Two long strings A and B, each having linear mass density 1.2×10−2kgm−1, are stretched by different tensions 4.8 N and 7.5 N respectively and are kept parallel to each other with their left ends at x = 0. Wave pulses are produced on the strings at the left ends at t = 0 on string A and at t = 20 ms on string B. When and where will the pulse on B overtake that on A ?

Explanation

To solve the problem of when and where the wave pulse on string B overtakes the wave pulse on string A, we will follow these steps:

### Step 1: Determine the velocities of the wave pulses on both strings.
The velocity of a wave on a string is given by the formula:

\[ v = \sqrt{\frac{T}{\mu}} \]

where \( T \) is the tension in the string and \( \mu \) is the linear mass density.

#### For String A:
- Tension \( T_1 = 4.8 \, \text{N} \)
- Linear mass density \( \mu = 1.2 \times 10^{-2} \, \text{kg/m} \)

Calculating the velocity \( v_1 \):

\[ v_1 = \sqrt{\frac{4.8}{1.2 \times 10^{-2}}} = \sqrt{400} = 20 \, \text{m/s} \]

#### For String B:
- Tension \( T_2 = 7.5 \, \text{N} \)

Calculating the velocity \( v_2 \):

\[ v_2 = \sqrt{\frac{7.5}{1.2 \times 10^{-2}}} = \sqrt{625} = 25 \, \text{m/s} \]

### Step 2: Set up the equations for the distances traveled by both pulses.
Let \( t \) be the time taken for the pulses to meet after the pulse on string A is generated. The pulse on string B is generated 20 ms (0.02 s) later.

#### Distance traveled by pulse on String A:
\[ x_A = v_1 \cdot t = 20t \]

#### Distance traveled by pulse on String B:
Since the pulse on string B starts 0.02 s later, the time for string B is \( t - 0.02 \):
\[ x_B = v_2 \cdot (t - 0.02) = 25(t - 0.02) \]

### Step 3: Set the distances equal to find when they meet.
At the point of overtaking, the distances will be equal:

\[ x_A = x_B \]

Substituting the expressions for \( x_A \) and \( x_B \):

\[ 20t = 25(t - 0.02) \]

### Step 4: Solve for \( t \).
Expanding the equation:

\[ 20t = 25t - 0.5 \]

Rearranging gives:

\[ 25t - 20t = 0.5 \]
\[ 5t = 0.5 \]
\[ t = \frac{0.5}{5} = 0.1 \, \text{s} \]

### Step 5: Calculate the distance where they meet.
Substituting \( t = 0.1 \) s into the equation for \( x_A \):

\[ x_A = 20t = 20 \times 0.1 = 2 \, \text{m} \]

### Final Answer:
The pulse on string B overtakes the pulse on string A at \( t = 0.1 \, \text{s} \) and at a distance of \( 2 \, \text{m} \).

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