Joint Entrance Examination

Graduate Aptitude Test in Engineering

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1

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)

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

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

2

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

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

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

3

MCQ (Single Correct Answer)

The total energy of particle, executing simple harmonic motion is

A

independent of $$x$$

B

$$ \propto \,{x^2}$$

C

$$ \propto \,x$$

D

$$ \propto \,{x^{1/2}}$$

At any instant the total energy is

$${1 \over 2}k{A^2} = \,\,$$ constant, where $$A=$$ amplitude

hence total energy is independent of $$x.$$

$${1 \over 2}k{A^2} = \,\,$$ constant, where $$A=$$ amplitude

hence total energy is independent of $$x.$$

4

MCQ (Single Correct Answer)

A particle at the end of a spring executes $$S.H.M$$ with a period $${t_1}$$. While the corresponding period for another spring is $${t_2}$$. If the period of oscillation with the two springs in series is $$T$$ then

A

$${T^{ - 1}} = t_1^{ - 1} + t_2^{ - 1}$$

B

$${T^2} = t_1^2 + t_2^2$$

C

$$T = {t_1} + {t_2}$$

D

$${T^{ - 2}} = t_1^{ - 2} + t_2^{ - 2}$$

For first spring, $${t_1} = 2\pi \sqrt {{m \over {{k_1}}}} ,$$

For second spring, $${t_2} = 2\pi \sqrt {{m \over {{k_2}}}} $$

when springs are in series then, $${k_{eff}} = {{{k_1}{k_2}} \over {{k_1} + {k_2}}}$$

$$\therefore$$ $$T = 2\pi \sqrt {{{m\left( {{k_1} + {k_2}} \right)} \over {{k_1}{k_2}}}} $$

$$\therefore$$ $$T = 2\pi \sqrt {{m \over {{k_2}}} + {m \over {{k_1}}}} $$

$$ = 2\pi \sqrt {{{t_2^2} \over {{{\left( {2\pi } \right)}^2}}} + {{t_1^2} \over {{{\left( {2\pi } \right)}^2}}}} $$

$$ \Rightarrow {T^2} = t_1^2 + t_2^2$$

For second spring, $${t_2} = 2\pi \sqrt {{m \over {{k_2}}}} $$

when springs are in series then, $${k_{eff}} = {{{k_1}{k_2}} \over {{k_1} + {k_2}}}$$

$$\therefore$$ $$T = 2\pi \sqrt {{{m\left( {{k_1} + {k_2}} \right)} \over {{k_1}{k_2}}}} $$

$$\therefore$$ $$T = 2\pi \sqrt {{m \over {{k_2}}} + {m \over {{k_1}}}} $$

$$ = 2\pi \sqrt {{{t_2^2} \over {{{\left( {2\pi } \right)}^2}}} + {{t_1^2} \over {{{\left( {2\pi } \right)}^2}}}} $$

$$ \Rightarrow {T^2} = t_1^2 + t_2^2$$

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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