Question

Two waves are represented by the equations, $y_1=a\sin (\omega t-kx+{\varphi }_1)$ and $y_2=a\sin (\omega t-kx+{\varphi }_2)$. If the amplitude of the resultant wave remains equal to that of superimposing waves, then the phase difference between the original waves is

• A. $\frac{\pi }{6}$
• B. $\frac{\pi }{3}$
• C. $\frac{2\pi }{3}$
• D. $\frac{\pi }{2}$

Answer 1

The correct option is C $\frac{2\pi }{3}$

Here,

$y_1=a\sin (\omega t-kx+{\varphi }_1)$

$y_2=a\sin (\omega t-kx+{\varphi }_2)$

So, phase difference, $\Delta \varphi ={\varphi }_2-{\varphi }_1$

We know that, for the same frequency,

Resultant amplitude,

$A=\sqrt{{A_1}^2+{A_2}^2+2A_1{A}_2\cos (\Delta }\varphi )$

From the data given in the question we can write that ,

$a=\sqrt{{\left(a\right)}^2+{\left(a\right)}^2+2a^2\cos (\Delta }\varphi )$

$\Rightarrow a^2=2a^2+2a^2\cos \left(\Delta \varphi \right)$

$\Rightarrow \cos \left(\Delta \varphi \right)=\frac{-1}{2}\Rightarrow \Delta \varphi =\frac{2\pi }{3}$

Hence, option (c) is the correct answer.

Why this question ? This understand tests the understanding of superposition principle of waves.