Question

Two waves $y_1=A\sin (\omega t-kx)$ and $y_2=A\sin (\omega t-kx+\varphi )$ are superimposed such that the resultant amplitude of oscillation is $\sqrt{2}A$, then the value of $\varphi$ is

• A. $\frac{\pi }{2}$
• B. $\frac{\pi }{4}$
• C. $\frac{\pi }{3}$
• D. $\frac{\pi }{6}$

Answer 1

The correct option is A $\frac{\pi }{2}$

The resultant amplitude of oscillation is given by,
$A_r=\sqrt{A_1^2+A_2^2+2A_1{A}_2\text{cos}\Delta \varphi }$
where $\Delta \varphi =\varphi$ is the phase difference between given waves.
Given:
$A_r=\sqrt{2}A$
$\Rightarrow \sqrt{2}A=\sqrt{A^2+A^2+2A^2\text{cos}\Delta \varphi }$
$\Rightarrow 2A^2=2A^2(1+\text{cos}\varphi )$
$\Rightarrow 1+\text{cos}\varphi =1$
$\Rightarrow \text{cos}\varphi =0$
$\Rightarrow \varphi =\frac{\pi }{2}\text{or}\frac{3\pi }{2}$
$\therefore$Option $\left(a\right)$ is correct.