Question

A mass of 0.5 kg is attached to a spring of force constant 200 N/m. What is the time period of oscillation?

• A. 0.25 s
• B. 0.50 s
• C. 1.00 s
• D. 2.00 s

Hint:

For a mass-spring system, the time period depends on the mass and the spring constant. The larger the mass, the greater the time period.

Answer 1

The Correct Option is A

The time period \( T \) of a mass-spring system is given by the formula: \[ T = 2\pi \sqrt{\frac{m}{k}} \] Where: - \( m \) is the mass, - \( k \) is the spring constant. Substituting the given values: - \( m = 0.5 \, \text{kg} \), - \( k = 200 \, \text{N/m} \). We get: \[ T = 2\pi \sqrt{\frac{0.5}{200}} = 2\pi \sqrt{0.0025} = 2\pi \times 0.05 \approx 0.25 \, \text{s} \] Thus, the time period of oscillation is 0.25 seconds.