Physics - SHM Question with Solution | TestHub

The displacement of a particle executing simple harmonic motion starting from the mean position is given by $x=A\sin (\omega t)$.

At time $t=\frac{T}{6}$, the phase angle is $\omega t=\frac{2\pi }{T}\times \frac{T}{6}=\frac{\pi }{3}=60^{\circ }$.

The potential energy is $PE=\frac{1}{2}m{\omega }^2{x}^2=\frac{1}{2}m{\omega }^2{A}^2{\sin }^2(60^{\circ })=\frac{1}{2}m{\omega }^2{A}^2{\left(\frac{\sqrt{3}}{2}\right)}^2=\frac{3}{4}\left(\frac{1}{2}m{\omega }^2{A}^2\right)$.

The kinetic energy is $KE=\frac{1}{2}m{v}^2=\frac{1}{2}m{\omega }^2{A}^2{\cos }^2(60^{\circ })=\frac{1}{2}m{\omega }^2{A}^2{\left(\frac{1}{2}\right)}^2=\frac{1}{4}\left(\frac{1}{2}m{\omega }^2{A}^2\right)$.

The ratio of potential energy to kinetic energy is $\frac{PE}{KE}=\frac{3/4}{1/4}=\frac{3}{1}$.

Answer: $3:1$