Physics - SHM Question with Solution | TestHub

The frequency of the particle executing S.H.M. is given by $f=\sqrt{\frac{Ka}{\pi m}}$

The angular frequency is $\omega =2\pi f=2\pi \sqrt{\frac{Ka}{\pi m}}$

Squaring both sides, we get ${\omega }^2=4{\pi }^2\left(\frac{Ka}{\pi m}\right)=\frac{4\pi Ka}{m}$

The potential energy of a particle in S.H.M. at a displacement $x$ is given by $U=\frac{1}{2}m{\omega }^2{x}^2$

Substituting the value of ${\omega }^2$, we get $U=\frac{1}{2}m\left(\frac{4\pi Ka}{m}\right)x^2$

$U=2\pi Ka{x}^2$

Answer: $2\pi Ka{x}^2$