\frac{1}{2}\left [ \frac{1}{2} m _{max}^{2} \right ]$
=$\frac{1}{2} K.E. _{max}$
(c) $v = a\omega \cos \omega t$
$V_{mean} = \frac{(a\omega -a\omega )}{2}$
Vmean = 0 ….. since vmax is not equal to vmean
$V_{rms} = \sqrt{\frac{v_{1}^{2}+v_{2}^{2}}{2} }= \frac{a\omega }{\sqrt{2}}$
Thus, $v_{rms }= \frac{v_{max}}{\sqrt{2}}$

Question:14.18

A particle is in linear simple harmonic motion between two points A and B, 10 cm apart (figure). Take the direction from A to B as the positive direction and choose the correct statements. __ _
AO = OB = 5 cm
BC= 8 cm

(a) The sign of velocity, acceleration and force on the particle when it is 3 cm away from A going towards B are positive
(b) The sign of velocity of the particle at C going towards B is negative
(c) The sign of velocity, acceleration and force on the particle when it is 4 cm away from B going towards A are negative
(d) The sign of acceleration and force on the particle when it is at point B is negative

Answer:

The answer is the option (a) The sign of velocity, acceleration & force on the particle when it is 3 cm away from A going towards B is positive, (c) The sign of velocity, acceleration & force on the particle when it is 4 cm away from B going towards A are negative, and (d) The sign of acceleration and force on the particle when it is at point B is negative.
Explanation: (a) The velocity of the particle increases up to 0 when it is 3 cm away from A and is going from A to B, i.e., in the positive direction. Thus, the velocity is positive. Also, acceleration in SHM is towards positive.
(b) Here the velocity is positive and not negative since the particle is going towards B.
(c) Here the particle is 4 cm away from B and is going towards A, i.e., the particle is going from B to A, i.e., in the negative direction. Hence, the velocity & acceleration towards mean position O is negative.
(d) Here, force and acceleration both are negative as the particle is at B, and they both are towards O.

NCERT Exemplar Class 11 Physics Solutions Chapter 14: Very Short Answer

Question:14.19

Displacement versus time curve for a particle executing SHM is shown in figure. Identify the points marked at which (i) velocity of the oscillator is zero, (ii) speed of the oscillator is maximum.

Answer:

(i) At A, C, E, and G the displacement is maximum. Hence, the velocity of the oscillator will also be maximum.
(ii) At B, D, F and H, the displacement of the oscillator is zero. Thus, there is no restoring force. Hence, the speed of the oscillator will be maximum.

Question:14.20

Two identical springs of spring constant k are attached to a block of mass m and to fixed supports as shown in figure. When the mass is displaced from equilibrium position by a distance x towards right, find the restoring force.

Answer:

If we displace the mass ‘m’ frpm its equilibrium position towards the right by distance ‘x’, The spring B will be compressed by distance x, & let kx be the force applied on the mass ‘m’ towards left. If we apply the force kx on the mass, A will be extended by distance x towards left; and apply force kx towards left. Thus, restoring force will act on the block as net force towards left.
F = kx + kx
= 2kx
Thus, restoring force towards the left is 2kx.

Question:14.21

What are the two basic characteristics of a simple harmonic motion?

Answer:

(i)The direction of acceleration is towards the mean position, and
$Acceleration \propto Displacement$ (from mean position)
(ii) The direction of force and displacement are opposite, thus, $F = -kx .$
$Restoring force \propto Displacement.$
These are the two basic characteristics of SHM.

Question:14.22

When will the motion of a simple pendulum be simple harmonic?

Answer:

Let us consider a pendulum whose,
Length = l
Mass of bob = m, viz., displaced by angle = $\theta$
Now, Restoring force $= F=$$-mg \sin \theta$
Also, if $\theta$ is small then
Sin $\theta$ = $\theta$
$=\frac{arc}{radius}$
$=\frac{x}{l}$
Thus,
$F=-mg\frac{x}{l}$
Or$F \propto (-x)$ …………. (since m, g & l are constants)
Thus, simple harmonic for the small-angle

Question:14.21

What is the ratio of maximum acceleration to the maximum velocity of a simple harmonic oscillator?

