Question

A body of mass \( 4 \) kg attached to a spring of force constant \( 64 \) N/m executes simple harmonic motion on a frictionless horizontal surface. The time period of oscillation is:

• A. \( \frac{\pi}{3} \) s
• B. \( \frac{\pi}{2} \) s
• C. \( \pi \) s
• D. \( \frac{3\pi}{2} \) s

Hint:

For simple harmonic motion, the time period is computed using: \[ T = 2\pi \sqrt{\frac{m}{k}} \] where \( k \) is the force constant and \( m \) is the mass.

Answer 1

The Correct Option is B

Step 1: Formula for Time Period in SHM The time period \( T \) of a simple harmonic oscillator is given by: \[ T = 2\pi \sqrt{\frac{m}{k}} \] where: - \( m = 4 \) kg (mass of the body), - \( k = 64 \) N/m (force constant of the spring). Step 2: Substituting Values \[ T = 2\pi \sqrt{\frac{4}{64}} \] \[ = 2\pi \sqrt{\frac{1}{16}} \] \[ = 2\pi \times \frac{1}{4} \] \[ = \frac{\pi}{2} \] Conclusion Thus, the correct answer is: \[ \frac{\pi}{2} \text{ s} \]