Question

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 :

• A. $\sin\, t +\frac{1}{2}\sin 2 t$
• B. $\sin\, t +\frac{1}{2}\cos 2 t$
• C. $\cos\, t -\frac{1}{2}\sin 2 t$
• D. $\sin\,t -\frac{1}{2}\sin\, 2 t$

Answer 1

Answer: (D)

$\sin\,t -\frac{1}{2}\sin\, 2 t$

Answer 2

At $t=0, \nu=0=\frac{d x}{d t}$
If $x=k\left(\sin t-\frac{1}{2} \sin 2 t\right), v=\frac{d x}{d t}=K(\cos t-\cos 2 t)$
At $t=0, \nu=0$