A string of length l is stretched by 1/30 and transverse waves in the string are found to travel at a speed `v_0`. Speed of transverse waves when the string is stretched by 1/15 will be :

Options

A. `(v_0)/(2)`
B. `2v_0`
C. `2sqrt2v_0`
D. `sqrt2v_0`

Answer

To solve the problem, we need to understand the relationship between the speed of transverse waves in a string and the change in length (stretch) of the string. The speed of transverse waves \( v \) in a string is given by the formula:

\[ v \propto \sqrt{\Delta L} \]

where \( \Delta L \) is the change in length of the string.

### Step-by-Step Solution:

1. **Identify the initial conditions**:
- The string is initially stretched by \( \Delta L_1 = \frac{1}{30} \) and the speed of the wave is \( v_1 = v_0 \).

2. **Identify the new conditions**:
- The string is then stretched by \( \Delta L_2 = \frac{1}{15} \) and we need to find the new speed \( v_2 \).

3. **Set up the relationship**:
- According to the relationship, we can write:
\[ \frac{v_2}{v_1} = \sqrt{\frac{\Delta L_2}{\Delta L_1}} \]

4. **Substitute the known values**:
- Substitute \( v_1 = v_0 \), \( \Delta L_1 = \frac{1}{30} \), and \( \Delta L_2 = \frac{1}{15} \):
\[ \frac{v_2}{v_0} = \sqrt{\frac{\frac{1}{15}}{\frac{1}{30}}} \]

5. **Simplify the fraction**:
- The fraction simplifies to:
\[ \frac{1}{15} \div \frac{1}{30} = \frac{1}{15} \times \frac{30}{1} = 2 \]

6. **Calculate the square root**:
- Now we can calculate the square root:
\[ \frac{v_2}{v_0} = \sqrt{2} \]

7. **Solve for \( v_2 \)**:
- Multiply both sides by \( v_0 \):
\[ v_2 = v_0 \sqrt{2} \]

### Final Answer:
The speed of transverse waves when the string is stretched by \( \frac{1}{15} \) is:
\[ v_2 = v_0 \sqrt{2} \]