Two long strings, A and B , each possessing a linear mass density...

The velocity of the wave pulse on string A is given by

$v_A = \sqrt{\dfrac{T_A}{\mu}} = \sqrt{\dfrac{4.8}{1.2 \times 10^{-2}}} = 20 \text{ m s}^{-1}$

The velocity of the wave pulse on string B is given by

$v_B = \sqrt{\dfrac{T_B}{\mu}} = \sqrt{\dfrac{7.5}{1.2 \times 10^{-2}}} = 25 \text{ m s}^{-1}$

Let the pulse on string B overtake the pulse on string A at time $t$. The distance traveled by the pulse on string A in time $t$ is

$x_A = v_A t = 20t$

The pulse on string B starts at $t = 20 \text{ ms} = 0.02 \text{ s}$. The distance traveled by the pulse on string B is

$x_B = v_B (t - 0.02) = 25(t - 0.02)$

Since both pulses meet at the same position, we equate $x_A$ and $x_B$:

$20t = 25(t - 0.02)$

$20t = 25t - 0.5$

$5t = 0.5$

$t = 0.1 \text{ s} = 100 \text{ ms}$

The position where they meet is

$x = 20 \times 0.1 = 2.0 \text{ m}$