A wire of length L, is hanging vertically from a rigid support. If a transverse wave pulse is generated at the free end of wire then which of the following statement is wrong

Options

A. Velocity at bottom end is zero
B. Velocity at top end is `sqrt(gL)`
C. Time taken to reach the top end is `2 sqrt(L/g)`
D. Acceleration of wave is g

Answer

To solve the problem, we need to analyze the statements regarding the behavior of a transverse wave pulse generated in a vertically hanging wire. Let's go through the statements one by one to identify which one is incorrect.

### Step 1: Analyze the first statement
**Statement:** Velocity at the bottom end is 0.

- At the bottom end of the wire, the displacement (x) is 0 because it is fixed to a support.
- The velocity of the wave can be expressed as \( v = \sqrt{g \cdot x} \).
- Substituting \( x = 0 \) gives \( v = \sqrt{g \cdot 0} = 0 \).

**Conclusion:** This statement is correct.

### Step 2: Analyze the second statement
**Statement:** Velocity at the top end is \( \sqrt{g \cdot L} \).

- At the top end of the wire, the displacement (x) is equal to the length of the wire (L).
- Using the same formula for velocity, we have \( v = \sqrt{g \cdot L} \).

**Conclusion:** This statement is also correct.

### Step 3: Analyze the third statement
**Statement:** Time taken to reach the top end is \( 2 \sqrt{\frac{L}{g}} \).

- The time taken for the wave to travel the length of the wire can be calculated using the formula \( t = \frac{L}{v} \).
- Substituting \( v = \sqrt{g \cdot L} \), we get \( t = \frac{L}{\sqrt{g \cdot L}} = \sqrt{\frac{L^2}{g \cdot L}} = \sqrt{\frac{L}{g}} \).
- However, this is the time for the wave to travel to the top, not double that value.

**Conclusion:** This statement is incorrect.

### Step 4: Analyze the fourth statement
**Statement:** Acceleration of the wave is g.

- The relationship between velocity and displacement is given by \( v^2 = g \cdot x \).
- Differentiating with respect to time gives \( 2v \frac{dv}{dt} = g \frac{dx}{dt} \).
- Using \( \frac{dx}{dt} = v \), we can express acceleration \( a \) as \( a = \frac{dv}{dt} = \frac{g}{2v} \).
- This indicates that the acceleration is not equal to g but rather \( \frac{g}{2} \).

**Conclusion:** This statement is also incorrect.

### Final Conclusion
The statement that is wrong is the one regarding the time taken to reach the top end, which is incorrectly stated as \( 2 \sqrt{\frac{L}{g}} \). The correct time is \( \sqrt{\frac{L}{g}} \).

### Summary of the Incorrect Statement
- **Incorrect Statement:** Time taken to reach the top end is \( 2 \sqrt{\frac{L}{g}} \).