Joint Entrance Examination

Graduate Aptitude Test in Engineering

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1

MCQ (Single Correct Answer)

If a simple pendulum has significant amplitude (up to a factor of $$1/e$$ of original ) only in the period between $$t = 0s\,\,to\,\,t = \tau \,s,$$ then $$\tau \,$$ may be called the average life of the pendulum When the spherical bob of the pendulum suffers a retardation (due to viscous drag) proportional to its velocity with $$b$$ as the constant of proportionality, the average life time of the pendulum is (assuming damping is small) in seconds :

A

$${{0.693} \over b}$$

B

$$b$$

C

$${1 \over b}$$

D

$${2 \over b}$$

The equation of motion for the pendulum, suffering retardation

$$I\alpha = - mg\left( {\ell \sin \theta } \right) - mbv\left( \ell \right)$$ where $$I = m{\ell ^2}$$

and $$\alpha = {d^2}\theta /d{t^2}$$

$$\therefore$$ $${{{d^2}\theta } \over {d{t^2}}} = - {g \over \ell }\tan \theta + {{bv} \over \ell }$$

on solving we get $$\theta = {\theta _0}\,{e^{{{bt} \over 2}\sin \left( {\omega t + \phi } \right)}}$$

According to questions $${{{\theta _0}} \over e} = {\theta _0}{e^{{{ - b\tau } \over 2}}}$$

$$\therefore$$ $$\tau = {2 \over b}$$

$$I\alpha = - mg\left( {\ell \sin \theta } \right) - mbv\left( \ell \right)$$ where $$I = m{\ell ^2}$$

and $$\alpha = {d^2}\theta /d{t^2}$$

$$\therefore$$ $${{{d^2}\theta } \over {d{t^2}}} = - {g \over \ell }\tan \theta + {{bv} \over \ell }$$

on solving we get $$\theta = {\theta _0}\,{e^{{{bt} \over 2}\sin \left( {\omega t + \phi } \right)}}$$

According to questions $${{{\theta _0}} \over e} = {\theta _0}{e^{{{ - b\tau } \over 2}}}$$

$$\therefore$$ $$\tau = {2 \over b}$$

2

MCQ (Single Correct Answer)

Two particles are executing simple harmonic motion of the same amplitude $$A$$ and frequency $$\omega $$ along the $$x$$-axis. Their mean position is separated by distance $${X_0}\left( {{X_0} > A} \right)$$. If the maximum separation between them is $$\left( {{X_0} + A} \right),$$ the phase difference between their motion is:

A

$${\pi \over 3}$$

B

$${\pi \over 4}$$

C

$${\pi \over 6}$$

D

$${\pi \over 2}$$

For $${X_0} + A$$ to be the maximum separation $$y$$ one body is at the mean position, the other should be at the extreme.

3

MCQ (Single Correct Answer)

A mass $$M,$$ attached to a horizontal spring, executes $$S.H.M.$$ with amplitude $${A_1}.$$ When the mass $$M$$ passes through its mean position then a smaller mass $$m$$ is placed over it and both of them move together with amplitude $${A_2}.$$ The ratio of $$\left( {{{{A_1}} \over {{A_2}}}} \right)$$ is :

A

$${{M + m} \over M}$$

B

$${\left( {{M \over {M + m}}} \right)^{{1 \over 2}}}$$

C

$${\left( {{{M + m} \over M}} \right)^{{1 \over 2}}}$$

D

$${M \over {M + m}}$$

The net force becomes zero at the mean point. Therefore, linear momentum must be conserved.

$$\therefore$$ $$M{v_1} = \left( {M + m} \right){v_2}$$

$$M{A_1}\sqrt {{k \over M}} = \left( {M + m} \right){A_2}\sqrt {{k \over {m + M}}} $$

$$\therefore$$ $$\left( {V = A\sqrt {{k \over M}} } \right)$$

$${A_1}\sqrt M = {A_2}\sqrt {M + m} $$

$$\therefore$$ $${{{A_1}} \over {{A_2}}} = \sqrt {{{m + M} \over M}} $$

$$\therefore$$ $$M{v_1} = \left( {M + m} \right){v_2}$$

$$M{A_1}\sqrt {{k \over M}} = \left( {M + m} \right){A_2}\sqrt {{k \over {m + M}}} $$

$$\therefore$$ $$\left( {V = A\sqrt {{k \over M}} } \right)$$

$${A_1}\sqrt M = {A_2}\sqrt {M + m} $$

$$\therefore$$ $${{{A_1}} \over {{A_2}}} = \sqrt {{{m + M} \over M}} $$

4

MCQ (Single Correct Answer)

If $$x,$$ $$v$$ and $$a$$ denote the displacement, the velocity and the acceleration of a particle executing simple harmonic motion of time period $$T,$$ then, which of the following does not change with time?

A

$$aT/x$$

B

$$aT + 2\pi v$$

C

$$aT/v$$

D

$${a^2}{T^2} + 4{\pi ^2}{v^2}$$

For an $$SHM,$$ the acceleration $$a = - {\omega ^2}x$$ where $${\omega ^2}$$ is a constant. Therefore $${a \over x}$$ is a constant. The time period $$T$$ is also constant. Therefore $${{aT} \over x}$$ is a constant.

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