Question

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 \(t = \frac{3T}{4}\)
(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 \(t = \frac{T}{2}\)

• A. (B), (C) and (D)
• B. (A), (B) and (C)
• C. (A) and (D)
• D. (A), (B) and (D)

Hint:

The graph starts at the positive maximum, so \(x = A\cos\omega t\). Evaluate displacement at \(3T/4\), \(T\), \(T/4\) and \(T/2\); force and acceleration vanish at \(x=0\) and are maximum at \(x=\pm A\).

Answer 1

The Correct Option is B

From the graph, displacement is maximum (positive) at \(t=0\), zero at \(T/4\), maximum negative at \(T/2\), zero at \(3T/4\), and maximum positive again at \(T\). So \(x = A\cos\omega t\).
(A) At \(t=3T/4\), \(x=0\), so force \(F=-kx = 0\). True.
(B) At \(t=T\), \(x=+A\), so \(|a| = \omega^2 A\) is maximum. True.
(C) At \(t=T/4\), \(x=0\), the particle is at the mean position where speed is maximum. True.
(D) At \(t=T/2\), \(x=-A\): K.E. = 0 and P.E. is maximum, so they are not equal. False.
True: (A), (B) and (C).
Answer: (2) (A), (B) and (C)