Question

A spring is stretched by \( 0.2 \, \text{m} \) when a mass of \( 0.5 \, \text{kg} \) is suspended to it. The time period of the spring when \( 0.5 \, \text{kg} \) mass is replaced with a mass of \( 0.25 \, \text{kg} \) is (Acceleration due to gravity \( = 10 \, \text{m/s}^2 \))

• A. \( 0.628 \, \text{s} \)
• B. \( 6.28 \, \text{s} \)
• C. \( 62.8 \, \text{s} \)
• D. \( 0.0628 \, \text{s} \)

Hint:

For a mass-spring system, the key is to first determine the spring constant (\( k \)) using the given initial conditions (e.g., equilibrium extension due to a known mass: \( kx = mg \)). Once \( k \) is known, it remains constant for that specific spring. Then, the time period of oscillation (\( T \)) for any mass (\( m \)) attached to this spring can be found using the formula \( T = 2\pi\sqrt{\frac{m}{k}} \). Always ensure units are consistent (SI units are generally preferred).

Answer 1

The Correct Option is A

Step 1: Determine the spring constant (k) of the spring.
When a mass is suspended from a spring, it stretches by a certain amount due to the gravitational force. At equilibrium, the spring force balances the gravitational force.
Spring force \( F_s = kx \), where \( k \) is the spring constant and \( x \) is the extension.
Gravitational force \( F_g = mg \).
At equilibrium, \( kx = mg \).
Given:
Mass \( m_1 = 0.5 \, \text{kg} \)
Extension \( x = 0.2 \, \text{m} \)
Acceleration due to gravity \( g = 10 \, \text{m/s}^2 \)
So, \( k \times (0.2 \, \text{m}) = (0.5 \, \text{kg}) \times (10 \, \text{m/s}^2) \)
\( 0.2k = 5 \)
\( k = \frac{5}{0.2} = \frac{50}{2} = 25 \, \text{N/m} \).
Step 2: Calculate the time period of oscillation for the new mass.
The time period \( T \) of a mass-spring system in simple harmonic motion is given by the formula:
\( T = 2\pi\sqrt{\frac{m}{k}} \)
Now, the \( 0.5 \, \text{kg} \) mass is replaced with a new mass \( m_2 = 0.25 \, \text{kg} \).
Using the calculated spring constant \( k = 25 \, \text{N/m} \):
\( T = 2\pi\sqrt{\frac{0.25 \, \text{kg}}{25 \, \text{N/m}}} \)
\( T = 2\pi\sqrt{\frac{1}{100}} \)
\( T = 2\pi \cdot \frac{1}{10} \)
\( T = \frac{2\pi}{10} = \frac{\pi}{5} \)
Step 3: Approximate the value of the time period.
Using the approximation \( \pi \approx 3.14 \):
\( T = \frac{3.14}{5} = 0.628 \, \text{s} \).