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 = 3T/4
(B) The acceleration is maximum at t = T
(C) The speed is maximum at t = T/4
(D) The P.E. is equal to K.E. of the oscillation at t = T/2

Solution

Physics sir · 3 steps

Read x off the graph at each instant; force, acceleration, speed and energy all follow from it.

1. Read the graph. The motion repeats after T, so the value at t = 0 equals the peak at t = T. The particle starts at x = +A, so x = A cos ω t. It passes x = 0 at T/4, reaches x = -A at T/2, passes x = 0 again at 3T/4 and is back at x = +A at T.
2. Link to x. In SHM, F = -kx and a = -ω² x, so both are zero at x = 0 and largest at x = ± A. Speed is largest at x = 0, where all the energy is kinetic. At x = ± A the particle stops, so all the energy is potential.

   Statement	Time	Displacement	Verdict
   (A) Force is zero	3T/4	x = 0	True, F = -kx = 0
   (B) Acceleration is maximum	T	x = +A	True, |a| = ω² A
   (C) Speed is maximum	T/4	x = 0	True, v = ω A
   (D) P.E. equals K.E.	T/2	x = -A	False, K.E. is zero and P.E. is maximum

3. So. Statements (A), (B) and (C) are true and (D) is false, which is option (b).

The idea to keep. In SHM, find x first. At x = 0: force and acceleration zero, speed maximum. At x = ± A: speed zero, force and acceleration maximum. P.E. equals K.E. only at x = ± A/√2.

Answer from the NTA final answer key. Solution by PrepWiser.