A heavy but unifrom rope of length L is suspended from a celling . A particle is dropped from the celling at the instant when the bottom end is given a transverse wave pulse. Where will the particle meet the pulse.

Options

A. at a distance `(2l)/ 3` form the bottom
B. at a distance `L/3` from the bottom
C. at a distance `(3L)/4` from the bottom
D. none of the above

Answer

To solve the problem of where the particle meets the transverse wave pulse in a heavy uniform rope suspended from the ceiling, we can follow these steps:

### Step 1: Understand the System
We have a rope of length \( L \) suspended from the ceiling. A particle is dropped from the ceiling at the same instant the bottom end of the rope is given a transverse wave pulse. We need to determine where the particle meets the pulse.

### Step 2: Analyze the Tension in the Rope
At a distance \( x \) from the bottom of the rope, the tension \( T_x \) in the rope can be expressed as:
\[ T_x = \frac{mg}{L} \cdot x \]
where \( m \) is the mass of the rope, \( g \) is the acceleration due to gravity, and \( L \) is the total length of the rope.

### Step 3: Calculate the Velocity of the Pulse
The velocity \( v_x \) of the wave pulse at a distance \( x \) can be given by:
\[ v_x = \sqrt{\frac{T_x}{\mu}} \]
where \( \mu \) is the mass per unit length of the rope, given by \( \mu = \frac{m}{L} \). Substituting for \( T_x \):
\[ v_x = \sqrt{\frac{\frac{mg}{L} \cdot x}{\frac{m}{L}}} = \sqrt{g x} \]

### Step 4: Set Up the Motion Equations
Let \( L_1 \) be the distance the wave pulse travels down the rope when the particle falls a distance \( L - L_1 \). The time \( t_0 \) taken for both to meet can be expressed in terms of the particle's motion:
\[ L - L_1 = \frac{1}{2} g t_0^2 \]

### Step 5: Time for the Pulse to Travel \( L_1 \)
The time taken for the pulse to travel a distance \( L_1 \) can be expressed as:
\[ t_0 = \int_0^{L_1} \frac{dx}{\sqrt{g x}} \]
This integral can be evaluated:
\[ t_0 = \int_0^{L_1} x^{-1/2} \, dx = 2\sqrt{L_1} \]

### Step 6: Relate the Two Times
Setting the two expressions for \( t_0 \) equal gives:
\[ \frac{1}{2} g t_0^2 = L - L_1 \]
Substituting \( t_0 = 2\sqrt{L_1} \):
\[ \frac{1}{2} g (2\sqrt{L_1})^2 = L - L_1 \]
This simplifies to:
\[ 2g L_1 = L - L_1 \]
Rearranging gives:
\[ 3L_1 = L \]
Thus:
\[ L_1 = \frac{L}{3} \]

### Step 7: Conclusion
The particle will meet the pulse at a distance \( L_1 = \frac{L}{3} \) from the bottom of the rope.