1. Write the SHM displacement using the given initial condition

Since the particle is at the equilibrium position at $t=0$, we can write its displacement as

$x = A\sin(\omega t)$

where

$\omega = \frac{2\pi}{T}.$

This satisfies $x=0$ at $t=0$.

2. Find the velocity

Velocity is

$v = \frac{dx}{dt} = A\omega \cos(\omega t).$

3. Write the kinetic energy

Kinetic energy is

$K = \frac{1}{2}mv^2 = \frac{1}{2}mA^2\omega^2\cos^2(\omega t).$

Let

$K_{\max} = \frac{1}{2}mA^2\omega^2.$

Then

$K(t) = K_{\max}\cos^2(\omega t).$

Using $\omega = \frac{2\pi}{T}$,

$K(t) = K_{\max}\cos^2\left(\frac{2\pi t}{T}\right).$

4. Understand the shape of the graph

The function $\cos^2(\omega t)$:

• is maximum at $t=0$, because $\cos^2 0 = 1$
• becomes zero at $\omega t = \frac{\pi}{2}, \frac{3\pi}{2}, \dots$ so $t = \frac{T}{4}, \frac{3T}{4}, \dots$
• again becomes maximum at $t = \frac{T}{2}, T, \dots$

So the kinetic-energy graph:

• starts from a maximum value at $t=0$
• remains always non-negative
• has period $\frac{T}{2}$ because $\cos^2(\omega t)$ repeats after phase change $\pi$.

5. Match with the correct option

Therefore the correct graph must be a $\cos^2$-type curve starting from maximum at $t=0$.

Hence, the correct option is A.