Answer:

Let us consider $x = A \sin \omega t$ to be an SHM
Now, $v = \frac{dx}{dt}$
$= A\omega \cos \omega t$
Now, for $v_{max}$,$\cos \omega t = -1$
Thus, $v_{max} = A \omega$
Now, $a = \frac{dv}{dt}$
$= -A\omega ^{2} \sin \omega t$
Now, for $a_{max}$,
$\sin \omega t = -1$
Thus, $a_{max} = A\omega ^{2}$
Now, $\frac{a_{max}}{v_{max}} = \frac{A\omega ^{2}}{A\omega }$
$=\omega$.

Question:14.24

What is the ratio between the distance travelled by the oscillator in one time period and amplitude?

Answer:

We already know that the distance travelled by an oscillator in a one-time period is equal to 4A,
Where A = amplitude of oscillation.
Thus, we know that the
Required ratio
$\frac{4A}{A}$
=4:1

Question:14.25

In figure, what will be the sign of the velocity of the point P’, which is the projection of the velocity of the reference particle P. P is moving in a circle of radius R in anti-clockwise direction

Answer:

At time = t, P’ is the foot perpendicular of the velocity vector of particle P.
This foot shifts from P to Q, i.e., towards the negative axis, when the particle moves from P to P₁.
Thus, there is –ve sign of $\theta$ motion of P’.

Question:14.26

Show that for a particle executing SHM, velocity and displacement have a phase difference of $\frac{\pi}{2}$.

Answer:

Let us consider $x = A \sin \omega t$ to be a SHM …….. (i)
Now, $v = \frac{dx}{dt}$
$= A\omega \cos \omega t$
$= A\omega \sin (90^{\circ} + \omega t)$ …….. [since $\sin (90^{\circ} + \theta ) = \cos \theta$]
Thus, $v = A\omega \sin \left (\omega t + \left ( \frac{\pi}{2} \right ) \right )$
From (i), we know that,
The phase of displacement = $\omega t$
& from (ii), we know that,
The phase of velocity = $\left (\omega t + \left ( \frac{\pi}{2} \right ) \right )$
Thus, the phase difference = $\omega t + \frac{\pi}{2} - \omega t$
$= \frac{\pi}{2}.$

Question:14.27

Draw a graph to show the variation of PE, KE and total energy of a simple harmonic oscillator with displacement.

Answer:

Let us consider that a mass is lying on a horizontal frictionless surface, where spring constant = k.
If we displace the mass by distance A from its mean position, then it will execute SHM.
At this stretched position, P.E. of mass = $\frac{1}{2} kA^{2}.$
Now, at maximum stretch, i.e.,
At x = A, K.E. = 0
Now, at $x = x\pm A$
P.E. = total energy
= $\frac{1}{2} kA^{2}.$
Now let’s consider that the mass is back at its mean position, now the restoring force acting on the particle will be
$P.E. = \frac{1}{2} kx^{2}$
= $\frac{1}{2} m\omega ^{2}x^{2}$
The restoring force constant of oscillator is$k = m\omega ^{2}$
When $x \pm A$
$P.E. = \frac{1}{2} m\omega ^{2}A^{2}$
& $K.E. = \frac{1}{2} mv^{2}$
$= \frac{1}{2} m\omega ^{2} (A^{2} -x^{2})$
Now, when $x = \pm A,K.E. = 0$ Now, at displacement ‘x’, T.E. will be,
E = K.E. + P.E.
= $\frac{1}{2} m\omega ^{2}A^{2}$
Thus, with displacement ‘x’, E is constant.

Question:14.28

The length of a second’s pendulum on the surface of earth is 1 m. What will be the length of a second’s pendulum on the moon?

Answer:

A second pendulum is a pendulum with time period (T) = 2 seconds
$T_{e} = 2\pi\sqrt{\frac{ l_{e}}{g_{e}}}$
Thus, $T_{e}^{2} = 4\pi ^{2}{\frac{l_{e}}{g_{e}}}$ …… (i)
Now, $T_m = 2\pi \sqrt{ \frac{l_{m}}{g_{m}}}$
Thus, $T_{m}^{2} = 4\pi ^{2} 6\frac{ lm}{g_e}$ ……. (ii) (g_m=g_e/6)
Now, for the second pendulum,
$\frac{T_{m}^{2}}{T_{e}^{2}}= {6g_e} \frac{l_{e}}{g_{e}l_m}$
$1 = \frac{6lm}{l_{e}}$
$1 = \frac{6l_{m}}{1m}$ ……. (since $l_{e}=1m$)
Thus, $l_{m} = \frac{1}{6} m$

NCERT Exemplar Class 11 Physics Solutions Chapter 14: Short Answer

Question:14.29

Find the time period of mass M when displaced from its equilibrium position and then released for the system as shown in figure.

