Question

A particle undergoing SHM has the equation $x=A\sin (\omega t+\varphi )$, where $x$ represents the displacement of the particle. The kinetic energy oscillates with time period

• A. $\frac{2\pi }{\omega }$
• B. $\frac{\pi }{\omega }$
• C. $\frac{4\pi }{\omega }$
• D. $\text{None of these}$

Answer 1

The correct option is B $\frac{\pi }{\omega }$

For a particle executing SHM, starting from mean position $(x=0)$ the displacement of the particle is given by
$x=a\sin \omega t$
For one complete oscillation, time period is $T=\frac{2\pi }{\omega }$

Kinetic energy of the particle executing SHM is given by
$K=\frac{1}{2}k{A}^2{\cos }^2\omega t=\frac{1}{4}k{A}^2(1+\cos 2\omega t)$
we can conclude that, kinetic energy oscillates with frequency of $2\omega .$

So, for one complete oscillation, kinetic energy varies with time as
$\Rightarrow T=\frac{2\pi }{2\omega }=\frac{\pi }{\omega }$
Thus, option (b) is the correct answer.