Question

Two waves represented by equation $y_1=A\sin (\omega t-kx+{\varphi }_1)$ and $y_2=A\sin (\omega t-kx+{\varphi }_2)$ are superimposed such that the amplitude of resultant wave is $A$. Find the phase difference between them.
Given: $({\varphi }_1>{\varphi }_2)$

• A. ${90}^{\circ }$
• B. ${120}^{\circ }$
• C. ${60}^{\circ }$
• D. ${135}^{\circ }$

Answer 1

The correct option is B ${120}^{\circ }$

Let phase difference $\Delta \varphi ={\varphi }_1-{\varphi }_2$
We know resultant amplitude is,
$A_r=\sqrt{A_1^2+A_2^2+2A_1{A}_2\cos \Delta \varphi }$
Given: $A_r=A$, substituting it in above formula gives
$\Rightarrow A=\sqrt{A^2+A^2+2A^2\text{cos}\Delta \varphi }$
$\Rightarrow A=\sqrt{2A^2(1+\text{cos}\Delta \varphi )}$
$\Rightarrow \frac{1}{\sqrt{2}}=\sqrt{1+\text{cos}\Delta \varphi }$
Squaring both sides,
$\Rightarrow 1+\text{cos}\Delta \varphi =\frac{1}{2}$
$\text{cos}\Delta \varphi =\frac{-1}{2}$
The negative value of $\cos \Delta \varphi$ indicates that it must be an angle of either second or third quadrant.
$\Rightarrow \Delta \varphi =\pi -\frac{\pi }{3}=\frac{2\pi }{3}\text{or}\Delta \varphi =\pi +\frac{\pi }{3}=\frac{4\pi }{3}$
$\therefore \Delta \varphi =\frac{2\pi }{3}={120}^{\circ }$, is the correct answer.