Answer:

If we pull mass M & then release it, it oscillates with the pulley up & down. Let x0 be the extension of the string when loaded with M, Due to acceleration and the same amount of forces the extension and compression of the spring from initial position is larger and smaller, respectively. Hence, we can neglect the gravitational force here.
Now let us apply force ‘F’ to pull M downwards by displacement x. As the string cannot be extended, its extension will be 2x.
Thus, the total extension = $(x_{0} + 2x)$
When we pull it downward by x
$T^{'}=k(x_{0} + 2x)$
When we do not pull M,
$T^{'}=kx_{0}$
F = 2T
Thus, $F=2kx_{0}$
& $F{'} = 2T{'}$
$F{'} = 2k (x_{0} + 2x)$
Now, restoring force,
Frest $= - (F^{'}-F)$
= $-[2k(x_{0} + 2x) - 2kx_{0}]$
= $-2k.2x$
Thus, $Ma = -4kx$
$a = -\frac{4k}{M }x$
Thus, $a \propto -x$
Therefore, it is a simple harmonic motion.
Now, $a = -\omega ^{2}x$
$\omega ^{2}=- \frac{a}{x}$
$= \frac{+4k}{M}$
$\omega = 2\sqrt{\frac{k}{M}}$
$T = \pi \sqrt{\frac{M}{k} }$

Question:14.3

Show that the motion of a particle represented by$y = \sin wt - \cos wt$ is simple harmonic with a period of $\frac{2\pi}{\omega}$.

Answer:

$y = \sin \omega t - \cos \omega t$
$= \sqrt{2}\ \left [sin \omega t \frac{1}{\sqrt{2}}- \cos \omega t \frac{1}{\sqrt{2}} \right ]$
$= \sqrt{2}\ sin\left [ \omega t- \frac{\pi }{4} \right ]$
Comparing this equation with standard SHM we get,
$\omega = \frac{2 \pi }{T}$
Or $t = \frac{2 \pi }{\omega}$

Question:14.31

Find the displacement of a simple harmonic oscillator at which its PE is half of the maximum energy of the oscillator.

Answer:

Let us consider an oscillator viz. at ,
displacement = x from its mean position,
Mass = m
$P.E. = \frac{1}{2} kx^{2}$
Force constant of oscillator, $k = m\omega ^{2}$
Thus, $P.E. = -m^{2}x^{2}$
At x = A, when K.E. = 0, P.E. will be maximum & it will be the total energy of the oscillator
$E = \frac{1}{2} m\omega ^{2}A^{2}$
At displacement ‘x’, $P.E. = \frac{1}{2}E$
Thus,
$\frac{1}{2} m\omega ^{2}x^{2}$= $\frac{1}{2}.\frac{1}{2} m\omega ^{2}A^{2}$
& $x = \pm \frac{A}{\sqrt{2}}$
Thus, when displacement = $\pm \frac{1}{\sqrt{2}}$ amplitude from mean position, P.E. will be half of the total energy.

Question:13.32

A body of mass m is situated in a potential field $U(x) = U_{0}(1- \cos \alpha x )$ when $U_{0}$ and $\alpha$ are constants. Find the time period of small oscillations.

Answer:

Here, dW = F.dx
So, if W = U, dU = F.dx
Or $F = -\frac{dUx}{dx}$ ….. (since, restoring force here is opposite to displacement)
$F = -\frac{d}{dx} [U_{0} (1 - \cos \alpha x) = - \frac{d}{dx} [U_{0} + U_{0} \cos \alpha x]$
$F = -\alpha U_{0} \sin \alpha x$
Now, $\alpha x$ is small for SHM, So sin $\alpha x$ will become $\alpha x$ …… (i)
Thus, $F = -\alpha U_{0} \alpha x$
Now, since, $U_{0}$ & $\alpha$ are constants,
$F \propto -x$
Thus, the motion will be SHM.
From (ii)
$k = \alpha ^{2}U_{0}$
$m \omega ^{2}= \alpha ^{2}U_{0}$
Thus, $\omega ^{2} = \alpha ^{2} .\frac{U_{0}}{m}$
$\left ( \frac{2\pi}{T} \right )^{2} = \alpha ^{2} .\frac{U_{0}}{m}$
$T = \frac{2\pi }{\alpha }\sqrt{\frac{m}{U_{0}}}$
Thus, considering (i) time period is valid for the small-angle $\alpha x$.

