Question

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

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

Answer 1

The correct option is D $\sin t-\frac{1}{2}\sin 2t$

According to the question,

Given; $F=Fsint\Rightarrow a=\frac{F}{m}sint\Rightarrow v=\frac{-F}{m}cost$

$x=\frac{-F}{m}sint\Rightarrow X=r\overrightarrow{x_1}+r\overrightarrow{x_2}=Asin2t-\frac{F}{m}sint$..............................$eq-i$

Now, we know that in $SHM$,

$a\left(acceleration\right)=-{\omega }^{2}Asin\omega t$ and above we find $a=\frac{F}{m}sint$

Resultant force, $F_{resultant}=sint-m{\omega }^{2}Asinwt$

$\Rightarrow F_R=sint-4mAsin2t$ (Given, $\omega =2radian/s$)

$\Rightarrow a_R=sint/m-4Asin2t$ (By integrating)

$\Rightarrow v_R=-cost/m+2Acos2t$ (By integrating)

$\Rightarrow x_R=-sint/m+Asin2t$

We can write, $x_R=\frac{-1}{m}(sint-Amsin2t)\Rightarrow x_R\propto (sint-Amsin2t)$

Only option $D$ is in the given form, Hence, option $D$ is correct.