Question

A body is executing simple harmonic motion with frequency ‘n’, the frequency of its potential energy is

• A. 4n
• B. n
• C. 2n
• D. 3n

Answer 1

The Correct Option is C

To solve this problem, we need to understand how the frequency of potential energy relates to the frequency of an object in simple harmonic motion (SHM).

In SHM, the displacement \( x \) of the body can be expressed as:

\(x = A \cos(\omega t + \phi)\)

where:

• \( A \) is the amplitude
• \( \omega \) is the angular frequency, which is related to the frequency \( n \) by the equation \( \omega = 2\pi n \)
• \( t \) is time
• \( \phi \) is the phase constant

The potential energy \( U \) in SHM is given by:

\(U = \frac{1}{2} k x^2\)

where \( k \) is the spring constant.

Substituting the expression for \( x \) in the potential energy formula, we get:

\(U = \frac{1}{2} k (A \cos(\omega t + \phi))^2\)

\(U = \frac{1}{2} k A^2 \cos^2(\omega t + \phi)\)

Recall the trigonometric identity:

\(\cos^2(\theta) = \frac{1 + \cos(2\theta)}{2}\)

Using this identity, the potential energy becomes:

\(U = \frac{1}{2} k A^2 \cdot \frac{1 + \cos(2\omega t + 2\phi)}{2}\)

Simplifying:

\(U = \frac{1}{4} k A^2 + \frac{1}{4} k A^2 \cos(2\omega t + 2\phi)\)

This shows that the potential energy oscillates with a frequency that is twice the frequency of the displacement.

Therefore, if the frequency of the SHM is \( n \), then the frequency of the potential energy is \( 2n \).

Hence, the correct answer is 2n.