A mass suspended on a vertical spring oscillates with a period of `0.5s`. When the mass is allowed to hang at rest, the spring is stretched by

Options

A. `3.1cm`
B. `6.2cm`
C. `9.3cm`
D. `12.4cm`

Explanation

To solve the problem step by step, we will use the information provided about the oscillation of the mass on the spring and the relationship between the period of oscillation, mass, and spring constant.

### Step 1: Understand the relationship between period, mass, and spring constant
The period \( T \) of a mass-spring system in simple harmonic motion is given by the formula:
\[ T = 2\pi \sqrt{\frac{m}{k}} \]
where:
- \( T \) is the period of oscillation,
- \( m \) is the mass attached to the spring,
- \( k \) is the spring constant.

### Step 2: Rearrange the formula to find \( \frac{m}{k} \)
We can rearrange the formula to express \( \frac{m}{k} \):
\[ T^2 = 4\pi^2 \frac{m}{k} \]
Thus,
\[ \frac{m}{k} = \frac{T^2}{4\pi^2} \]

### Step 3: Substitute the known value of the period
Given that \( T = 0.5 \, \text{s} \), we can substitute this value into the equation:
\[ \frac{m}{k} = \frac{(0.5)^2}{4\pi^2} \]
Calculating \( (0.5)^2 \):
\[ (0.5)^2 = 0.25 \]
Now substituting this into the equation:
\[ \frac{m}{k} = \frac{0.25}{4\pi^2} \]

### Step 4: Calculate \( \frac{m}{k} \)
Using \( \pi \approx 3.14 \):
\[ \pi^2 \approx 9.86 \]
Thus,
\[ 4\pi^2 \approx 39.44 \]
Now substituting this value:
\[ \frac{m}{k} = \frac{0.25}{39.44} \approx 0.00635 \]

### Step 5: Relate the stretch of the spring to the weight of the mass
When the mass is at rest, the force due to gravity \( mg \) is balanced by the spring force \( k\delta \):
\[ k\delta = mg \]
From this, we can express \( \delta \):
\[ \delta = \frac{mg}{k} \]

### Step 6: Substitute \( m \) in terms of \( k \) and \( \delta \)
Using \( \frac{m}{k} \) from Step 4:
\[ \delta = \frac{g \cdot k \cdot \frac{m}{k}}{k} = g \cdot \frac{m}{k} \]
Substituting the value of \( \frac{m}{k} \):
\[ \delta = g \cdot 0.00635 \]

### Step 7: Use \( g \approx 9.81 \, \text{m/s}^2 \)
Now substituting the value of \( g \):
\[ \delta = 9.81 \cdot 0.00635 \approx 0.0625 \, \text{m} \]

### Step 8: Convert to centimeters
To convert meters to centimeters:
\[ \delta = 0.0625 \, \text{m} \times 100 = 6.25 \, \text{cm} \]

### Final Answer
The stretch in the spring when the mass is allowed to hang at rest is approximately:
\[ \delta \approx 6.25 \, \text{cm} \]