The displacement time graph of a particle executing S.H.M. is given in 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 $t = \frac{3 T}{4}$
(B) The acceleration is maximum at t = T
(C) The speed is maximum at $t = \frac{T}{4}$
(D) The P.E. is equal to K.E. of the oscillation at $t = \frac{T}{2}$

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(A), (B) and (C)

(A), (B) and (D)

(B), (C) and (D)

(A) and (D)

answer is B.

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Detailed Solution

In SHM $a = - ω^{2} x$
So when x is max $\Rightarrow$ acceleration is maximum
When x is zero $\Rightarrow$ acceleration is zero $\Rightarrow$ Force is zero & velocity is max.
PE & KE are equal when $x = \pm \frac{A}{\sqrt{2}}$

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