A string of length 2x is stretched by 0.1 x and the velocity of a transverse wave along it is v. When it is stretched by 0.4x , the velocity of the wave is

Options

A. `sqrt(5/6)v`
B. `sqrt(11/7)v`
C. `sqrt(32/7)v`
D. `sqrt(27/6)v`

Answer

To solve the problem step by step, we will analyze the relationship between the tension in the string, the change in length, and the velocity of the transverse wave.

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

### Step 2: Determine the mass per unit length
The mass per unit length \( \mu \) can be expressed as:
\[ \mu = \frac{M}{L} \]
where \( M \) is the mass of the string and \( L \) is its length.

### Step 3: Calculate the initial conditions
Initially, the string has a length of \( 2x \) and is stretched by \( 0.1x \). Therefore, the new length \( L_1 \) is:
\[ L_1 = 2x + 0.1x = 2.1x \]
The tension \( T_1 \) can be expressed as proportional to the extension:
\[ T_1 = k \cdot 0.1x \]
where \( k \) is a constant of proportionality.

### Step 4: Calculate the initial wave velocity
Substituting the values into the wave velocity formula:
\[ v_1 = \sqrt{\frac{T_1}{\mu}} = \sqrt{\frac{k \cdot 0.1x}{\frac{M}{2.1x}}} = \sqrt{\frac{k \cdot 0.1x \cdot 2.1x}{M}} = \sqrt{\frac{0.21kx^2}{M}} \]

### Step 5: Determine the new conditions
Now, when the string is stretched by \( 0.4x \), the new length \( L_2 \) is:
\[ L_2 = 2x + 0.4x = 2.4x \]
The new tension \( T_2 \) is:
\[ T_2 = k \cdot 0.4x \]

### Step 6: Calculate the new wave velocity
Substituting the new values into the wave velocity formula:
\[ v_2 = \sqrt{\frac{T_2}{\mu}} = \sqrt{\frac{k \cdot 0.4x}{\frac{M}{2.4x}}} = \sqrt{\frac{k \cdot 0.4x \cdot 2.4x}{M}} = \sqrt{\frac{0.96kx^2}{M}} \]

### Step 7: Find the ratio of the velocities
To find the ratio \( \frac{v_1}{v_2} \):
\[ \frac{v_1}{v_2} = \sqrt{\frac{0.21kx^2/M}{0.96kx^2/M}} = \sqrt{\frac{0.21}{0.96}} = \sqrt{\frac{21}{96}} = \sqrt{\frac{7}{32}} \]

### Step 8: Solve for \( v_2 \)
From the ratio, we can express \( v_2 \) in terms of \( v_1 \):
\[ v_2 = v_1 \cdot \sqrt{\frac{32}{7}} \]
Given that \( v_1 = v \):
\[ v_2 = v \cdot \sqrt{\frac{32}{7}} \]

### Final Answer
Thus, the velocity of the wave when the string is stretched by \( 0.4x \) is:
\[ v_2 = \sqrt{\frac{32}{7}} \cdot v \]