The extension in a string obeying Hooke's law is x. The speed of transverse wave in the stretched string is v. If the extension in the string is increased to 1.5x, the speed of transverse wave will be

Options

A. 1.22v
B. 0.61v
C. 1.50v
D. 0.75v

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the relationship between wave speed and tension
The speed of a transverse wave in a stretched 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. Since the mass per unit length \( \mu \) is constant, the speed \( v \) is directly proportional to the square root of the tension \( T \).

**Hint:** Remember that the tension in the string is related to the extension of the string according to Hooke's law.

### Step 2: Relate tension to extension
According to Hooke's law, the tension \( T \) in the string is directly proportional to the extension \( x \):
\[ T \propto x \]
This means that if the extension increases, the tension also increases proportionally.

**Hint:** Think about how the tension changes when the extension of the string changes.

### Step 3: Set up the relationship for the two cases
Let:
- \( x_1 = x \) (initial extension)
- \( v_1 = v \) (initial wave speed)
- \( x_2 = 1.5x \) (new extension)
- \( v_2 \) (new wave speed)

From the relationship established, we can write:
\[ \frac{v_2}{v_1} = \sqrt{\frac{x_2}{x_1}} \]

**Hint:** You will use the ratio of the extensions to find the new speed.

### Step 4: Substitute the values into the equation
Substituting the values of \( x_1 \) and \( x_2 \):
\[ \frac{v_2}{v} = \sqrt{\frac{1.5x}{x}} \]

**Hint:** Simplify the fraction under the square root.

### Step 5: Simplify the equation
The \( x \) terms cancel out:
\[ \frac{v_2}{v} = \sqrt{1.5} \]

**Hint:** Calculate the square root of 1.5 to find the relationship between \( v_2 \) and \( v \).

### Step 6: Solve for \( v_2 \)
Now, multiply both sides by \( v \):
\[ v_2 = v \cdot \sqrt{1.5} \]

### Step 7: Calculate \( \sqrt{1.5} \)
The approximate value of \( \sqrt{1.5} \) is about 1.22. Therefore:
\[ v_2 \approx 1.22v \]

**Hint:** Use a calculator or approximate values to find \( \sqrt{1.5} \).

### Conclusion
The speed of the transverse wave when the extension is increased to \( 1.5x \) will be approximately:
\[ v_2 \approx 1.22v \]

Thus, the final answer is that the speed of the transverse wave will increase by a factor of \( \sqrt{1.5} \) when the extension is increased to \( 1.5x \).