Two long strings, A and B , each possessing a linear mass density of 1.2 10⁻² kg m ⁻¹ , are stretched by tensions of 4.8 N and 7.5 N respectively. They are kept parallel to each other with their left ends positioned at x = 0 . Wave pulses are initiated at these left ends at t = 0 on string A and at t = 20 ms on string B . Determine when and where the pulse on string B overtakes the pulse on string A .

Step-by-step solution

The velocity of the wave pulse on string A is given by v_A = T_A = 4.8 1.2 10⁻² = 20 m s ⁻¹ The velocity of the wave pulse on string B is given by v_B = T_B = 7.5 1.2 10⁻² = 25 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 ms = 0.02 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 eq