Question:14.33

A mass of 2 kg is attached to the spring of spring constant 50 Nm^-1. The block is pulled to a distance of 5cm from its equilibrium position at x = 0 on a horizontal frictionless surface from rest at t = 0. Write the expression for its displacement at anytime t.

Answer:

Here, the value of x is-
x = 5 sin 5t.

Question:14.34

Consider a pair of identical pendulums, which oscillate with equal amplitude independently such that when one pendulum is at its extreme position making an angle of $2^{\circ}$ to the right with the vertical, the other pendulum makes an angle of $1^{\circ}$ to the left of the vertical. What is the phase difference between the pendulums?

Answer:

Now, $\Theta_{1} = \Theta_{0} \sin (\omega t + \delta_{1})$
& $\Theta_{2} = \Theta_{0} \sin (\omega t + \delta_{2})$
Now, for the first pendulum,
$\Theta = 2^{\circ}$
Thus, $\sin (\omega t + \delta _{1}) = 1$
& for second pendulum,
$\Theta = -1^{\circ}$
Thus, $\sin (\omega t + \delta _{2}) = -1$
Thus, $(\omega t + \delta _{1}) = 90^{\circ}$ & $(\omega t + \delta _{2}) = 30^{\circ}$
Therefore,
$\delta _{1}+\delta _{2}=120^{\circ}$

NCERT Exemplar Class 11 Physics Solutions Chapter 14: Long Answer

Question:14.35

A person normally weighing 50 kg stands on a massless platform which oscillates up and down harmonically at a frequency of 2.0 s^-1 and an amplitude 5.0 cm. A weighing machine on the platform gives the persons weight against time.
(a) Will there be any change in weight of the body, during the oscillation?
(b) If answer to part (a) is yes, what will be the maximum and minimum reading in the machine and at which position?

Answer:

Due to normal reaction ‘N’, there will be weight in weight machine.
Let us consider the top positions of the platform, both the forces, due to the weight of the person and oscillator acts downwards.
Thus, the motion will be downwards.
Let us consider, acceleration = a
ma = mg – N …….. (i)
Now, when the platform moves upwards from its lowest position,
ma = N – mg …….. (ii)
Now, acceleration of oscillator is
$a = \omega ^{2}A$
From (i),
$N = mg - m \omega ^{2}A$
Where,
Amplitude = A
Angular frequency = ω &
Mass of oscillator = m
$\omega =2 \pi v$
$= 4 \pi rad /s$
A = 5cm
$= 5 \times 10^{-2}m$
m = 50 kg
$N = 50 \times 9.8 - 50 \times 4\pi \times 4\pi \times 5 \times 10^{-2}$
$= 50[ 9.8 -16\pi ^{2} \times 5\times 10^{-2}]$
$= 50 \times 1.91$
$= 95.5 N$
Thus, the minimum weight is 95.5 N.
From (ii),
N – mg = ma
Now, for upward motion from the lowest point of the oscillator,
$N = m (a + g)$
$= m [ 9.81 + \omega ^{2}A] = \omega ^{2}A$
$= 50 [9.81 + 7.89]$
$= 50 [17.7]$
$= 885 N$
Therefore, during oscillation, there is a change in the weight of the body.
Also, maximum weight = 885 N, when the platform moves to upward direction from lowest direction & minimum weight = 95.5 N, when the platform moves to downward direction from the highest point.

Question:14.36

A body of mass m is attached to one end of a massless spring which is suspended vertically from a fixed point. The mass is held in hand so that the spring is neither stretched nor compressed. Suddenly the support of the hand is removed. The lowest position attained by the mass during oscillation is 4cm below the point, where it was held in hand.
(a) What is the amplitude of oscillation?
(b) Find the frequency of oscillation?

