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JEE Mains Previous Years Questions with Solutions

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1

AIEEE 2004

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

Explanation

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

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

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

AIEEE 2004

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

Explanation

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, $${k_{eff}} = {{{k_1}{k_2}} \over {{k_1} + {k_2}}}$$

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

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

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

$$ \Rightarrow {T^2} = t_1^2 + t_2^2$$
3

AIEEE 2004

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

Explanation

At any instant the total energy is

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

hence total energy is independent of $$x.$$
4

AIEEE 2004

MCQ (Single Correct Answer)
The bob of a simple pendulum executes simple harmonic motion in water with a period $$t,$$ while the period of oscillation of the bob is $${t_0}$$ in air. Neglecting frictional force of water and given that the density of the bob is $$\left( {4/3} \right) \times 1000\,\,kg/{m^3}.$$ What relationship between $$t$$ and $${t_0}$$ is true
A
$$t = 2{t_0}$$
B
$$t = {t_0}/2$$
C
$$t = {t_0}$$
D
$$t = 4{t_0}$$

Explanation

$$t = 2\pi \sqrt {{\ell \over {{g_{eff}}}}} ;\,{t_o}\,\, = 2\pi \sqrt {{\ell \over g}} $$



$$m{g_{eff}} = mg - B = my - V \times 100 \times g$$

$$\therefore$$ $${g_{eff}} = g - {{100} \over {\left( {m/v} \right)}}g$$

$$ = g - {{1000} \over {{4 \over 3} \times 1000}}g = {g \over 4}$$

$$\therefore$$ $$t = 2\pi \sqrt {{\ell \over {g/4}}} \,\,\,\,\,\,\,\,\,\,\,t = 2{t_0}$$

Questions Asked from Simple Harmonic Motion

On those following papers in MCQ (Single Correct Answer)
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