Read the displacement-time graph qualitatively

From the standard schematic for SHM used in such questions, the particle starts from mean position at $t=0$ with positive displacement increasing, so we can take $x = A\sin\omega t$ where the time period is $T = \frac{2\pi}{\omega}.$

Find displacement at the given instants

Using $x=A\sin\omega t$:

• At $t=\dfrac{T}{4}$, $\omega t = \omega \frac{T}{4} = \frac{2\pi}{T}\cdot \frac{T}{4}=\frac{\pi}{2}$ so $x = A\sin\frac{\pi}{2}=A$ i.e. the particle is at an extreme position.

• At $t=\dfrac{T}{2}$, $\omega t=\pi$ so $x=A\sin\pi=0$ i.e. the particle is at mean position.

• At $t=\dfrac{3T}{4}$, $\omega t=\frac{3\pi}{2}$ so $x=A\sin\frac{3\pi}{2}=-A$ i.e. again an extreme position.

• At $t=T$, $\omega t=2\pi$ so $x=A\sin 2\pi=0$ i.e. mean position.

Use SHM relations

For SHM, $F=-kx=-m\omega^2 x$ $a=-\omega^2 x$ $v=\omega\sqrt{A^2-x^2}$

Also,

Check each statement

(A) The force is zero at $t=\dfrac{3T}{4}$

At $t=\dfrac{3T}{4}$, we found $x=-A$. Therefore, $F=-kx=-k(-A)=kA\neq 0.$ So (A) is false.

(B) The acceleration is maximum at $t=T$

At $t=T$, $x=0$, hence $a=-\omega^2 x = 0.$ Acceleration magnitude is maximum at extreme positions $x=\pm A$, not at mean position. So (B) is false.

(C) The speed is maximum at $t=\dfrac{T}{4}$

At $t=\dfrac{T}{4}$, $x=A$ (extreme position), so $v=\omega\sqrt{A^2-A^2}=0.$ Maximum speed occurs at mean position $x=0$, not at the extreme. So (C) is false.

(D) The P.E. is equal to K.E. at $t=\dfrac{T}{2}$

At $t=\dfrac{T}{2}$, $x=0$. Hence, $\text{P.E.}=\frac{1}{2}k(0)^2=0,$ $\text{K.E.}=\frac{1}{2}kA^2.$ They are not equal. So (D) is false.

Conclusion from the above interpretation

Under the usual interpretation $x=A\sin\omega t$, none of the four statements is true.

Why this conflicts with the options

The stored answer is option B, which corresponds to $(A)$, $(B)$ and $(C)$ being true. That is inconsistent with standard SHM if the graph is the usual sine-type displacement-time graph.

The only way option-based statements could differ is if the unseen figure has a different phase labeling than assumed, but even then the combination in option B is not compatible with standard SHM properties unless the time markings refer to different points than the natural quarter-period positions.

Final judgment

Based on standard SHM relations, the stored answer appears incorrect or the figure/options are mismatched.