Question

The mass of a particle is \( 1 \) kg and it is moving along the \( x \)-axis. The period of its oscillation is \( \frac{\pi}{2} \). Its potential energy at a displacement of \( 0.2 \) m is:

• A. \( 0.24 \) J
• B. \( 0.48 \) J
• C. \( 0.32 \) J
• D. \( 0.16 \) J

Hint:

The potential energy in simple harmonic motion is given by \( U = \frac{1}{2} k x^2 \). The spring constant \( k \) can be found using \( \omega = \sqrt{\frac{k}{m}} \).

Answer 1

The Correct Option is C

Step 1: Understanding potential energy in simple harmonic motion
The potential energy in simple harmonic motion (SHM) is given by: \[ U = \frac{1}{2} k x^2. \] where: - \( k \) is the force constant (spring constant),
- \( x \) is the displacement from equilibrium.
Step 2: Finding the spring constant \( k \)
The angular frequency \( \omega \) is related to the period \( T \) by: \[ \omega = \frac{2\pi}{T}. \] Given that \( T = \frac{\pi}{2} \), we find: \[ \omega = \frac{2\pi}{\frac{\pi}{2}} = 4. \] Since \( \omega = \sqrt{\frac{k}{m}} \), we substitute \( m = 1 \) kg: \[ 4 = \sqrt{k}. \] Squaring both sides: \[ k = 16. \] Step 3: Calculating the Potential Energy
Using \( k = 16 \) and \( x = 0.2 \) m: \[ U = \frac{1}{2} \times 16 \times (0.2)^2. \] \[ U = 8 \times 0.04. \] \[ U = 0.32 { J}. \] Step 4: Conclusion
Thus, the potential energy at a displacement of \( 0.2 \) m is: \[ 0.32 { J}. \]