JEE Main 2015 · 10 Apr · Shift 1 · Physics Q12

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

\sin t+\dfrac12\sin2t

\sin t+\dfrac12\cos2t

\cos t-\dfrac12\sin2t

\sin t-\dfrac12\sin2t

Answer and worked solution

Official answer: D

$\ddot x+4x=\sin t$ , with particular solution $x_p=\frac13\sin t$ . With $x(0)=0$ and $\dot x(0)=0$ :
$x=\frac13\sin t-\frac16\sin2t=\frac13\left(\sin t-\frac12\sin2t\right).$

Topics:Damped and forced oscillations

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JEE Main 2015 · 10 Apr · Shift 1, Physics question 12