Answer:

(a) As no deforming force acts on the spring when mass ‘m’ is supported by hand extension in the spring. Let m reach its new position at displacement = x unit, from the previous one, then,
P.E. of the spring or mass = gravitational P.E. lost by man
P.E. = mgx
But due to spring,
$\frac{1}{2} kx^{2}k = \omega ^{2}A$
Thus, $x = \frac{2mg}{k}$
When extension is $x_{0}$, the spring will be at the mean position by the block
$F=kx_{0}$
F = mg
Thus, $x_{0}=\frac{mg}{k}$ ……. (ii)
Now, from (i) & (ii)
$x=2\left (\frac{mg}{k} \right )$
=$2x_{0} = 4cm$
Thus, $x_{0} = 2cm$
$x - x_{0} = 4-2 = 2cm.$
Thus, from the mean position, the amplitude of the oscillator is maximum.
(b) Now, we know that time period (T) does not depend on amplitude,
$T = 2\pi \sqrt{\frac{m}{k}}$
From (i)
$\frac{2mg}{k} = x$
$\frac{m}{k} = \frac{x}{2g}$
$= \frac{4 \times 10^{-2}}{2\times 9.8}$
Or, $\frac{k}{m}= \frac{2\times 9.8}{4\times 10^{-2}}$
Now, $v = \frac{1}{2}\pi \sqrt{\frac{k}{m}}$
$= \frac{1}{2}\times 3.14\sqrt{\frac{2 \times 9.8}{4 \times 10^{-2}} }$
$= \frac{10 \times 2.21}{6.28}$
$= 3.52 Hz$
Since the total extension in the spring is 4 cm when released & amplitude is 2 cm, the oscillator will not rise above 4cm, & hence, it will oscillate below the released position.

Question:14.37

A cylindrical log of wood of height h and area of cross-section A floats in water. It is pressed and then released. Show that the log would execute S.H.M. with a time period.
$T = 2 \pi \sqrt{\frac{m}{A \rho g}}$
where m is mass of the body and ρ is density of the liquid.

Answer:

If we press a log downward into the liquid, a buoyant force acts on it, and due to inertia it moves upwards from its mean position & comes down again due to gravity.
Thus, the restoring force on the block = Buoyant force (B.F.) – mg
Volume of liquid displaces by block = V
When it floats,
mg = BF
Or, $mg = V\rho g$
$mg = Ax_{0} \rho g$ …….. (i)
Area of crossection = A
Height of liquid block =$x_{0}$
When pressed in water, the total height of the block in water = $(x + x_{0})$
Thus, net restoring force = $[A(x + x_{0})]\rho g -mg$
Frest $= -Ax\rho g$ …….. ( since BF is upward & x is downward)
Frest proportional to-x
Hence, the motion here is SHM.
Now, $kA\rho g$
$a = -\omega ^{2}x\omega$
$= \frac{k}{m}$
Thus, $T= 2 \pi \sqrt{\frac{m}{k}}$
Frest $= -A\rho gx$
$ma= -A\rho gx$
$a=\frac{ -A\rho gx}{m}$
$-\omega ^{2}x=\frac{ -A\rho gx}{m}$
$K = A\rho g$
$\left (\frac{2\pi 2}{T} \right )= \frac{A\rho g}{m}$
Thus,
$\frac{T}{2\pi} = \frac{m}{A\rho g}$
Thus,
$T = 2 \pi \sqrt{\frac{m}{A \rho g}}$

Question:14.38

One end of a V-tube containing mercury is connected to a suction pump and the other end to atmosphere. The two arms of the tube are inclined to horizontal at an angle of $45^{\circ}$ each. A small pressure difference is created between two columns when the suction pump is removed. Will the column of mercury in V-tube execute simple harmonic motion? Neglect capillary and viscous forces. Find the time period of oscillation.

