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$\left(KE\right)$-time $\left(t\right)$ graph of the particle will look like:

• A.
• B.
• C.
• D.

Answer 1

The correct option is A

$T$ = Time period of particle
Kinetic energy in SHM $=\frac{1}{2}m{\omega }^2(A^2-x^2)=\frac{1}{2}m{\omega }^2{A}^2{\cos }^2\omega t$

At time $t=0$, particle is at mean position.
$x=0$ and $v=v_{max}=A\omega$
$\therefore KE={\left(KE\right)}_{max}=\frac{1}{2}m{A}^2{\omega }^2$
At extreme position:
at $t=\frac{T}{4},\omega t=\frac{\pi }{2},x=A,v=v_{min}=0$
$\therefore KE={\left(KE\right)}_{min}=0$
Hence graph (a) correct depicts kinetic energy-time graph.

Tip: In simple harmonic motion, kinetic energy at mean position will be maximum and at extreme position, it will be zero.