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

Options

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

Answer

To solve the problem, we need to analyze the forces acting on the mass tied to the spring during horizontal circular motion. We will use the concepts of centripetal force and Hooke's law.

### Step-by-Step Solution:

1. **Identify Variables**:
- Let \( l \) be the original length of the spring.
- The elongation when the angular velocity is \( \omega \) is given as \( x_1 = 1 \) cm.
- The elongation when the angular velocity is doubled (\( 2\omega \)) is \( x_2 = 5 \) cm.

2. **Centripetal Force and Spring Force**:
- When the spring is elongated by \( x_1 \), the effective radius \( r_1 \) becomes \( l + x_1 = l + 1 \) cm.
- The centripetal force required for circular motion is provided by the spring's restoring force.
- The centripetal force \( F_c \) is given by:
\[ F_c = m \cdot \frac{v^2}{r} = m \cdot (l + 1) \cdot \omega^2 \]
- The restoring force from the spring is given by Hooke's law:
\[ F_s = k \cdot x_1 = k \cdot 1 \]
- Setting these equal gives us the first equation:
\[ m \cdot (l + 1) \cdot \omega^2 = k \]

3. **Second Condition with Doubled Angular Velocity**:
- When the angular velocity is doubled, the effective radius becomes \( r_2 = l + x_2 = l + 5 \) cm.
- The centripetal force now becomes:
\[ F_c' = m \cdot (l + 5) \cdot (2\omega)^2 = m \cdot (l + 5) \cdot 4\omega^2 \]
- The restoring force is:
\[ F_s' = k \cdot x_2 = k \cdot 5 \]
- Setting these equal gives us the second equation:
\[ m \cdot (l + 5) \cdot 4\omega^2 = 5k \]

4. **Divide the Two Equations**:
- From the first equation:
\[ k = m \cdot (l + 1) \cdot \omega^2 \]
- Substitute \( k \) into the second equation:
\[ m \cdot (l + 5) \cdot 4\omega^2 = 5 \cdot m \cdot (l + 1) \cdot \omega^2 \]
- Cancel \( m \) and \( \omega^2 \) from both sides:
\[ 4(l + 5) = 5(l + 1) \]

5. **Solve for \( l \)**:
- Expand and simplify:
\[ 4l + 20 = 5l + 5 \]
- Rearranging gives:
\[ 20 - 5 = 5l - 4l \]
\[ 15 = l \]

6. **Conclusion**:
- The original length of the spring is \( l = 15 \) cm.