Answer:

Let initial height = h₀ of liquid in both columns; due to pressure difference, if the liquid in arm A is pressed by x, then liquid in arm B will rise by x.
Let us consider dx as an element of height, then,
Mass $dm = A.dx.\rho,$where, Area of a cross-section of tube = A
PE of left dm element columns = (dm) gh
& PE of dm in left columns = $dm = A.dx.\rho gx$
Thus, total PE in left column = $\int_{0}^{h}A\rho gx dx=A\rho g\left [ \frac{x^{2}}{2} \right ]^{h_{0}}_{0}$
$= A\rho g\frac{ h_{1}^{2}}{2}$
From the figure
$\sin 45^{\circ} = \frac{h_{1}}{l}$
$h_{1} = h_{2} = l \sin 45^{\circ} =\frac{1}{\sqrt{2}}$
$h_{1}^{2} = h_{2}^{2} =\frac{ l^{2}}{2}$
Thus, PE in left column = PE in right column = $A\rho g \frac{ l^{2}}{4}$
Thus, total potential energy = $A\rho g \frac{ l^{2}}{2}$
Let the element move towards right by y units due to the pressure difference, then,
Liquid column in left arm = (l – y)
Liquid column in right arm = (l + y)
PE in left arm =$A\rho g (l - y)^{2} \sin^{2} 45^{\circ}$
PE in right arm = $A\rho g (l + y)^{2} \sin^{2} 45^{\circ}$
Total $PE = A\rho g \left ( \frac{1}{2} \right )^{2} [(l-y)^{2} + (l + y)^{2}]$
Final $PE =\frac{ A\rho g}{\sqrt{2}} [l^{2} + y^{2} - 2ly + l^{2} + y^{2} + 2ly]$
$=\frac{ A\rho g}{2} (2l^{2}+2y^{2})$
Change PE = Final PE – Initial PE
$=\frac{ A\rho g}{2} (2l^{2}+2y^{2})-A\rho g \frac{l^{2}}{2}$
$=\frac{ A\rho g}{2} (2l^{2}+2y^{2})$
If there is change in velocity(v) of liquid column,
$\Delta KE = \frac{1}{2}mv^{2}$
$m = (A.2l) \rho$
Thus, $\Delta KE = A\rho lv^{2}$
Now, change in total energy = $\frac{A\rho g}{2 }(l^{2} + 2y^{2}) + - A\rho lv^{2}$
Total change in energy,
$\Delta PE + \Delta KE = 0$
Thus,$\frac{A\rho g}{2 }(l^{2} + 2y^{2}) + - A\rho lv^{2} =0$
$\frac{A\rho}{2} [g(l^{2} + 2y^{2}) + 2lv^{2}] = 0$
$\frac{A\rho}{2}$ is not equal to 0, thus,
$[g(l^{2} + 2y^{2}) + 2lv^{2}] = 0$
Now, let us differentiate w.r.t. t,
$g \left [0 + 2 \times 2\times2 \frac{dy}{dt} \right ] + 2l.2v.\frac{dv}{dt} = 0$
$\frac{d^{2}y}{dt^{2}} + \frac{g}{l} .y= 0$
Since 4v is not equal to 0,
$\frac{d^{2}y}{dt^{2}} + \omega ^{2}y= 0$
Thus, $\omega ^{2}=\frac{g}{l}$
$\frac{2\pi }{T}=\sqrt{ \frac{g}{l}}$
Thus,
$T=2\pi\sqrt{ \frac{l}{g}}$

Question:14.39

A tunnel is dug through the centre of the Earth. Show that a body of mass ‘m’ when dropped from rest from one end of the tunnel will execute simple harmonic motion.

Answer:

We know that,
Acceleration due to gravity inside the earth = g’
$g^{'} = g \left (1 - \frac{d}{R} \right )$
= $= g \left (R - \frac{d}{R} \right )$
Now, R-d = y
Thus, $g^{'} = \frac{g.y}{R}$
On both of them, force at depth ‘d’ will be –
$F = -mg^{'}$
$\frac{-mg. y}{R}$
$F \propto (-y)$
Thus, in the tunnel, the motion of the body is SHM.
We can say,
$ma = -mg^{'}$
$a =\frac{ -g}{R }y$
$\omega ^{2}y =\frac{ -g}{R}y$
Thus, $\frac{2\pi }{T} =\sqrt{ \frac{g}{R}}$
Or, $T =2\pi\sqrt{ \frac{R}{g}}$.

Question:14.40

A simple pendulum of time period 1s and length l is hung from a fixed support at O, such that the bob is at a distance H vertically above A on the ground (Figure). The amplitude is $\theta _{0}$. The string snaps at $\theta =\frac{\theta _{0}}{2}$. Find the time taken by the bob to hit the ground. Also, find the distance from A where bob hits the ground. Assume θ0 to be small so that $\sin \theta _{0}{\cong } \theta _{0}$ and cos $\theta _{0}{\cong } 1$.

