Question: 25. 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
Answer: Option (2)
Explanation:
In simple harmonic motion, the displacement of the body is given by
x = A\sin(\omega t), where \omega = 2\pi n is the angular frequency.
The potential energy of a body executing SHM is given by U = \frac{1}{2}kx^{2}.
Substituting the expression for displacement, U = \frac{1}{2}kA^{2}\sin^{2}(\omega t).
Using the identity \sin^{2}(\omega t) = \frac{1}{2}[1 - \cos(2\omega t)],
the potential energy becomes
U = \frac{1}{4}kA^{2}[1 - \cos(2\omega t)].
This shows that the potential energy varies with angular frequency 2\omega.
Since the frequency f is related to angular frequency by
f = \frac{\omega}{2\pi}, the frequency of variation of potential energy is
f_{U} = \frac{2\omega}{2\pi} = 2n.
Therefore, the frequency of the potential energy is twice the frequency of the SHM,
and the correct answer is Option (2).