Question

A particle is executing simple harmonic motion with a time period $T$. At time $t=0$, it is at its position of equilibrium. The kinetic energy-time graph of the particle will look like

• A.
• B.
• C.
• D.

Answer 1

The correct option is A

Position of particle excecuting SHM.

$x=A\sin wt$

Its velocity function is $v=\frac{dx}{dt}=A\omega \cos wt$

Kinetic energy K.E. $=\frac{1}{2}m{v}^2=\frac{1}{2}m{A}^2{\omega }^2{\cos }^2\omega t$

$\Rightarrow K.E.\propto {\cos }^2\omega t$

But at $t=\frac{T}{4}$, $\cos \left(\omega \frac{r}{4}\right)=0\Rightarrow K.E=0$