Question

The magnitude of force acting on a particle during its motion in a plane $xy$ according to the law $x=a\sin \omega t$ and $y=b\cos \omega t$ where $a,b$ and $\omega$ are constants is given as $\overrightarrow{F}=-xm{\omega }^2\overrightarrow{r}$. Find $x$.

Answer 1

$x=a\sin \omega t$
$\frac{dx}{dt}=a\omega \cos \omega t$
$\frac{d^{2}x}{d{t}^2}=-a{\omega }^2\sin \omega t=-{\omega }^{2}x$
$\therefore F_x=m\frac{d^{2}x}{d{t}^2}=-m{\omega }^{2}x$
As $y=b\cos \omega t$ so
similarly, $F_y=-m{\omega }^{2}y$
$\therefore \overrightarrow{F}=F_x\hat{i}+F_y\hat{j}=-m{\omega }^{2}x\hat{i}-m{\omega }^{2}y\hat{j}=-m{\omega }^2(x\hat{i}+y\hat{j})=-m{\omega }^2\overrightarrow{r}$ thus, $x=1$