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 ?

Explanation

(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 root yields `( 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.