Question

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

• A. \(\frac{1}{2}Kax^2\)
• B. \(πKax^2\)
• C. \(2πKax^2\)
• D. \(2Kax^2\)

Hint:

Find omega from the frequency, then use U = (1/2) m omega^2 x^2.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
The potential energy of a particle in SHM at displacement \(x\) is \(U = \dfrac12m\omega^2x^2\). We are given the frequency \(f\), so we first get \(\omega = 2\pi f\).

Step 2: Key Formula or Approach:
\[ \omega = 2\pi f,\qquad U = \frac12m\omega^2x^2 \]

Step 3: Detailed Explanation:
\[ f = \sqrt{\frac{Ka}{\pi m}} \Rightarrow \omega = 2\pi\sqrt{\frac{Ka}{\pi m}} \]
\[ \omega^2 = 4\pi^2\cdot\frac{Ka}{\pi m} = \frac{4\pi Ka}{m} \]
Substitute:
\[ U = \frac12m\cdot\frac{4\pi Ka}{m}\,x^2 = 2\pi Ka\,x^2 \]
Mass cancels out. Option (A) \(\tfrac12Kax^2\) forgets the factor \(4\pi\). Option (B) \(\pi Kax^2\) has half the right factor, which appears if \(\omega^2\) is taken as \(2\pi Ka/m\). Option (D) lacks the factor \(\pi\).

Final Answer:
The potential energy is \(2\pi Ka\,x^2\), option (C). \[ \boxed{2\pi Kax^2 \text{ (C)}} \]