Question

The potential energy of a particle of mass $4 kg$ in motion along the $x$-axis is given by $U=4(1-\cos 4 x) J$. The time period of the particle for small oscillation $(\sin \theta=\theta)$ is $\left(\frac{\pi}{K}\right)$ s. The value of $K$ is

Answer 1

$U =4(1-\cos 4 x ) $
$ F =-\frac{ dU }{ dx }=-4(+\sin 4 x ) 4=-16 \sin (4 x ) $
$ \text { For small } \theta$
$ \sin \theta \approx \theta $
$ F =-64 x $
$ a =-64 x / m =-16 x$
$ \omega^2=16$
$ T =\frac{2 \pi}{\omega}=\frac{\pi}{2}$