A loop of a string of mass length `mu` and redius R is rotated about an axis passing through centre perpendicular to the plane with an angular velocity `omega`.A small disturbance is created in the loop having the same sense of rotation. The linear speed of the disturbance observer is

Options

A. `omegaR`
B. `2omegaR`
C. `3omegaR`
D. Zero

Answer

To find the linear speed of the disturbance observer in a rotating loop of a string, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the System**:
- We have a loop of string with mass per unit length \( \mu \) and radius \( R \) rotating about an axis through its center with angular velocity \( \omega \).

2. **Identifying the Mass Element**:
- Consider a small arc of the loop with an angle \( d\theta \). The length of this arc is \( dl = R d\theta \).
- The mass of this small arc is given by \( dm = \mu \cdot dl = \mu R d\theta \).

3. **Centripetal Force**:
- The centripetal force required to keep the mass element \( dm \) in circular motion is provided by the tension in the string.
- The component of tension acting towards the center can be approximated as \( T \cdot \sin(d\theta/2) \). For small angles, \( \sin(d\theta/2) \approx \frac{d\theta}{2} \).
- Thus, the centripetal force can be approximated as \( T \cdot \frac{d\theta}{2} \).

4. **Setting up the Equation**:
- The centripetal force is also given by \( dm \cdot \omega^2 R \).
- Therefore, we can write the equation:
\[ T \cdot \frac{d\theta}{2} = \mu R d\theta \cdot \omega^2 \]

5. **Simplifying the Equation**:
- Cancel \( d\theta \) from both sides:
\[ T \cdot \frac{1}{2} = \mu R \omega^2 \]
- Rearranging gives:
\[ T = 2 \mu R \omega^2 \]

6. **Finding the Wave Velocity**:
- The wave velocity \( v \) in the string is given by:
\[ v = \sqrt{\frac{T}{\mu}} = \sqrt{\frac{2 \mu R \omega^2}{\mu}} = \sqrt{2 R \omega^2} = \omega \sqrt{2R} \]

7. **Linear Speed of the Disturbance**:
- The linear speed of the disturbance observed is the sum of the wave velocity \( v \) and the linear speed of the string itself, which is \( \omega R \).
- Thus, we have:
\[ v_p = v + v_s = \omega \sqrt{2R} + \omega R \]
- This simplifies to:
\[ v_p = \omega R + \omega \sqrt{2R} = \omega (R + \sqrt{2R}) \]

### Final Answer:
The linear speed of the disturbance observer is \( v_p = \omega (R + \sqrt{2R}) \).