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 D

Step 1. Given data

A particle is executing simple harmonic motion within a time period $T$.

Step 2. Kinetic energy time-graph

1. The kinetic energy of an object is the energy due to its motion.
2. SHM is defined as a motion in which the restoring force is directly proportional to the displacement of the body from the mean position.

Position of a particle executing SHM,

$x=A\sin \omega t$ [ $\omega$ is angular frequency, $t$ is time]

Velocity function, $v=\frac{dx}{dt}$ [$v$ is velocity]

$v=A\omega \cos \omega t$

Kinetic energy, $K.E.=\frac{1}{2}m{v}^2$

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

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

Step 3. Check the kinetic energy at $t=\frac{T}{4}$

But, at $t=\frac{T}{4}$, [$T$ is time period of the particle]

$\begin{array}{rcl}{\cos }^2\omega \left(\frac{T}{4}\right)& =& {\cos }^2\omega \left(\frac{2\mathrm{\pi }}{4\omega }\right)\left[\because T=\frac{2\mathrm{\pi }}{\omega }\right]\\ & =& {\cos }^2\left(\frac{\mathrm{\pi }}{2}\right)\\ & =& 0\left[\because \cos \frac{\mathrm{\pi }}{2}=0\right]\end{array}$

$\Rightarrow$ $K.E.=0$

Step 4. Check the kinetic energy at $t=\frac{T}{2}$

at $t=\frac{T}{2}$,

$\begin{array}{rcl}{\cos }^2\omega \left(\frac{T}{2}\right)& =& {\cos }^2\omega \left(\frac{2\mathrm{\pi }}{2\omega }\right)\left[\because T=\frac{2\mathrm{\pi }}{\omega }\right]\\ & =& {\cos }^2\left(\mathrm{\pi }\right)\\ & =& 1\left[\because \mathrm{cos\pi }=-1\right]\end{array}$

$\Rightarrow$ $K.E.$ is maximum

Step 5. check the kinetic energy at $t=T$

at $t=T$,

$\begin{array}{rcl}{\cos }^2\omega \left(T\right)& =& {\cos }^2\omega \left(\frac{2\mathrm{\pi }}{\omega }\right)\left[\because T=\frac{2\mathrm{\pi }}{\omega }\right]\\ & =& {\cos }^2\left(2\mathrm{\pi }\right)\\ & =& 1\left[\because \cos 2\mathrm{\pi }=1\right]\end{array}$

$\Rightarrow$ $K.E.$ is maximum

Hence, option $\left(D\right)$ is correct.