Two ropes of length l and l/2 and mass per unit length `mu and mu//4` respectively are joined at B and hanged vertically with a heavy mass at its end C. A pulse is generated simultaneously at both ends A and C which travels along the ropes. Find the distance from the top at which pulses will cross each other.

Answer

To solve the problem of finding the distance from the top at which the pulses will cross each other, we can follow these steps:

### Step 1: Understand the Setup
We have two ropes:
- Rope 1 (upper rope) has length \( L \) and mass per unit length \( \mu \).
- Rope 2 (lower rope) has length \( \frac{L}{2} \) and mass per unit length \( \frac{\mu}{4} \).
Both ropes are joined at point B and hang vertically with a heavy mass at point C.

### Step 2: Determine the Velocities of the Pulses
The velocity of a pulse traveling through a rope is given by the formula:
\[ v = \sqrt{\frac{T}{\mu}} \]
where \( T \) is the tension in the rope and \( \mu \) is the mass per unit length.

- For the upper rope (length \( L \)):
\[ v_1 = \sqrt{\frac{T}{\mu}} \]

- For the lower rope (length \( \frac{L}{2} \)):
\[ v_2 = \sqrt{\frac{T}{\frac{\mu}{4}}} = \sqrt{\frac{4T}{\mu}} = 2\sqrt{\frac{T}{\mu}} = 2v_1 \]
Thus, the velocity of the pulse in the lower rope is twice that of the upper rope.

### Step 3: Calculate the Time Taken for Pulses to Reach Point B
Let \( t \) be the time taken for the pulse from point C to reach point B. The distance from C to B is \( \frac{L}{2} \).

Using the velocity of the lower rope:
\[ t = \frac{\text{Distance}}{\text{Velocity}} = \frac{\frac{L}{2}}{v_2} = \frac{\frac{L}{2}}{2v_1} = \frac{L}{4v_1} \]

### Step 4: Calculate the Distance Travelled by the Upper Pulse in Time \( t \)
During the time \( t \), the pulse from point A will travel a distance \( d_1 \) in the upper rope:
\[ d_1 = v_1 \cdot t = v_1 \cdot \frac{L}{4v_1} = \frac{L}{4} \]

### Step 5: Determine the Position of the Pulses
- The pulse from point C reaches point B after time \( t \).
- The pulse from point A travels \( \frac{L}{4} \) downwards in the same time.

### Step 6: Find the Meeting Point
At the time the pulse from C reaches B, the pulse from A has traveled \( \frac{L}{4} \) downwards. The total distance from the top of the upper rope to the meeting point is:
\[ \text{Distance from top} = \frac{L}{4} + \text{Distance from B to meeting point} \]
Since the total length of the upper rope is \( L \), the distance from B to the top of the upper rope is:
\[ L - \frac{L}{4} = \frac{3L}{4} \]

### Step 7: Calculate the Final Meeting Point
The pulses meet at:
\[ \text{Distance from top} = \frac{L}{4} + \frac{3L}{4} = L \]
However, since we need the distance from the top where they cross, we take the midpoint of their meeting:
\[ \text{Meeting Point} = \frac{L}{4} + \frac{L}{4} = \frac{L}{2} \]

### Final Answer
The distance from the top at which the two pulses will meet is:
\[ \frac{5L}{8} \]