Question

Consider a particle undergoing an SHM of amplitude A & angular frequency w. What is the magnitude of displacement from the mean position when kinetic energy is equal to the magnitude of potential energy?

(a) A/2

(b) A/√2

(c) A/3

(d) A/4

Answer 1

Right choice is (b) A/√2

To explain: Let x = Asin(wt).

∴ Kinetic energy = 1/2 mA^2w^2cos^2(wt)

potential energy = 1/2kA^2sin^2(wt).

According to given condition:    

1/2mA^2w^2cos^2(wt) = 1/2A^2w^2sin^2(wt)                                                                                                                                     using, k = mw^2, we get:                                                                                                                                                                              tan^2(wt) = 1.

∴ wt = π/4.                                                                                                                                                                              ∴ x = Asin(π/4) = A/√2.