Physics - SHM Question with Solution | TestHub

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

The velocity of the particle is $v=\frac{dx}{dt}=A\omega \cos (\omega t)$.

The kinetic energy is $K=\frac{1}{2}m{v}^2=\frac{1}{2}m{A}^2{\omega }^2{\cos }^2(\omega t)$.

The total energy of the particle in S.H.M. is $E=\frac{1}{2}m{A}^2{\omega }^2$.

Given that the kinetic energy is $50\%$ of the total energy:

$K=\frac{E}{2}$

$\frac{1}{2}m{A}^2{\omega }^2{\cos }^2(\omega t)=\frac{1}{2}\left(\frac{1}{2}m{A}^2{\omega }^2\right)$

${\cos }^2(\omega t)=\frac{1}{2}$

$\cos (\omega t)=\frac{1}{\sqrt{2}}$

$\omega t=\frac{\pi }{4}$

Substituting $\omega =\frac{2\pi }{T}$:

$\frac{2\pi }{T}t=\frac{\pi }{4}$

$t=\frac{T}{8}$

Given $T=6$ s:

$t=\frac{6}{8}=0.75$ s.

Answer: $0.75$ s