A string has a linear mass density `mu`. The ends of the string are joined to form a closed loop and is given the shape of a circular ring of radius R. this ring is now rotated about its axis with an angular velocity `omega`
Q. The velocity of the transverse wave set up in the string, with respect to an observer at the centre of

Options

A. a.`omegaR`
B. b.`2omegaR`
C. c.`3omegaR`
D. d.`4omegaR`

Answer

To find the velocity of the transverse wave set up in the string with respect to an observer at the center of the circular ring, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the System**:
- We have a string with linear mass density \( \mu \) that is formed into a circular ring of radius \( R \).
- The ring is rotating about its axis with an angular velocity \( \omega \).

2. **Identifying Forces**:
- Consider a small segment of the string with mass \( dm \) and length \( dl \).
- The tension \( T \) in the string acts along the string, and we need to analyze the forces acting on this small segment.

3. **Setting Up the Forces**:
- The tension creates a centripetal force that must balance the outward force due to the rotation of the string.
- The radial component of the tension can be expressed as \( T \cos(\theta) \) and the vertical component as \( T \sin(\theta) \).

4. **Centripetal Force Requirement**:
- The outward force due to rotation is given by \( dm \cdot R \cdot \omega^2 \).
- For small angles, we can approximate \( \sin(\theta) \approx \theta \) and \( \cos(\theta) \approx 1 \).

5. **Relating Tension and Mass**:
- The mass \( dm \) can be expressed as \( dm = \mu \cdot dl \).
- The length \( dl \) can be approximated as \( R \cdot d\theta \) for a small angle \( d\theta \).

6. **Equating Forces**:
- The net force in the radial direction gives us:
\[ dm \cdot R \cdot \omega^2 = 2T \sin(\theta) \]
- Substituting \( dm \) gives:
\[ \mu \cdot R \cdot d\theta \cdot R \cdot \omega^2 = 2T \theta \]

7. **Solving for Tension**:
- Rearranging the above equation leads to:
\[ T = \frac{\mu R^2 \omega^2}{2} \]

8. **Finding the Velocity of the Wave**:
- The velocity of a wave on a string is given by:
\[ v = \sqrt{\frac{T}{\mu}} \]
- Substituting the expression for \( T \):
\[ v = \sqrt{\frac{\frac{\mu R^2 \omega^2}{2}}{\mu}} = \sqrt{\frac{R^2 \omega^2}{2}} = \frac{R \omega}{\sqrt{2}} \]

9. **Relative Velocity with Respect to Observer**:
- The velocity of the string with respect to the observer at the center is \( R \omega \).
- Therefore, the total velocity of the wave with respect to the observer is:
\[ v_{observer} = v + R \omega = \frac{R \omega}{\sqrt{2}} + R \omega = R \omega \left(1 + \frac{1}{\sqrt{2}}\right) \]

10. **Final Expression**:
- The final expression for the velocity of the transverse wave with respect to the observer at the center of the ring is:
\[ v_{observer} = R \omega \left(1 + \frac{1}{\sqrt{2}}\right) \]