Joint Entrance Examination

Graduate Aptitude Test in Engineering

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1

MCQ (Single Correct Answer)

The displacement of an object attached to a spring and executing simple harmonic motion is given by $$x = 2 \times {10^{ - 2}}$$ $$cos$$ $$\pi t$$ metre. The time at which the maximum speed first occurs is

A

$$0.25$$ $$s$$

B

$$0.5$$ $$s$$

C

$$0.75$$ $$s$$

D

$$0.125$$ $$s$$

Here, $$x = 2 \times {10^{ - 2}}\cos \,\pi \,t$$

$$\therefore$$ $$v = {{dx} \over {dt}} = 2 \times {10^{ - 2}}\,\pi \sin \pi t$$

For the first time, the speed to be maximum,

$$\sin \pi t = 1$$ or, $$\sin \pi t = \sin {\pi \over 2}$$

$$ \Rightarrow \pi t = {\pi \over 2}\,\,\,$$ or, $$\,\,\,\,t = {1 \over 2} = 0.5\,\sec .$$

$$\therefore$$ $$v = {{dx} \over {dt}} = 2 \times {10^{ - 2}}\,\pi \sin \pi t$$

For the first time, the speed to be maximum,

$$\sin \pi t = 1$$ or, $$\sin \pi t = \sin {\pi \over 2}$$

$$ \Rightarrow \pi t = {\pi \over 2}\,\,\,$$ or, $$\,\,\,\,t = {1 \over 2} = 0.5\,\sec .$$

2

MCQ (Single Correct Answer)

A point mass oscillates along the $$x$$-axis according to the law $$x = {x_0}\,\cos \left( {\omega t - \pi /4} \right).$$ If the acceleration of the particle is written as $$a = A\,\cos \left( {\omega t + \delta } \right),$$ then

A

$$A = {x_0}{\omega ^2},\,\,\delta = 3\pi /4$$

B

$$A = {x_0},\,\,\delta = - \pi /4$$

C

$$A = {x_0}{\omega ^2},\,\,\delta = \pi /4$$

D

$$A = {x_0}{\omega ^2},\,\,\delta = - \pi /4$$

Here,

$$x = {x_0}\cos \left( {\omega t - \pi /4} \right)$$

$$\therefore$$ Velocity, $$v = {{dx} \over {dt}} = - {x_0}\omega \sin \left( {\omega t - {\pi \over 4}} \right)$$

Acceleration,

$$a = {{dv} \over {dt}} = - {x_0}{\omega ^2}\cos \left( {\omega t - {\pi \over 4}} \right)$$

$$ = {x_0}{\omega ^2}\cos \left[ {\pi + \left( {\omega t - {\pi \over 4}} \right)} \right]$$

$$ = {x_0}{\omega ^2}\cos \left( {\omega t + {{3\pi } \over 4}} \right)$$ $$\,\,\,\,\,\,\,\,\,\,\,...\left( 1 \right)$$

Acceleration, $$a = A\cos \left( {\omega t + \delta } \right)$$ $$\,\,\,\,\,\,\,\,\,\,\,...\left( 2 \right)$$

Comparing the two equations, we get

$$A = {x_0}{\omega ^2}$$ and $$\delta = {{3\pi } \over 4}.$$

$$x = {x_0}\cos \left( {\omega t - \pi /4} \right)$$

$$\therefore$$ Velocity, $$v = {{dx} \over {dt}} = - {x_0}\omega \sin \left( {\omega t - {\pi \over 4}} \right)$$

Acceleration,

$$a = {{dv} \over {dt}} = - {x_0}{\omega ^2}\cos \left( {\omega t - {\pi \over 4}} \right)$$

$$ = {x_0}{\omega ^2}\cos \left[ {\pi + \left( {\omega t - {\pi \over 4}} \right)} \right]$$

$$ = {x_0}{\omega ^2}\cos \left( {\omega t + {{3\pi } \over 4}} \right)$$ $$\,\,\,\,\,\,\,\,\,\,\,...\left( 1 \right)$$

Acceleration, $$a = A\cos \left( {\omega t + \delta } \right)$$ $$\,\,\,\,\,\,\,\,\,\,\,...\left( 2 \right)$$

Comparing the two equations, we get

$$A = {x_0}{\omega ^2}$$ and $$\delta = {{3\pi } \over 4}.$$

3

MCQ (Single Correct Answer)

A particle of mass $$m$$ executes simple harmonic motion with amplitude a and frequency $$v.$$ The average kinetic energy during its motion from the position of equilibrium to the end is

A

$$2{\pi ^2}\,m{a^2}{v^2}$$

B

$${\pi ^2}\,m{a^2}{v^2}$$

C

$${1 \over 4}\,m{a^2}{v^2}$$

D

$$4{\pi ^2}m{a^2}{v^2}$$

$$K = {1 \over 2}m{a^2}{\omega ^2}{\sin ^2}\omega t$$

$$\therefore$$ average $$K.E. = < K > = < {1 \over 2}m{\omega ^2}{a^2}{\sin ^2}\omega t > $$

$$ = {1 \over 2}m\omega {}^2{a^2} < {\sin ^2}\omega t > $$

$$ = {1 \over 2}m{\omega ^2}{a^2}\left( {{1 \over 2}} \right)$$

$$\left( \, \right.$$ as $$\left. { < {{\sin }^2}\theta > = {1 \over 2}} \right)$$

$$ = {1 \over 4}m{\omega ^2}{a^2} = {1 \over 4}m{a^2}{\left( {2\pi v} \right)^2}$$

$$\left( \, \right.$$ $$\left. {\omega = 2\pi v} \right)$$

or, $$\,\,\,\,\, < K > = {\pi ^2}m{a^2}{v^2}$$

4

MCQ (Single Correct Answer)

Two springs, of force constant $${k_1}$$ and $${k_2}$$ are connected to a mass $$m$$ as shown. The frequency of oscillation of the mass is $$f.$$ If both $${k_1}$$ and $${k_2}$$ are made four times their original values, the frequency of oscillation becomes

A

$$2f$$

B

$$f/2$$

C

$$f/4$$

D

$$4f$$

The two springs are in parallel.

$$f = {1 \over {2\pi }}\sqrt {{{{K_1} + {K_2}} \over m}} \,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( i \right)$$

$$f' = {1 \over {2\pi }}\sqrt {{{4{K_1} + 4{K_2}} \over m}} $$

$$ = {1 \over {2\pi }}\sqrt {{{4\left( {{K_1} + 4{K_2}} \right)} \over m}} $$

$$ = 2\left( {{1 \over {2\pi }}\sqrt {{{{K_1} + {K_2}} \over m}} } \right)$$

$$ = 2f\,\,\,\,\,\,\,\,\,\,\,\,\,\,$$ from eqn. $$\left( i \right)$$

$$f = {1 \over {2\pi }}\sqrt {{{{K_1} + {K_2}} \over m}} \,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( i \right)$$

$$f' = {1 \over {2\pi }}\sqrt {{{4{K_1} + 4{K_2}} \over m}} $$

$$ = {1 \over {2\pi }}\sqrt {{{4\left( {{K_1} + 4{K_2}} \right)} \over m}} $$

$$ = 2\left( {{1 \over {2\pi }}\sqrt {{{{K_1} + {K_2}} \over m}} } \right)$$

$$ = 2f\,\,\,\,\,\,\,\,\,\,\,\,\,\,$$ from eqn. $$\left( i \right)$$

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