Joint Entrance Examination

Graduate Aptitude Test in Engineering

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1

MCQ (Single Correct Answer)

Two simple harmonic motions are represented by the equations $${y_1} = 0.1\,\sin \left( {100\pi t + {\pi \over 3}} \right)$$ and $${y_2} = 0.1\,\cos \,\pi t.$$ The phase difference of the velocity of particle $$1$$ with respect to the velocity of particle $$2$$ is

A

$${\pi \over 3}$$

B

$${{ - \pi } \over 6}$$

C

$${\pi \over 6}$$

D

$${{ - \pi } \over 3}$$

$${v_1} = {{d{y_1}} \over {dt}} = 0.1 \times 100\pi \cos \left( {100\pi t + {\pi \over 3}} \right)$$

$${v_2} = {{d{y_2}} \over {dt}} = - 0.1\pi sin\pi t = 0.1\pi cos\left( {\pi t + {\pi \over 2}} \right)$$

$$\therefore$$ Phase diff. $$ = {\phi _1} - {\phi _2} = {\pi \over 3} - {\pi \over 2} = {{2\pi - 3\pi } \over 6} = {\pi \over 6}$$

$${v_2} = {{d{y_2}} \over {dt}} = - 0.1\pi sin\pi t = 0.1\pi cos\left( {\pi t + {\pi \over 2}} \right)$$

$$\therefore$$ Phase diff. $$ = {\phi _1} - {\phi _2} = {\pi \over 3} - {\pi \over 2} = {{2\pi - 3\pi } \over 6} = {\pi \over 6}$$

2

MCQ (Single Correct Answer)

The function $${\sin ^2}\left( {\omega t} \right)$$ represents

A

a periodic, but not $$SHM$$ with a period $${\pi \over \omega }$$

B

a periodic, but not $$SHM$$ with a period $${{2\pi } \over \omega }$$

C

a $$SHM$$ with a period $${\pi \over \omega }$$

D

a $$SHM$$ with a period $${{2\pi } \over \omega }$$

y = sin^{2}$$\omega $$t

= $${{1 - \cos 2\omega t} \over 2}$$

$$ = {1 \over 2} - {1 \over 2}\cos \,2\omega t$$

$$ \therefore $$ Angular speed = 2$$\omega $$

$$ \therefore $$ Period (T) = $${{2\pi } \over {angular\,speed}}$$ = $${{2\pi } \over {2\omega }}$$ = $${\pi \over \omega }$$

So it is a periodic function.

As y = sin^{2}$$\omega $$t

$${{dy} \over {dt}}$$ = 2$$\omega $$sin$$\omega $$t cos$$\omega $$t = $$\omega $$ sin2$$\omega $$t

$${{{d^2}y} \over {d{t^2}}}$$ = $$2{\omega ^2}$$ cos2$$\omega $$t which is not proportional to -y.

Hence it is is not SHM.

= $${{1 - \cos 2\omega t} \over 2}$$

$$ = {1 \over 2} - {1 \over 2}\cos \,2\omega t$$

$$ \therefore $$ Angular speed = 2$$\omega $$

$$ \therefore $$ Period (T) = $${{2\pi } \over {angular\,speed}}$$ = $${{2\pi } \over {2\omega }}$$ = $${\pi \over \omega }$$

So it is a periodic function.

As y = sin

$${{dy} \over {dt}}$$ = 2$$\omega $$sin$$\omega $$t cos$$\omega $$t = $$\omega $$ sin2$$\omega $$t

$${{{d^2}y} \over {d{t^2}}}$$ = $$2{\omega ^2}$$ cos2$$\omega $$t which is not proportional to -y.

Hence it is is not SHM.

3

MCQ (Single Correct Answer)

In forced oscillation of a particle the amplitude is maximum for a frequency $${\omega _1}$$ of the force while the energy is maximum for a frequency $${\omega _2}$$ of the force; then

A

$${\omega _1} < {\omega _2}$$ when damping is small and $${\omega _1} > {\omega _2}$$ when damping is large

B

$${\omega _1} > {\omega _2}$$

C

$${\omega _1} = {\omega _2}$$

D

$${\omega _1} < {\omega _2}$$

The maximum of amplitude and energy is obtained when the frequency is equal to the natural frequency (resonance condition)

$$\therefore$$ $${\omega _1} = {\omega _2}$$

$$\therefore$$ $${\omega _1} = {\omega _2}$$

4

MCQ (Single Correct Answer)

A particle of mass $$m$$ is attached to a spring (of spring constant $$k$$) and has a natural angular frequency $${\omega _0}.$$ An external force $$F(t)$$ proportional to $$\cos \,\omega t\left( {\omega \ne {\omega _0}} \right)$$ is applied to the oscillator. The time displacement of the oscillator will be proportional to

A

$${1 \over {m\left( {\omega _0^2 + {\omega ^2}} \right)}}$$

B

$${1 \over {m\left( {\omega _0^2 - {\omega ^2}} \right)}}$$

C

$${m \over {\omega _0^2 - {\omega ^2}}}$$

D

$${m \over {\omega _0^2 + {\omega ^2}}}$$

Given that, initial angular velocity = $${\omega _0}$$

and at any instant time t, angular velocity = $$\omega $$

So when displacement is x then the resultant acceleration

f = $$\left( {\omega _0^2 - {\omega ^2}} \right)x$$

So the external force, F = $$m\left( {\omega _0^2 - {\omega ^2}} \right)x$$ ............(i)

But given that $$F \propto \cos \omega t$$

From (i) we get,

$$m\left( {\omega _0^2 - {\omega ^2}} \right)x \propto \cos \omega t$$ .........(ii)

From equation of SHM we know,

$$x = A\sin \left( {\omega t + \phi } \right)$$

When t = 0 then x = A

$$\therefore$$ A = $$A\sin \left( \phi \right)$$

$$ \Rightarrow A = {\pi \over 2}$$

$$\therefore$$ $$x = A\sin \left( {\omega t + {\pi \over 2}} \right) = A\cos \omega t$$

Putting value of x in (ii), we get

$$m\left( {\omega _0^2 - {\omega ^2}} \right)A\cos \omega t \propto \cos \omega t$$

$$ \Rightarrow A \propto {1 \over {m\left( {\omega _0^2 - {\omega ^2}} \right)}}$$

and at any instant time t, angular velocity = $$\omega $$

So when displacement is x then the resultant acceleration

f = $$\left( {\omega _0^2 - {\omega ^2}} \right)x$$

So the external force, F = $$m\left( {\omega _0^2 - {\omega ^2}} \right)x$$ ............(i)

But given that $$F \propto \cos \omega t$$

From (i) we get,

$$m\left( {\omega _0^2 - {\omega ^2}} \right)x \propto \cos \omega t$$ .........(ii)

From equation of SHM we know,

$$x = A\sin \left( {\omega t + \phi } \right)$$

When t = 0 then x = A

$$\therefore$$ A = $$A\sin \left( \phi \right)$$

$$ \Rightarrow A = {\pi \over 2}$$

$$\therefore$$ $$x = A\sin \left( {\omega t + {\pi \over 2}} \right) = A\cos \omega t$$

Putting value of x in (ii), we get

$$m\left( {\omega _0^2 - {\omega ^2}} \right)A\cos \omega t \propto \cos \omega t$$

$$ \Rightarrow A \propto {1 \over {m\left( {\omega _0^2 - {\omega ^2}} \right)}}$$

On those following papers in MCQ (Single Correct Answer)

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Units & Measurements *keyboard_arrow_right*

Motion *keyboard_arrow_right*

Laws of Motion *keyboard_arrow_right*

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