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A simple harmonic oscillator of angular ...

A simple harmonic oscillator of angular frequency `2 "rad" s^(-1)` is acted upon by an external force `F = sint N`. If the oscillator is at rest in its equilibrium position at `t = 0`, its position at later times is proportional to :-

A

`sin t - 1/2 sin 2t`

B

`sin t + 1/2 cos 2t`

C

`sin t + 1/2 sin 2t`

D

`cos t - 1/2 sin 2t`

Text Solution

Verified by Experts

The correct Answer is:
A

`F = F sin t rArr a = F/m sin t rArr v = - (F)/(m) cos t`
`x = - (F)/(m) sin t rArr X= vec(rx)_(1) + vec(rx)_(2) = A sin 2t - (F)/(m) sin t`
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