 JEE Mains Previous Years Questions with Solutions

4.5
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1

AIEEE 2004

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

Explanation

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

$\therefore$ ${\omega _1} = {\omega _2}$
2

AIEEE 2004

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

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

$\Rightarrow A \propto {1 \over {m\left( {\omega _0^2 - {\omega ^2}} \right)}}$
3

AIEEE 2004

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

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

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

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

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

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