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

In simple harmonic motion (SHM), the potential energy (PE) of a body oscillating at a frequency depends on the displacement from its equilibrium position. The potential energy function is given by:

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

where \(k\) is the spring constant and \(x\) is the displacement.  

The displacement \( x \) varies as:

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

where:

• \( A \) is the amplitude,
• \( \omega = 2\pi n \) is the angular frequency,
• \( \phi \) is the phase constant.

The potential energy can be expressed as:

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

This can be rewritten in terms of trigonometric identities:

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

Thus, the expression becomes:

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

The term \(\cos(2\omega t + 2\phi)\) suggests that the potential energy oscillates with a frequency of \(2\omega\).

Therefore, since \(\omega = 2\pi n\), the frequency of the potential energy is \(2n\).

Hence, the correct answer is 2n.