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, we need to find the distance from the top at which the pulses from both ends of the ropes cross each other. We will follow these steps:

### Step 1: Determine the velocities of the pulses in each rope

The velocity of a wave in 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.

1. **For the first rope (length \( l \), mass per unit length \( \mu \))**:
- The velocity \( v_1 \) is given by:
\[ v_1 = \sqrt{\frac{T}{\mu}} \]

2. **For the second rope (length \( l/2 \), mass per unit length \( \mu/4 \))**:
- The velocity \( v_2 \) is given by:
\[ v_2 = \sqrt{\frac{T}{\mu/4}} = \sqrt{\frac{4T}{\mu}} = 2\sqrt{\frac{T}{\mu}} = 2v_1 \]

### Step 2: Set up the equations for the distances traveled by the pulses

Let \( x \) be the distance from the top of the first rope where the pulses cross each other. The distance traveled by the pulse from the top of the first rope to the crossing point is \( x \), and the distance traveled by the pulse from the bottom of the second rope to the crossing point is \( l - x \).

- Time taken by the pulse from the first rope to reach the crossing point:
\[ t_1 = \frac{x}{v_1} \]

- Time taken by the pulse from the second rope to reach the crossing point:
\[ t_2 = \frac{l - x}{v_2} \]

### Step 3: Set the times equal to each other

Since both pulses are generated simultaneously, we can set the times equal to each other:

\[ \frac{x}{v_1} = \frac{l - x}{v_2} \]

Substituting \( v_2 = 2v_1 \):

\[ \frac{x}{v_1} = \frac{l - x}{2v_1} \]

### Step 4: Solve for \( x \)

We can simplify the equation by multiplying both sides by \( 2v_1 \):

\[ 2x = l - x \]

Now, add \( x \) to both sides:

\[ 3x = l \]

Thus, we find:

\[ x = \frac{l}{3} \]

### Step 5: Conclusion

The distance from the top at which the pulses will cross each other is:

\[ \boxed{\frac{l}{3}} \]