A string of length 2L, obeying Hooke's Law, is stretched so that its extension is L. The speed of the tranverse wave travelling on the string is v. if the string is futher stretched so that the extension in the string becomes 4L. The speed of transverse wave travelling on the string will be

Options

A. `sqrt(2)`V
B. V
C. `2sqrt(2)V`
D. 2V

Explanation

To solve the problem, we will analyze the relationship between tension, extension, and the speed of transverse waves on a string.

### Step-by-Step Solution:

1. **Understanding the Initial Conditions**:
- The original length of the string is \(2L\).
- The string is stretched such that the extension is \(L\).
- Therefore, the total length of the string after the first stretch is:
\[ L_1 = 2L + L = 3L \]

2. **Calculating the Tension in the First Case**:
- According to Hooke's Law, the tension \(T_1\) in the string is proportional to the extension. If the extension is \(L\), we can express the tension as:
\[ T_1 \propto L \]
- For the first case, we can denote the tension as:
\[ T_1 = k \cdot L \quad \text{(where \(k\) is a constant)} \]

3. **Speed of the Transverse Wave in the First Case**:
- The speed \(v_1\) of the transverse wave on the string is given by:
\[ v_1 = \sqrt{\frac{T_1}{\mu}} \]
- Here, \(\mu\) is the mass per unit length of the string.

4. **Understanding the Second Conditions**:
- The string is further stretched so that the extension becomes \(4L\).
- The total length of the string after this second stretch is:
\[ L_2 = 2L + 4L = 6L \]

5. **Calculating the Tension in the Second Case**:
- For the second case, the tension \(T_2\) is proportional to the new extension \(4L\):
\[ T_2 \propto 4L \]
- Thus, we can express the tension as:
\[ T_2 = k \cdot 4L \]

6. **Speed of the Transverse Wave in the Second Case**:
- The speed \(v_2\) of the transverse wave in the second case is given by:
\[ v_2 = \sqrt{\frac{T_2}{\mu}} \]
- Substituting for \(T_2\):
\[ v_2 = \sqrt{\frac{k \cdot 4L}{\mu}} \]

7. **Finding the Ratio of Speeds**:
- We can now find the ratio of the speeds \(v_2\) to \(v_1\):
\[ \frac{v_2}{v_1} = \sqrt{\frac{T_2}{T_1}} = \sqrt{\frac{4L}{L}} = \sqrt{4} = 2 \]
- Therefore, we can express \(v_2\) in terms of \(v_1\):
\[ v_2 = 2v_1 \]

8. **Conclusion**:
- The final speed of the transverse wave when the extension becomes \(4L\) is:
\[ v_2 = 2v \]