A string of length 2L, obeying hooke's law, is stretched so that its extension is L. the speed of the transverse wave travelling on the string is v. If the string is further stretched so that the extension in the string becomes 4L. The speed of transverse wave travelling on the string will be

Options

A. `(1)/sqrt(2)V`
B. `sqrt(2)v`
C. `(1)/(2)v`
D. `2sqrt(2)v`

Explanation

To solve the problem, we need to determine the speed of a transverse wave on a string when it is stretched to a new extension. We will follow these steps:

### Step 1: Understand the relationship between wave speed, tension, and linear mass density
The speed \( v \) of a transverse wave on a string is given by the formula:
\[ v = \sqrt{\frac{T}{\mu}} \]
where \( T \) is the tension in the string and \( \mu \) is the linear mass density.

### Step 2: Calculate the initial tension and linear mass density
Initially, the string has an extension \( X_1 = L \). The tension \( T_1 \) in the string can be expressed as:
\[ T_1 = K \cdot X_1 = K \cdot L \]
where \( K \) is the spring constant of the string.

The linear mass density \( \mu \) is given by:
\[ \mu = \frac{M}{L_{\text{total}}} \]
where \( L_{\text{total}} = 2L + L = 3L \). Thus,
\[ \mu = \frac{M}{3L} \]

### Step 3: Substitute values into the wave speed formula
Substituting \( T_1 \) and \( \mu \) into the wave speed formula:
\[ v = \sqrt{\frac{K \cdot L}{\frac{M}{3L}}} = \sqrt{\frac{3K \cdot L^2}{M}} \]

### Step 4: Calculate the new tension and linear mass density for the second case
Now, the string is stretched so that the extension becomes \( X_2 = 4L \). The new tension \( T_2 \) is:
\[ T_2 = K \cdot X_2 = K \cdot 4L \]

The new linear mass density \( \mu' \) is given by:
\[ \mu' = \frac{M}{L_{\text{total}}'} = \frac{M}{2L + 4L} = \frac{M}{6L} \]

### Step 5: Substitute values into the wave speed formula for the new case
Now, substituting \( T_2 \) and \( \mu' \) into the wave speed formula for the new case:
\[ v' = \sqrt{\frac{T_2}{\mu'}} = \sqrt{\frac{K \cdot 4L}{\frac{M}{6L}}} = \sqrt{\frac{24K \cdot L^2}{M}} \]

### Step 6: Relate the new speed to the original speed
Now we have:
- Original speed: \( v = \sqrt{\frac{3K \cdot L^2}{M}} \)
- New speed: \( v' = \sqrt{\frac{24K \cdot L^2}{M}} \)

To find the relationship between \( v' \) and \( v \):
\[ \frac{v'}{v} = \frac{\sqrt{24K \cdot L^2 / M}}{\sqrt{3K \cdot L^2 / M}} = \sqrt{\frac{24}{3}} = \sqrt{8} = 2\sqrt{2} \]
Thus,
\[ v' = 2\sqrt{2} \cdot v \]

### Conclusion
The speed of the transverse wave traveling on the string when the extension becomes \( 4L \) is:
\[ v' = 2\sqrt{2} \cdot v \]

### Final Answer
The speed of the transverse wave will be \( 2\sqrt{2}v \).

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