For a particle performing S.H.M. the displacement - time graph is shown.
For that particle the force - time graph is correctly shown in graph

For a particle in simple harmonic motion with displacement $y(t) = A \cos(\omega t)$ starting from maximum positive displacement at $t=0$, the force can be derived from the relationship between acceleration and displacement.

The restoring force is given by Newton's second law using the standard S.H.M. acceleration expression $a = -\omega^2 y$. Substituting yields $F = m(-\omega^2 y) = -m\omega^2 y$.

Inserting the displacement function gives $F(t) = -m\omega^2 A \cos(\omega t)$. Letting $F_\text{max} = m\omega^2 A$, this simplifies to $F(t) = -F_\text{max} \cos(\omega t)$.

This describes a negative cosine function, which starts at its minimum value $-F_\text{max}$ at $t=0$. Among the options, graph (d) matches this behavior precisely, as it is sinusoidal and begins at a negative maximum.