Question

At the moment $t=0$ a particle of mass $m$ starts moving due to a force $\overrightarrow{F}=\overrightarrow{F_0}\sin \omega t$, where $F_0$ and $\omega$ are constants. The distance covered by the particle as a function of $t$ is $s=(\omega t-\sin \omega t)\frac{x{F}_0}{m{\omega }^2}$. Find $x$.

Answer 1

We have $\overrightarrow{F}=\overrightarrow{F_0}\sin \omega t$
or $m\frac{d\overrightarrow{v}}{dt}=\overrightarrow{F_0}\sin \omega t$ or $md\overrightarrow{v}=\overrightarrow{F_0}\sin \omega tdt$
On integrating,
$\overrightarrow{m}v=\frac{-\overrightarrow{F_0}}{\omega }\cos \omega t+C$, (where $C$ is integration constant)
When $t=0$, $v=0$, so $C=\frac{\overrightarrow{F_0}}{m\omega }$
Hence, $\overrightarrow{v}=\frac{-\overrightarrow{F_0}}{m\omega }\cos \omega t+\frac{\overrightarrow{F_0}}{m\omega }$
As $|\cos \omega t\le 1$ so, $v=\frac{F_0}{m\omega }(1-\cos \omega t)$
Thus $s={\int }_0^{t}vdt$
$=\frac{F_{0}t}{m\omega }-\frac{F_0\sin \omega t}{m{\omega }^2}=\frac{F_0}{m{\omega }^2}(\omega t-\sin \omega t)$

$\therefore x=1$