A body of mass m is tied to one end of a spring and whirled round in a horizontal circle with a constant angular velocity .The elongation is 1 cm . If the angular velocity is doubled , the elongation in the spring is 5 cm. what is the original length of the spring ?

Options

A. 13 cm
B. 14 cm
C. 15 cm
D. 16 cm

Explanation

To solve the problem, we need to analyze the forces acting on the mass tied to the spring when it is whirled in a horizontal circle. We will set up equations based on the elongation of the spring and the forces involved.

### Step-by-Step Solution:

1. **Identify the Variables:**
- Let the original length of the spring be \( L_0 \).
- The elongation when the angular velocity is \( \omega \) is \( \Delta L_1 = 1 \, \text{cm} = 0.01 \, \text{m} \).
- The elongation when the angular velocity is doubled \( (2\omega) \) is \( \Delta L_2 = 5 \, \text{cm} = 0.05 \, \text{m} \).

2. **Determine the Total Lengths:**
- When the angular velocity is \( \omega \):
\[ r_1 = L_0 + \Delta L_1 = L_0 + 0.01 \]
- When the angular velocity is \( 2\omega \):
\[ r_2 = L_0 + \Delta L_2 = L_0 + 0.05 \]

3. **Set Up the Force Equations:**
- The centrifugal force acting on the mass is given by:
\[ F_c = m r \omega^2 \]
- The spring force is given by Hooke's law:
\[ F_s = k \Delta L \]
- For the first case:
\[ m (L_0 + 0.01) \omega^2 = k (0.01) \]
- For the second case:
\[ m (L_0 + 0.05) (2\omega)^2 = k (0.05) \]

4. **Simplify the Equations:**
- From the first case:
\[ m (L_0 + 0.01) \omega^2 = k (0.01) \quad \text{(1)} \]
- From the second case:
\[ m (L_0 + 0.05) (4\omega^2) = k (0.05) \quad \text{(2)} \]

5. **Express \( k \) from Equation (1):**
- Rearranging equation (1):
\[ k = \frac{m (L_0 + 0.01) \omega^2}{0.01} \]

6. **Substitute \( k \) into Equation (2):**
- Substitute \( k \) from equation (1) into equation (2):
\[ m (L_0 + 0.05) (4\omega^2) = \frac{m (L_0 + 0.01) \omega^2}{0.01} (0.05) \]
- Cancel \( m \) and \( \omega^2 \) (assuming \( m \neq 0 \) and \( \omega \neq 0 \)):
\[ 4 (L_0 + 0.05) = \frac{(L_0 + 0.01)(0.05)}{0.01} \]
- Simplifying gives:
\[ 4 (L_0 + 0.05) = 5 (L_0 + 0.01) \]

7. **Expand and Rearrange:**
- Expanding both sides:
\[ 4L_0 + 0.2 = 5L_0 + 0.05 \]
- Rearranging gives:
\[ 5L_0 - 4L_0 = 0.2 - 0.05 \]
\[ L_0 = 0.15 \, \text{m} = 15 \, \text{cm} \]

### Final Answer:
The original length of the spring \( L_0 \) is **15 cm**.