If two waves, each of amplitude zz, produce a resultant wave of amplitude zz, then what is the phase difference between them? Is there a specific equation or formula I can use?

Seems like the phase difference is between 0 and 180 because 0 (inphase) would imply a resultant of 2z and 180 would give a resultant of 0.....

1 Answer
Nov 21, 2017

For the amplitude of the superposition of two waves to have the same amplitude as the original waves, their phase difference must be 120^o = (2\pi)/3120o=2π3 radiansradians

Explanation:

Consider the superposition of two sinusoidal waves of identical amplitude \psi_0ψ0, separated just by a phase shift \phiϕ:

The mathematical expressions for the waves \psi_1(x)ψ1(x) and \psi_2(x)ψ2(x) are:
\psi_1(x)=\psi_0\sin(kx); \qquad \qquad \psi_2(x)=\psi_0\sin(kx+\phi)

The mathematical expression for the superposition of these two waves is :
\psi_{12}(x) = [2\psi_0\cos(\phi/2)]\sin(kx + \phi/2) ...... (derived below)

This shows that the superposition is another sinusoidal wave of amplitude 2\psi_0\cos(\phi/2).
This amplitude is the same as \psi_0 for \phi = 120^o = (2\pi)/3 radians.

Derivation of the expression for the superposition of two waves:

\psi_{12}(x) = \psi_1(x) + \psi_2(x);
\psi_{12}(x) = \psi_0\sin(kx) + \psi_0\sin(kx + \phi)

Use the trigonometric identity (TI3) to expand \sin(kx+\phi) term,
\psi_{12}(x) = \psi_0\sin(kx) + \psi_0[\sin(kx)\cos\phi + \cos(kx)\sin\phi]
\psi_{12}(x) = \psi_0[\sin(kx)(1+\cos\phi) + \cos(kx)\sin\phi]

Use trigonometric identity (TI1) to rewrite the (1 + \cos\phi) term and the identity (TI2) to rewrite the \sin\phi term, in forms which would allow further simplification

\psi_{12}(x)= \psi_0[\sin(kx)(2\cos^2(\phi/2)) + \cos(kx)(2\sin(\phi/2)\cos(\phi/2))]

\psi_{12}(x) = 2\psi_0\cos(\phi/2)[\sin(kx)\cos(\phi/2) + \cos(kx)\sin(\phi/2)]
Use the trigonometric identity (TI3) again to simplify the terms inside the square bracket
\psi_{12}(x) = 2\psi_0\cos(\phi/2)\sin(kx + \phi/2)

Useful trigonometric identities:
[ 1 + \cos(a)] = 2\cos^2(a/2) ...... (TI1)
\sin(a) = 2\sin(a/2)\cos(a/2) ...... (TI2)
\sin(a+b) = \sin(a)\cos(b) + \cos(a)\sin(b) ...... (TI3)