A spring is stretched by 5 cm by a force 10 N. The time period of the oscillations when a mass of 2 kg is suspended by it is

Physics Problem: Time Period of a Mass-Spring System

A spring is stretched by \( 5 \, \text{cm} \) by a force of \( 10 \, \text{N} \). The time period of the oscillations when a mass of \( 2 \, \text{kg} \) is suspended by it is:

• (a) \( 0.0628 \, \text{s} \)
• (b) \( 6.28 \, \text{s} \)
• (c) \( 3.14 \, \text{s} \)
• (d) \( 0.628 \, \text{s} \)

Detailed Solution

The spring constant \( k \) can be calculated using Hooke's Law:

\[ F = kx \]

where \( F = 10 \, \text{N} \) and \( x = 5 \, \text{cm} = 0.05 \, \text{m} \).

\[ k = \frac{F}{x} = \frac{10}{0.05} = 200 \, \text{N/m} \]

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

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

where \( m = 2 \, \text{kg} \) and \( k = 200 \, \text{N/m} \).

Substituting the values:

\[ T = 2\pi \sqrt{\frac{2}{200}} = 2\pi \sqrt{\frac{1}{100}} = 2\pi \times 0.1 = 0.2\pi \, \text{s} \]

Using \( \pi \approx 3.14 \):

\[ T \approx 0.2 \times 3.14 = 0.628 \, \text{s} \]

Thus, the time period of the oscillations is \( 0.628 \, \text{s} \).

Answer: (d) \( 0.628 \, \text{s} \)