Physics - SHM Question with Solution | TestHub

The equation of displacement for a particle starting from the equilibrium position is given by $x=A\sin (\omega t)$.

Given $t=\frac{T}{12}$ and $\omega =\frac{2\pi }{T}$, the displacement is:

$x=A\sin \left(\frac{2\pi }{T}\times \frac{T}{12}\right)=A\sin \left(\frac{\pi }{6}\right)=\frac{A}{2}$

The potential energy (PE) of the particle is:

$PE=\frac{1}{2}m{\omega }^2{x}^2=\frac{1}{2}m{\omega }^2{\left(\frac{A}{2}\right)}^2=\frac{1}{8}m{\omega }^2{A}^2$

The total energy (TE) of the particle is:

$TE=\frac{1}{2}m{\omega }^2{A}^2$

The kinetic energy (KE) is:

$KE=TE-PE=\frac{1}{2}m{\omega }^2{A}^2-\frac{1}{8}m{\omega }^2{A}^2=\frac{3}{8}m{\omega }^2{A}^2$

The ratio of kinetic energy to potential energy is:

$\frac{KE}{PE}=\frac{\frac{3}{8}m{\omega }^2{A}^2}{\frac{1}{8}m{\omega }^2{A}^2}=\frac{3}{1}$

Answer: $3:1$