Question

What phase difference between two otherwise identical traveling waves, moving in the same direction along a stretched string, will result in the combined wave having an amplitude 0.70 times that of the common amplitude of the two combining waves?

Answer 1

`\phi=2.43\ rad=139.2^(\circ)`

Step-by-step explanation:

The resultant wave of two identical traveling waves of amplitude
`y_M` with a phase difference
`\phi` between them is:

`Y(x,t)=y_Msin(kx-\omega t+\phi)+y_Msin(kx-\omega t)`

Using the trigonometric formula

`sin(a)+sin(b)=2sin((a+b)/(2))cos((a-b)/(2))`

we have:

`Y(x,t)=2y_Msin((2kx-2\omega t+\phi)/(2))cos((\phi)/(2))=2y_Mcos((\phi)/(2))sin(kx-\omega t+(\phi)/(2))`

where we can see that the amplitude of the combined wave is
`Y_M=2y_Mcos((\phi)/(2))`, which we want to be equal to
`0.7y_M`, so we do:

`2y_Mcos((\phi)/(2))=0.7y_M`

`cos((\phi)/(2))=0.35`

`\phi=2Arccos(0.35)=2.43\ rad`