an object oscillating in simple harmonic motion has a time period t. the first graph shows how its…

Explanation:

Step1: Recall kinetic - energy formula for SHM

The displacement of an object in simple harmonic motion is given by $x = A\sin(\omega t+\varphi)$. The velocity $v=\frac{dx}{dt}=A\omega\cos(\omega t + \varphi)$. The kinetic energy $E_{k}=\frac{1}{2}mv^{2}=\frac{1}{2}m(A\omega)^{2}\cos^{2}(\omega t+\varphi)$.

Step2: Analyze the properties of $\cos^{2}(\omega t+\varphi)$

We know that $\cos^{2}(\omega t+\varphi)=\frac{1 + \cos(2(\omega t+\varphi))}{2}$. The period of $\cos^{2}(\omega t+\varphi)$ is $\frac{T}{2}$ (since the period of $\cos(2\omega t)$ is $\frac{2\pi}{2\omega}=\frac{T}{2}$, where $T = \frac{2\pi}{\omega}$). When the displacement $x = 0$, the velocity is maximum and the kinetic energy is maximum. When the displacement $x=\pm A$, the velocity is $0$ and the kinetic energy is $0$.

Step3: Match with the graphs

We need a graph that has a period of $\frac{T}{2}$, is zero at the points where the displacement is maximum (amplitude) and maximum at the points where the displacement is zero.

Answer:

A. Option Text (assuming graph A has the correct period and shape as described above)