A body of mass 5 kg hangs from a spring and oscillates with a time period of 2 Pi seconds. If the ball is removed, the length of the spring will decrease by

Physics Problem: Decrease in Length of a Spring

A body of mass \( 5 \, \text{kg} \) hangs from a spring and oscillates with a time period of \( 2\pi \) seconds. If the ball is removed, the length of the spring will decrease by:

• (a) \( \frac{g}{k} \) metres
• (b) \( \frac{k}{g} \) metres
• (c) \( 2\pi \) metres
• (d) \( g \) metres

Detailed Solution

The time period \( T \) of a mass-spring system is given by:

\[ T = 2\pi \sqrt{\frac{m}{k}} \]

where \( m \) is the mass and \( k \) is the spring constant.

Given \( T = 2\pi \) seconds and \( m = 5 \, \text{kg} \), we can solve for \( k \):

\[ 2\pi = 2\pi \sqrt{\frac{5}{k}} \]

\[ \sqrt{\frac{5}{k}} = 1 \]

\[ \frac{5}{k} = 1 \implies k = 5 \, \text{N/m} \]

When the mass is removed, the spring returns to its natural length. The extension \( x \) of the spring due to the mass is given by Hooke's Law:

\[ mg = kx \]

\[ x = \frac{mg}{k} \]

Substituting \( m = 5 \, \text{kg} \), \( g = 9.8 \, \text{m/s}^2 \), and \( k = 5 \, \text{N/m} \):

\[ x = \frac{5 \times 9.8}{5} = 9.8 \, \text{metres} = g \]

Answer: (d)