The displacement-time graph of a particle executing SHM is given in the figure (sketch is schematic and not to scale). Which of the following statements is/are true for this motion?
(A) The force is zero at $t = \dfrac{3T}{4}$
(B) The magnitude of acceleration is maximum at $t = T$
(C) The speed is maximum at $t = \dfrac{T}{4}$
(D) The P.E. is equal to K.E. of the oscillation at $t = \dfrac{T}{2}$
Answer: (A) (A), (B) and (C)
From the graph the displacement is zero at $T/4$, $3T/4$ and $5T/4$ and is extreme at $T/2$ (minimum) and $T$ (maximum).
(A) At $3T/4$, $x = 0$, so $F = -kx = 0$. True.
(B) At $T$, $|x|$ is maximum, so $|a| = \omega^{2}|x|$ is maximum. True.
(C) At $T/4$, $x = 0$, so the speed is maximum. True.
(D) At $T/2$ the particle is at an extreme: KE $= 0$, PE is maximum. False.
Solution by Sreeraj P, M.Sc Physics