Question 169: Oscillations

Question

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 $$\cos \,\omega t\left( {\omega  
e {\omega _0}} ight)$$ is applied to the oscillator. The time displacement of the oscillator will be proportional to

Accepted Answer

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 = $$\left( {\omega _0^2 - {\omega ^2}} ight)x$$
So the external force, F = $$m\left( {\omega _0^2 - {\omega ^2}} ight)x$$    ............(i)
But given that $$F \propto \cos \omega t$$
From (i) we get,
$$m\left( {\omega _0^2 - {\omega ^2}} ight)x \propto \cos \omega t$$  .........(ii)
From equation of SHM we know,
$$x = A\sin \left( {\omega t + \phi } ight)$$
When t = 0 then x = A
$$	herefore$$ A = $$A\sin \left( \phi  ight)$$
$$ \Rightarrow A = {\pi  \over 2}$$
$$	herefore$$ $$x = A\sin \left( {\omega t + {\pi  \over 2}} ight) = A\cos \omega t$$
Putting value of x in (ii), we get
$$m\left( {\omega _0^2 - {\omega ^2}} ight)A\cos \omega t \propto \cos \omega t$$
$$ \Rightarrow A \propto {1 \over {m\left( {\omega _0^2 - {\omega ^2}} ight)}}$$

Answer

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

Answer

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

Answer

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

Answer

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

Oscillations

Choose the correct answer: