Question

A rectangular block of mass 5 kg attached to an horizontal spiral spring executes simple harmonic motion of amplitude 1 m and time period 3.14s. The maximum force exerted by spring on block is _____N.

Answer 1

Correct Answer: 20

Step 1: Calculate the angular frequency \( \omega \). The angular frequency for simple harmonic motion is given by: \[ \omega = \frac{2\pi}{T} \] Substituting the time period \( T = 3.14 \, \text{s} \): \[ \omega = \frac{2 \times \frac{22}{7}}{3.14} = 2 \, \text{rad/s} \] Step 2: Calculate the maximum acceleration \( a_{\text{max}} \). The maximum acceleration in simple harmonic motion is given by: \[ a_{\text{max}} = \omega^2 A \] Substituting \( \omega = 2 \, \text{rad/s} \) and \( A = 1 \, \text{m} \): \[ a_{\text{max}} = 2^2 \times 1 = 4 \, \text{m/s}^2 \] Step 3: Calculate the maximum force \( F_{\text{max}} \). The maximum force is given by: \[ F_{\text{max}} = m a_{\text{max}} \] Substituting \( m = 5 \, \text{kg} \) and \( a_{\text{max}} = 4 \, \text{m/s}^2 \): \[ F_{\text{max}} = 5 \times 4 = 20 \, \text{N} \] Thus, the maximum force exerted by the spring is \( 20 \, \text{N} \).