Question

The potential energy of a particle performing linear S.H.M is \( 0.1 \pi^2 x^2 \) joules. If the mass of the particle is \( 20 \, \mathrm{g} \), find the frequency of S.H.M:

• A. \( 0.4 \, \mathrm{Hz} \)
• B. \( 0.6 \, \mathrm{Hz} \)
• C. \( 1.581 \, \mathrm{Hz} \)
• D. \( 2.0 \, \mathrm{Hz} \)

Hint:

N/A

Answer 1

The Correct Option is C

Step 1: Start with the formula for potential energy in S.H.M.
The potential energy \( PE \) is given by: \[ PE = \frac{1}{2} m \omega^2 x^2, \] where: \( m \) is the mass of the particle, \( \omega \) is the angular frequency, \( x \) is the displacement. From the question, the potential energy is given as: \[ PE = 0.1 \pi^2 x^2. \] Equating: \[ \frac{1}{2} m \omega^2 x^2 = 0.1 \pi^2 x^2. \] Cancel \( x^2 \) from both sides: \[ \frac{1}{2} m \omega^2 = 0.1 \pi^2. \] Substitute \( m = 20 \, \mathrm{g} = 0.02 \, \mathrm{kg} \): \[ \frac{1}{2} (0.02) \omega^2 = 0.1 \pi^2. \] Simplify: \[ 0.01 \omega^2 = 0.1 \pi^2. \] \[ \omega^2 = 10 \pi^2. \] Step 2: Solve for angular frequency and frequency.
Take the square root: \[ \omega = \sqrt{10 \pi^2} = \pi \sqrt{10}. \] The frequency \( f \) is given by: \[ f = \frac{\omega}{2 \pi}. \] Substitute \( \omega = \pi \sqrt{10} \): \[ f = \frac{\pi \sqrt{10}}{2 \pi} = \frac{\sqrt{10}}{2}. \] Using \( \sqrt{10} \approx 3.162 \): \[ f = \frac{3.162}{2} = 1.581 \, \mathrm{Hz}. \] Thus, the frequency of S.H.M is \( \mathbf{1.581 \, \mathrm{Hz}} \).