A uniform rope having linear density `lambda` hangs vertically from the ceiling and its lower end is free. A disturbance produced at the free end has a speed `v_0` at point P midway on the rope . Then the time taken by the disturbance pulse to reach the ceiling is

Options

A. `(4mu_0)/g`
B. `(2sqrt2V_0)/g`
C. `(2sqrt2V_0^2)/(lambdag)`
D. `(2sqrt2V_0)/(lambdag)`

Answer

To solve the problem of determining the time taken by a disturbance pulse to reach the ceiling of a vertically hanging rope, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Setup**:
- We have a uniform rope with linear density \( \lambda \) hanging vertically from the ceiling. The lower end of the rope is free.
- A disturbance is produced at the free end of the rope, and it travels upward.

2. **Identifying the Point of Interest**:
- Let point \( P \) be the midpoint of the rope. The total length of the rope is \( L \), so the distance from the free end to point \( P \) is \( \frac{L}{2} \).

3. **Calculating Tension in the Rope**:
- Consider a small segment of the rope at a distance \( x \) from the free end. The mass of this segment is \( \lambda \cdot x \).
- The tension \( T \) at this point due to the weight of the rope below it is given by:
\[ T = \lambda \cdot x \cdot g \]
- Here, \( g \) is the acceleration due to gravity.

4. **Finding the Wave Speed**:
- The speed of a transverse wave in the rope is given by:
\[ v = \sqrt{\frac{T}{\mu}} \]
- Where \( \mu \) is the linear mass density of the rope, which is \( \lambda \).
- Substituting for \( T \):
\[ v = \sqrt{\frac{\lambda \cdot x \cdot g}{\lambda}} = \sqrt{x \cdot g} \]

5. **Setting the Speed at Midpoint**:
- At point \( P \) (the midpoint), the distance \( x \) is \( \frac{L}{2} \). The speed of the wave at this point is given as \( v_0 \):
\[ v_0 = \sqrt{\frac{L}{2} \cdot g} \]
- Squaring both sides:
\[ v_0^2 = \frac{L \cdot g}{2} \]
- Rearranging gives:
\[ L = \frac{2 v_0^2}{g} \]

6. **Setting Up the Time Calculation**:
- The speed of the wave can also be expressed as:
\[ v = \frac{dx}{dt} \]
- Thus, we can write:
\[ dt = \frac{dx}{\sqrt{x \cdot g}} \]

7. **Integrating to Find Total Time**:
- We need to integrate from \( 0 \) to \( L \):
\[ t = \int_0^L \frac{dx}{\sqrt{x \cdot g}} = \frac{1}{\sqrt{g}} \int_0^L \frac{dx}{\sqrt{x}} \]
- The integral \( \int \frac{dx}{\sqrt{x}} \) evaluates to \( 2\sqrt{x} \):
\[ t = \frac{1}{\sqrt{g}} \left[ 2\sqrt{x} \right]_0^L = \frac{2\sqrt{L}}{\sqrt{g}} \]

8. **Substituting for \( L \)**:
- Now substituting \( L = \frac{2 v_0^2}{g} \):
\[ t = \frac{2\sqrt{\frac{2 v_0^2}{g}}}{\sqrt{g}} = \frac{2 \cdot \sqrt{2} \cdot v_0}{g} \]

9. **Final Result**:
- Thus, the time taken by the disturbance pulse to reach the ceiling is:
\[ t = \frac{2\sqrt{2} v_0}{g} \]