A simple harmonic oscillator of angular frequency 2 rad s ⁻¹ is acted… | JEE Main 2015 (10 Apr, Shift 1)

Q15JEE Main 2015 · 10 APR · SHIFT 1Medium

Damped, Forced Oscillations & Superposition

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:

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Answer: (d)

Equation: $\ddot{x} + 4x = \frac{1}{m}\sin t$ (natural $\omega = 2$ ).
Particular solution: $x_p = \frac{\sin t}{m(4-1)} = \frac{\sin t}{3m}$ .
Add free part $C\sin 2t + D\cos 2t$ ; $x(0) = 0 \Rightarrow D = 0$ .
$\dot{x}(0) = 0 \Rightarrow \frac{1}{3m} + 2C = 0 \Rightarrow C = -\frac{1}{6m}$ .
$x = \frac{1}{3m}\left(\sin t - \frac{1}{2}\sin 2t\right)$ .
So the answer is (d) $\sin t - \frac{1}{2}\sin 2t$ .

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Damped, Forced Oscillations & Superposition question from JEE Main 2015 (10 Apr, Shift 1)

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