Question

A block of mass 3 kg executes simple harmonic motion under the restoring force of a spring. The amplitude and the time period of the motion are 0.1 m and 3.14 s respectively. The maximum force exerted by the spring on the block is

• A. 1.2 N
• B. 3 N
• C. 12 N
• D. 30 N
• E. 90 N

Hint:

For a spring in simple harmonic motion, the maximum force is related to the spring constant and the amplitude. Use the formula \(F_{{max}} = k \cdot A\).

Answer 1

The Correct Option is A

Step 1: Use the formula for maximum force in simple harmonic motion. The maximum force exerted by the spring is given by the formula: \[ F_{{max}} = kA \] where \( k \) is the spring constant and \( A \) is the amplitude. 
Step 2: Relate the spring constant \( k \) to the time period \( T \). The time period \( T \) of a block undergoing simple harmonic motion is related to the spring constant \( k \) and the mass \( m \) by the formula: \[ T = 2\pi \sqrt{\frac{m}{k}} \] Rearranging for \( k \): \[ k = \frac{4\pi^2 m}{T^2} \] Substitute the given values \( m = 3 \, {kg} \) and \( T = 3.14 \, {s} \): \[ k = \frac{4\pi^2 (3)}{(3.14)^2} \approx 12 \, {N/m} \] 
Step 3: Calculate the maximum force. Now that we know \( k = 12 \, {N/m} \) and \( A = 0.1 \, {m} \), we can calculate the maximum force: \[ F_{{max}} = kA = 12 \times 0.1 = 1.2 \, {N} \] Thus, the maximum force exerted by the spring on the block is \( 1.2 \, {N} \).