The graph shows variation of displacement of a particle performing...

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

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

At $t = \dfrac{3T}{4}$, 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 $t = \dfrac{T}{2}$, displacement $x = 0$ (mean position). Velocity is maximum at the mean position, not zero.

At $t = \dfrac{T}{4}$, 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 $\dfrac{3T}{4}$.