# If two waves, each of amplitude z, produce a resultant wave of amplitude z, 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.....

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 ${120}^{o} = \frac{2 \setminus \pi}{3}$ $r a \mathrm{di} a n s$

#### Explanation:

Consider the superposition of two sinusoidal waves of identical amplitude $\setminus {\psi}_{0}$, separated just by a phase shift $\setminus \phi$:

The mathematical expressions for the waves $\setminus {\psi}_{1} \left(x\right)$ and $\setminus {\psi}_{2} \left(x\right)$ 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 :
$\setminus {\psi}_{12} \left(x\right) = \left[2 \setminus {\psi}_{0} \setminus \cos \left(\setminus \frac{\phi}{2}\right)\right] \setminus \sin \left(k x + \setminus \frac{\phi}{2}\right)$ ...... (derived below)

This shows that the superposition is another sinusoidal wave of amplitude $2 \setminus {\psi}_{0} \setminus \cos \left(\setminus \frac{\phi}{2}\right)$.
This amplitude is the same as $\setminus {\psi}_{0}$ for $\setminus \phi = {120}^{o} = \frac{2 \setminus \pi}{3}$ $r a \mathrm{di} a n s$.

Derivation of the expression for the superposition of two waves:

\psi_{12}(x) = \psi_1(x) + \psi_2(x);
$\setminus {\psi}_{12} \left(x\right) = \setminus {\psi}_{0} \setminus \sin \left(k x\right) + \setminus {\psi}_{0} \setminus \sin \left(k x + \setminus \phi\right)$

Use the trigonometric identity (TI3) to expand $\setminus \sin \left(k x + \setminus \phi\right)$ term,
$\setminus {\psi}_{12} \left(x\right) = \setminus {\psi}_{0} \setminus \sin \left(k x\right) + \setminus {\psi}_{0} \left[\setminus \sin \left(k x\right) \setminus \cos \setminus \phi + \setminus \cos \left(k x\right) \setminus \sin \setminus \phi\right]$
$\setminus {\psi}_{12} \left(x\right) = \setminus {\psi}_{0} \left[\setminus \sin \left(k x\right) \left(1 + \setminus \cos \setminus \phi\right) + \setminus \cos \left(k x\right) \setminus \sin \setminus \phi\right]$

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

$\setminus {\psi}_{12} \left(x\right) = \setminus {\psi}_{0} \left[\setminus \sin \left(k x\right) \left(2 \setminus {\cos}^{2} \left(\setminus \frac{\phi}{2}\right)\right) + \setminus \cos \left(k x\right) \left(2 \setminus \sin \left(\setminus \frac{\phi}{2}\right) \setminus \cos \left(\setminus \frac{\phi}{2}\right)\right)\right]$

$\setminus {\psi}_{12} \left(x\right) = 2 \setminus {\psi}_{0} \setminus \cos \left(\setminus \frac{\phi}{2}\right) \left[\setminus \sin \left(k x\right) \setminus \cos \left(\setminus \frac{\phi}{2}\right) + \setminus \cos \left(k x\right) \setminus \sin \left(\setminus \frac{\phi}{2}\right)\right]$
Use the trigonometric identity (TI3) again to simplify the terms inside the square bracket
$\setminus {\psi}_{12} \left(x\right) = 2 \setminus {\psi}_{0} \setminus \cos \left(\setminus \frac{\phi}{2}\right) \setminus \sin \left(k x + \setminus \frac{\phi}{2}\right)$

Useful trigonometric identities:
$\left[1 + \setminus \cos \left(a\right)\right] = 2 \setminus {\cos}^{2} \left(\frac{a}{2}\right)$ ...... (TI1)
$\setminus \sin \left(a\right) = 2 \setminus \sin \left(\frac{a}{2}\right) \setminus \cos \left(\frac{a}{2}\right)$ ...... (TI2)
$\setminus \sin \left(a + b\right) = \setminus \sin \left(a\right) \setminus \cos \left(b\right) + \setminus \cos \left(a\right) \setminus \sin \left(b\right)$ ...... (TI3)