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 finding 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 hanging vertically from the ceiling with linear density \( \lambda \).
- The lower end of the rope is free, and a disturbance is produced at this end.
- We need to determine the time it takes for this disturbance to travel to the ceiling.

2. **Identifying the Variables**:
- Let the length of the rope be \( L \).
- The linear density of the rope is given as \( \lambda \).
- The speed of the disturbance at point \( P \) (midway) is \( v_0 \).

3. **Finding the Tension in the Rope**:
- At a distance \( x \) from the ceiling, the tension \( T \) in the rope due to the weight of the rope below that point is:
\[ T = \lambda g x \]
- Here, \( g \) is the acceleration due to gravity.

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

5. **Using the Given Condition**:
- At point \( P \) (midway), where \( x = \frac{L}{2} \), the speed of the wave is given as \( v_0 \):
\[ v_0 = \sqrt{g \frac{L}{2}} \]
- Squaring both sides gives:
\[ v_0^2 = \frac{gL}{2} \]
- Rearranging gives:
\[ L = \frac{2v_0^2}{g} \]

6. **Setting Up the Time Calculation**:
- The relationship between distance \( dx \), time \( dt \), and velocity \( v \) is:
\[ v = \frac{dx}{dt} \]
- Substituting \( v = \sqrt{g x} \):
\[ \sqrt{g x} = \frac{dx}{dt} \Rightarrow dt = \frac{dx}{\sqrt{g x}} \]

7. **Integrating to Find Total Time**:
- To find the total time \( t \) taken for the disturbance to travel from \( x = 0 \) to \( x = L \):
\[ t = \int_0^L \frac{dx}{\sqrt{g x}} \]
- This can be simplified:
\[ t = \frac{1}{\sqrt{g}} \int_0^L x^{-1/2} dx = \frac{1}{\sqrt{g}} \left[ 2\sqrt{x} \right]_0^L = \frac{2\sqrt{L}}{\sqrt{g}} \]

8. **Substituting for L**:
- From our earlier result, we substitute \( L = \frac{2v_0^2}{g} \):
\[ t = \frac{2\sqrt{\frac{2v_0^2}{g}}}{\sqrt{g}} = \frac{2 \cdot \sqrt{2} \cdot v_0}{g} \]

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