Question

The acceleration of a particle, starting from rest, varies with time according to the relation $a=-s{\omega }^2\sin \omega t$. The displacement of this particle at time $t$ will be:

• A. $s\sin \omega t$
• B. $s\omega \cos \omega t$
• C. $s\omega \sin \omega t$
• D. $-\frac{1}{2}\left(s{\omega }^2\sin \omega t\right)t^2$

Answer 1

The correct option is A $s\sin \omega t$

Given, a= -$s{\omega }^{2}sin\omega t$

$\frac{dv}{dt}=-s{\omega }^{2}sin\omega t$

=>$dv$=-$=-s{\omega }^{2}sin\omega tdt$

Integrating both side, from time $t=0,t$

${\int }_0^{v}dv={\int }_0^t(-s{\omega }^{2}sin\omega t)dt$

$v=s\omega .cos\omega (t-0)+c_1=s\omega .cos\omega t+c_1$

We know that, $v=\frac{ds}{dt}$

$\frac{ds}{dt}$= $s\omega .cos\omega t+c_1$

$ds$=$s\omega .cos\omega tdt+c_1$

Integrating te above w.r.t time,

${\int }_0^{s}ds={\int }_0^t(s\omega .cos\omega \omega t)dt$

$s=s.sin\omega t+c1+c2=s.sin\omega t+c$

Eliminating integrating constant, c,we get,

displacement $S=s.sin\omega t$