Question

A particle of mass 4 mg is executing simple harmonic motion along x-axis with an angular frequency of 40 rad/s. If the potential energy of the particle is \( V(x) = a + bx^2 \), where \( V(x) \) is in joule and \( x \) is in meter, then the value of \( b \) is:

• A. \( 800 \times 10^{-6} \) J/m\(^2\)
• B. \( 1600 \times 10^{-6} \) J/m\(^2\)
• C. \( 3200 \times 10^{-6} \) J/m\(^2\)
• D. \( 6400 \times 10^{-6} \) J/m\(^2\)

Hint:

The potential energy function in SHM is \( V(x) = \frac{1}{2} k x^2 \). - Use \( \omega^2 = k/m \) to determine \( k \).

Answer 1

The Correct Option is C

Step 1: Use SHM potential energy formula
Potential energy for SHM is: \[ V(x) = \frac{1}{2} k x^2. \] Given \( V(x) = a + bx^2 \), comparing with the standard equation, we get: \[ b = \frac{1}{2} k. \] Step 2: Compute \( k \) using angular frequency
The angular frequency is: \[ \omega = \sqrt{\frac{k}{m}}. \] Squaring both sides: \[ k = m \omega^2. \] Substituting \( m = 4 \times 10^{-6} \) kg and \( \omega = 40 \) rad/s, \[ k = (4 \times 10^{-6}) (40)^2. \] \[ k = 6.4 \times 10^{-3} \text{ N/m}. \] Step 3: Compute \( b \)
\[ b = \frac{1}{2} \times 6.4 \times 10^{-3} = 3.2 \times 10^{-3} = 3200 \times 10^{-6} \text{ J/m}^2. \] Thus, the correct answer is \( 3200 \times 10^{-6} \) J/m\(^2\).