Question

Two waves of the same frequency and same amplitude are superimposed to produce a resultant disturbance wave of the same amplitude. What is the phase difference between the two original waves?

Answer 1

Given that,
Frequency of first wave = Frequency of second wave.
Amplitude of first wave = Amplitude of second wave $=A$
Let $\varphi$ is the phase difference between the two waves, then the resultant amplitude$\left(A_R\right)$ of the wave is given by
$A_R=\sqrt{A_1^2+A_2^2+2A_1{A}_2\cos \varphi }$
Substitute $A_1=A_2=A$
$\Rightarrow A_R=\sqrt{A^2+A^2+2A^2\cos \varphi }$
$\Rightarrow A_R=A=\sqrt{2A^2(1+\cos \varphi )}$
$A=\sqrt{4A^2{\cos }^2\frac{\varphi }{2}}$
$A=2A\cos \frac{\varphi }{2}$
$\cos \frac{\varphi }{2}=\frac{1}{2}$
$\cos \frac{\varphi }{2}=\cos \frac{\pi }{3}$
$\varphi =\frac{2\pi }{3}$

Why this question? Bottom lines: Always remember that if two wave of same frequency and different amplitude is superpose then the resultant amplitude after superposition of wave is $A_R=\sqrt{{A_1}^2+{A_2}^2+2A_1{A}_{{}^2}\cos \varphi }$ where $\varphi$ is phase difference between wave.