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 a transverse wave set up in the string, with respect to the string is

Options

A. `omegaR`
B. `2omegaR`
C. `3omegaR`
D. `4omegaR`

Answer

To find the velocity of a transverse wave set up in a rotating string that forms a circular ring, we can follow these steps:

### Step 1: Understand the setup
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 \( \omega \).

### Step 2: Determine the tension in the string
When the ring rotates, a tension \( T \) is developed in the string due to the centripetal force required to keep the string in circular motion.

1. Consider a small segment of the ring with an angle \( d\theta \) at radius \( R \).
2. The length of this segment is \( R d\theta \).
3. The mass of this segment \( dm \) can be expressed as:
\[ dm = \mu \cdot (R d\theta) = \mu R d\theta \]

### Step 3: Analyze the forces acting on the segment
The tension \( T \) in the string has components that provide the necessary centripetal force for the segment:

1. The vertical component of the tension acting towards the center provides the centripetal force:
\[ F_c = 2T \sin\left(\frac{d\theta}{2}\right) \]
For small angles, \( \sin\left(\frac{d\theta}{2}\right) \approx \frac{d\theta}{2} \).

2. Therefore, we can approximate:
\[ F_c \approx 2T \cdot \frac{d\theta}{2} = T d\theta \]

### Step 4: Set up the centripetal force equation
The centripetal force required for the mass \( dm \) moving with velocity \( v \) is given by:
\[ F_c = dm \cdot \frac{v^2}{R} \]
Substituting \( dm \):
\[ F_c = \mu R d\theta \cdot \frac{v^2}{R} = \mu d\theta v^2 \]

### Step 5: Equate the forces
Setting the two expressions for centripetal force equal gives:
\[ T d\theta = \mu d\theta v^2 \]
Dividing both sides by \( d\theta \) (assuming \( d\theta \neq 0 \)):
\[ T = \mu v^2 \]

### Step 6: Relate tension to angular velocity
The tension \( T \) can also be expressed in terms of the angular velocity \( \omega \):
\[ T = \mu R \omega^2 \]

### Step 7: Substitute to find the wave velocity
Now, we can substitute \( T \) back into the wave velocity formula:
\[ v_t = \sqrt{\frac{T}{\mu}} = \sqrt{\frac{\mu R \omega^2}{\mu}} = \sqrt{R \omega^2} = R \omega \]

### Final Result
The velocity of the transverse wave set up in the string, with respect to the string, is:
\[ v_t = R \omega \]

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