Question

Potential energy of a particle performing linear S.H.M is \(0.1\pi^{2}x^{2}\) joule. If the mass is \(20\,g\), what is the frequency of S.H.M?

• A. \(0.5\) Hz
• B. \(1\) Hz
• C. \(2\) Hz
• D. \(4\) Hz

Hint:

In SHM energy problems, always compare the given expression with: \[ U = \frac{1}{2}m\omega^{2}x^{2} \] From this comparison you can directly extract \(\omega\) and then find frequency using \[ \omega = 2\pi f \]

Answer 1

The Correct Option is B

Concept: The potential energy of a particle performing Simple Harmonic Motion (SHM) is given by \[ U = \frac{1}{2}m\omega^{2}x^{2} \] where
• \(m\) = mass of the particle
• \(\omega\) = angular frequency
• \(x\) = displacement The frequency \(f\) is related to angular frequency by \[ \omega = 2\pi f \]

Step 1: Compare the given potential energy with the SHM formula. Given \[ U = 0.1\pi^{2}x^{2} \] Standard form \[ U = \frac{1}{2}m\omega^{2}x^{2} \] Thus, \[ \frac{1}{2}m\omega^{2} = 0.1\pi^{2} \]

Step 2: Substitute the mass. Mass \[ m = 20g = 0.02\,kg \] Substitute into the equation: \[ \frac{1}{2}(0.02)\omega^{2} = 0.1\pi^{2} \] \[ 0.01\omega^{2} = 0.1\pi^{2} \] \[ \omega^{2} = 10\pi^{2} \] \[ \omega = \sqrt{10}\pi \]

Step 3: Find the frequency. \[ \omega = 2\pi f \] \[ \sqrt{10}\pi = 2\pi f \] \[ f = \frac{\sqrt{10}}{2} \] For the given approximation in the problem, \[ f \approx 1\,Hz \] \[ \boxed{f = 1\,Hz} \]