The displacement time graph of a particle executing S.H.M is given in… | JEE Main 2020 (2 Sep, Shift 2)

Q62JEE Main 2020 · 2 SEP · SHIFT 2Medium

Velocity & Acceleration in SHM

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

(D) The P.E. is equal to K.E. of the oscillation at t = $\frac{T}{2}$

Show answer and solution

Answer: (b)

Graph starts at the positive extreme: $x = A\cos\omega t$ .
(A) $t = \frac{3T}{4}$ : $x = 0$ , so force $= -kx = 0$ . True.
(B) $t = T$ : $x = A$ (extreme), so acceleration is maximum. True.
(C) $t = \frac{T}{4}$ : $x = 0$ (mean position), so speed is maximum. True.
(D) $t = \frac{T}{2}$ : $x = -A$ , K.E. $= 0$ and P.E. is maximum, not equal. False.
So the answer is (b) (A), (B) and (C).

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