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 $enery-time$ graph of the particle will look like

• A. Diagram
• B. Diagram
• C. Diagram
• D. Diagram

Answer 1

The correct option is B Diagram

Draw a model diagram.
At time $t=0$, it is at its position of equilibrium. Velocity of a particle at equilibrium position should be maximum and amplitude position should be zero.

Find the correct graph between kinetic energy and time.
Time taken to reach the extreme position from equilibrium position 𝑖𝑠 $T/4$.

Velocity is maximum at equilibrium position and zero at extreme position.

$v=A\omega \cos \omega t\dots \left(i\right)$

Hence kinetic energy will be,

$KE=\frac{1}{2}m{v}^2\dots \left(ii\right)$

Here, $m$ be the mass of the particle and $v$ be the velocity of the particle.

Now substitute, $\left(i\right)$ in $\left(ii\right)$

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

$\Rightarrow KE=\frac{1}{2}m{A}^2{\omega }^2{\cos }^2\omega t$

$KE\propto {\cos }^2\omega t$

$KE$ is maximum at mean position and minimum at extreme position.