Question

A particle of mass ' \(m\) ' is executing S.H.M. about the origin on \(x\)-axis with frequency \(\sqrt{\frac{ka}{\pi m}}\), where ' \(k\) ' is a constant and ' \(a\) ' is the amplitude of S.H.M. If ' \(x\) ' is a displacement of a particle, at time ' \(t\) ', potential energy of the particle will be

• A. \(\frac{1}{2} kax^2\)
• B. \(\pi kax^2\)
• C. \(2\pi kax^2\)
• D. \(2 kax^2\)

Hint:

SHM: $U = \frac{1}{2} m\omega^2 x^2$

Answer 1

The Correct Option is C

Concept: \[ \omega = 2\pi f \quad \text{and} \quad U = \frac{1}{2} m \omega^2 x^2 \]

Step 1: Given frequency. \[ f = \sqrt{\frac{ka}{\pi m}} \] \[ \omega = 2\pi f = 2\pi \sqrt{\frac{ka}{\pi m}} \]

Step 2: Square $\omega$. \[ \omega^2 = 4\pi^2 \cdot \frac{ka}{\pi m} = \frac{4\pi ka}{m} \]

Step 3: Substitute in PE. \[ U = \frac{1}{2} m \cdot \frac{4\pi ka}{m} x^2 = 2\pi kax^2 \]

Step 4: Conclusion.
$U = 2\pi kax^2$ Final Answer: Option (C)