A 0.02 kg weight produces an extension of 0.02 m in a vertical spring. A mass of 0.1kg is suspended at its bottam and is left after putting down. What is the period of oscillation ?

Explanation

To solve the problem, we will follow these steps:

### Step 1: Identify the given data
We have the following information:
- Mass \( m_1 = 0.02 \, \text{kg} \) produces an extension \( y = 0.02 \, \text{m} \).
- New mass \( m_2 = 0.1 \, \text{kg} \) is suspended from the spring.

### Step 2: Calculate the spring constant \( k \)
The extension produced by a mass hanging from a spring is given by Hooke's Law:
\[ F = k \cdot x \]
where \( F \) is the force applied (weight of the mass), \( k \) is the spring constant, and \( x \) is the extension of the spring.

The force due to the weight \( m_1 \) is:
\[ F = m_1 \cdot g = 0.02 \, \text{kg} \cdot 9.8 \, \text{m/s}^2 = 0.196 \, \text{N} \]

Now, using the extension \( y = 0.02 \, \text{m} \):
\[ k = \frac{F}{y} = \frac{0.196 \, \text{N}}{0.02 \, \text{m}} = 9.8 \, \text{N/m} \]

### Step 3: Calculate the period of oscillation for the new mass
The formula for the period of oscillation \( T \) of a mass \( m \) attached to a spring with spring constant \( k \) is given by:
\[ T = 2\pi \sqrt{\frac{m}{k}} \]

Substituting \( m_2 = 0.1 \, \text{kg} \) and \( k = 9.8 \, \text{N/m} \):
\[ T = 2\pi \sqrt{\frac{0.1 \, \text{kg}}{9.8 \, \text{N/m}}} \]

Calculating the square root:
\[ \sqrt{\frac{0.1}{9.8}} \approx \sqrt{0.010204} \approx 0.101 \, \text{s} \]

Now substituting this value back into the period formula:
\[ T \approx 2\pi \cdot 0.101 \approx 0.634 \, \text{s} \]

### Final Answer
The period of oscillation \( T \) is approximately \( 0.63 \, \text{s} \).

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