Question

A particle executes simple harmonic motion with a frequency v. The frequency with which the kinetic energy oscillates is

(a) v/2

(b) v

(c) 2 v

(d) zero.

Answer 1

The correct answer is

(c) 2 v

EXPLANATION: 

Between one extreme position to another i.e. in half the time period the K.E. increases from zero to maximum (at the mean position) and again to zero thus completing one cycle. Therefore the time period of oscillation of the K.E. = T/2 where T is the time period of the SHM. Since the frequency of SHM ν =1/T, the frequency of oscillation of the K.E. ν'=1/(T/2)=2*(1/T)=2ν. Hence the option (c).