Q.The graph shows the variation of displacement of a particle performing S.H.M. with time t, starting from the mean position (x = 0 at t = 0) and rising first towards the positive extreme, i.e. x = A sin(2\pi t/T), so that the particle is at the mean position at t = 0, T/2, T, ... and at an extreme position at t = T/4, 3T/4, ... . Which of the following statements is correct from the graph? (A) The acceleration is maximum at time T.
(B) The force is maximum at time 3T/4.
(C) The velocity is zero at time T/2.
(D) The kinetic energy is equal to total energy at time T/4.

[!TLDR] Starting from the mean position (x=0 at t=0): at t=T, x=0 so force=0 (not max); at T/2, x=0 so velocity is max (not zero); at T/4, x=A so KE=0 not equal to total E. At $3T/4$, $x=-A$ (an extreme), so th …

With $x=A\sin(2\pi t/T)$ (starting at the mean position, moving positive), check each option in turn. (A) At $t=T$, $x=A\sin(2\pi)=0$ (mean position), so acceleration $a=-\omega^2x=0$, not maximum -- FALSE. (B) At $t=3T/4$, $x=A\sin(3\pi/2)=-A$ (the negative extreme), and since $|F|=k|x|$ is maximum at the extremes, the force IS maximum in magnitude there -- TRUE. (C) At $t=T/2$, $x=A\sin(\pi)=0$ (mean position), where velocity is actually MAXIMUM, not zero -- FALSE. (D) At $t=T/4$, $x=A\sin(\pi/2)=A$ (an extreme), w …

Evaluate $x=A\sin(2\pi t/T)$ at each named instant and use it to check the corresponding ve …

Assuming t=T/2 must be an extreme position (by analogy with t=T/4) rather than correctly checking it is act …