A body is executing simple harmonic motion with frequency 'n', the frequency of its potential energy is:
(1) n (2) 2n
(3) 3n (4) 4n
Solution
In SHM, the potential energy is maximum at extreme positions and minimum at mean position. There are two extreme positions in one cycle and two mean positions. So, the time period should be halved or frequency should be doubled.
Alternate Solution
Potential Energy $U = \frac{1}{2}m{\omega ^2}{A^2}{\sin ^2}(\omega t + \phi )$
Or $U \propto {\sin ^2}(\omega t + \phi )$
Sin function alternates above and below x-axes whereas square of Sin function cannot be negative. So, the region below x-axis of Sin function is folded upwards above x-axes when it is squared. Thus the frequency is doubled or the time period is halved.
Answer: (2)