Two pulses travelling on the same string are described by y_(1) = (5)/(( 3x - 4 t)^(2)+ 2) and y_(2) = ( -5) /(( 3x + 4t - 6)^(2) + 2) (a). In which direction does each pulse travel ? (b). At what instant do the two cancel everywhere ? ( c). At what point do the two pulses always cancel ?

Two pulses travelling on the same string are described by y_(1) = (5)

1.	Two pulses travelling on the same string are described by y_(1) = (5)/(( 3x - 4 t)^(2)+ 2) and y_(2) = ( -5) /(( 3x + 4t - 6)^(2) + 2) (a). In which direction does each pulse travel ? (b). At what instant do the two cancel everywhere ? ( c). At what point do the two pulses always cancel ?
Answer» Solution :(a). At constant phase , `phi = 3x - 4t` will be constant . Then ` x = ( phi + 4t)//3` will change : the wave moves . As `t` increases in this EQUATION , `x` increases , so the first wave moves to the right . In the same way , in the second case ` x = ( phi - 4t + 6)//3`. As `t` increases , `x` MUST decreases , so the second wave moves to the left. (b) We required that ` y_(1) + y_(2) = 0` `( 5)/(( 3x - 4t)^(2)) + ( -5)/(( 3x + 4t -6)^(2) + 2) = 0` This can be written as `( 3x - 4t)^(2) = ( 3 x + 4 t -6)^(2)` Solving for the positive root , `t = 0.750 s` ( c) The negative rootyields `( 3x - 4t) = -( 3x + 4t - 6)` The time terms cancel , leaving ` x = 1.00 m`. At this POINT , the waves always cancel. The TOTAL wave is not a standing wave , but we could call teh point at ` x = 1.00 m`a mode of the superposition.