JEE MAIN - Physics (2004 - No. 63)
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
$${1 \over {m\left( {\omega _0^2 + {\omega ^2}} \right)}}$$
$${1 \over {m\left( {\omega _0^2 - {\omega ^2}} \right)}}$$
$${m \over {\omega _0^2 - {\omega ^2}}}$$
$${m \over {\omega _0^2 + {\omega ^2}}}$$
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
$$m\left( {\omega _0^2 - {\omega ^2}} \right)A\cos \omega t \propto \cos \omega t$$
$$ \Rightarrow A \propto {1 \over {m\left( {\omega _0^2 - {\omega ^2}} \right)}}$$
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
$$m\left( {\omega _0^2 - {\omega ^2}} \right)A\cos \omega t \propto \cos \omega t$$
$$ \Rightarrow A \propto {1 \over {m\left( {\omega _0^2 - {\omega ^2}} \right)}}$$
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