Question

A particle undergoing simple harmonic motion has time dependent displacement given by $x\left(t\right)=A\sin \frac{\pi t}{90}$. The ratio of kinetic to potential energy of this particle at $t=210$ s will be :

• A. $3$
• B. $\frac{1}{9}$
• C. $\frac{1}{3}$
• D. $1$

Answer 1

The correct option is C $\frac{1}{3}$

Kinetic Energy, $K=\frac{1}{2}m{\omega }^2{A}^2{\cos }^2\omega t$

Potential Energy, $U=\frac{1}{2}m{\omega }^2{A}^2{\sin }^2\omega t$

$\frac{K}{U}={\cot }^2\omega t={\cot }^2\frac{\pi }{90}\left(210\right)={\cot }^2(2\pi +\frac{\pi }{3})=\frac{1}{3}$

Hence ratio is $\frac{1}{3}$.

The correct option is C.