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

Answer:

$\phi=2.43\ rad=139.2^{\circ}$

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

$sin(a)+sin(b)=2sin(\frac{a+b}{2})cos(\frac{a-b}{2})$

we have:

$Y(x,t)=2y_Msin(\frac{2kx-2\omega t+\phi}{2})cos(\frac{\phi}{2})=2y_Mcos(\frac{\phi}{2})sin(kx-\omega t+\frac{\phi}{2})$

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

$2y_Mcos(\frac{\phi}{2})=0.7y_M$

$cos(\frac{\phi}{2})=0.35$

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