Answer:

Let us consider the diagram at which,
t = 0
$\Theta = \frac{\Theta _{0}}{2} ,\Theta = \Theta _{0} \cos \omega t$
Now, at t = t₁,
$\Theta =\frac{\Theta _{0}}{2}$

Thus, $\frac{\Theta _{0}}{2} = \Theta _{0} \cos \frac{2\pi }{T}t_{1}$
…….. (Given: T = 1sec)
Thus, $\cos 2\pi t_{1} = \frac{1}{2} = \frac{cos \pi}{3}$
$2\pi t_{1} = \frac{\pi}{3}$
Or, $t_{1} = \frac{1}{6}$
$\Theta = \Theta _{0} \cos 2\pi t$
$\frac{d\Theta }{dt} = -\Theta _{0} 2\pi \sin 2\pi t$

Now, at $t = \frac{1}{6}$
$\Theta =\frac{\Theta _{0}}{2}$
$\frac{d\Theta }{dt} = -\Theta _{0} 2\pi \sin 2\pi \frac{1}{6}$
$= -\Theta _{0} 2\pi \frac{3}{\sqrt{2}}$
$= -\Theta _{0} \pi \sqrt{3}$
Now, $\omega= -\Theta _{0} \pi \sqrt{3}$
i.e., $\frac{v}{l}= -\Theta _{0} \pi \sqrt{3}$
Thus, $v= - \sqrt{3}\Theta _{0} \pi l$
By the –ve sign it is clear that the bob’s motion is towards left.
Now,
$vx = v \cos \frac{\Theta_{0}}{2}$
=$= - \sqrt{3}\Theta _{0} \pi l \cos \frac{\Theta _{0}}{2}$
$vy = v \sin \frac{\Theta _{0}}{2} = - 3\pi \Theta _{0}l \sin \frac{\Theta _{0}}{2}$
Let H’ be the vertical distance covered by vy,
$s = ut + \frac{1}{2} gt^{2}$
$H^{'}=vyt= \frac{1}{2} gt^{2}$
$\frac{1}{2}gt^{2} + 3\pi \Theta _{0}l \sin \frac{\Theta _{0}}{2} \times t - H^{'} = 0$
Thus,$\frac{1}{2}gt^{2} + 3\pi \Theta _{0}l \frac{\Theta _{0}}{2} \times t - H^{'} = 0$ ……. (Given: $\sin \frac{\Theta _{0}}{2} =\frac{\Theta _{0}}{2}$)
Now, by quadratic formula,
$t = \frac{-B\pm \sqrt{B^{2}-4AC}}{2A}$
$t = \frac{-\frac{\sqrt{3}\pi \theta _{0}^{2}l}{4}\pm \sqrt{(3\pi ^{2}\theta _{0}^{4}l^{2})^{2}-\frac{4.1}{2.g}H^{'}}}{\frac{2g}{2}}$
We will neglect $\theta _{0}^{4}$ and $\theta _{0}^{2}$ since $\theta _{0}$ is very small.
Thus, $t = \frac{+\sqrt{2gH^{'}}}{g } H^{'} = H + H^{"}$
Thus, $\sqrt{\frac{2H}{g}}$
H’’<<H’ as $\frac{\Theta _{0}}{2}$ is very small.
Thus, H = H’
Thus, vxt = distance covered in horizontal direction
$X =\sqrt{ 3}\pi \theta _{0}l.\sqrt{\frac{2H}{g}}$
(Given: $\cos \frac{\theta _{0}}{2}=1$)
Thus, $X = \Theta _{0}l\pi \sqrt{\frac{6H}{g}}$
The bob was at a distance of $l\sin \frac{\theta _{0}}{2}=l \frac{\theta _{0}}{2}$ from A at the time of snapping.
Thus, the distance of bob from A where it meets the ground is
$=l \frac{\Theta _{0}}{2} - X$
$=l \frac{\Theta _{0}}{2} -\Theta _{0}l\pi \sqrt{\frac{6H}{g}}$
$= l\Theta _{0}\left [ \frac{1}{2} - \pi\sqrt{ \frac{6H}{g}} \right ]$