Physics - SHM Question with Solution | TestHub

From the given graph, the displacement of the particle is $x=A\sin (\omega t)$, where $\omega =\frac{2\pi }{T}$.

At $t=T$, displacement $x=0$. Acceleration $a=-{\omega }^{2}x=0$. Thus, acceleration is zero.

At $t=\frac{3T}{4}$, displacement $x=-A$ (extreme position). The restoring force is $F=-kx$. The magnitude of force is $|F|=k|-A|=kA$, which is the maximum value. Thus, the force is maximum.

At $t=\frac{T}{2}$, displacement $x=0$ (mean position). Velocity is maximum at the mean position, not zero.

At $t=\frac{T}{4}$, displacement $x=A$ (extreme position). Velocity is zero, so kinetic energy is zero. Total energy is entirely potential energy.

Answer: The force is maximum at time $\frac{3T}{4}$.