Physics - Sound Wave Question with Solution | TestHub

The phase difference $\varphi$ is related to the path difference $\Delta x$ by the relation:

$\varphi =\frac{2\pi }{\lambda }\Delta x$

Given $\Delta x=\frac{\lambda }{3}$, we get:

$\varphi =\frac{2\pi }{\lambda }\times \frac{\lambda }{3}=\frac{2\pi }{3}=120^{\circ }$

The resultant amplitude $A_R$ of two interfering waves is given by:

$A_R=\sqrt{A_1^2+A_2^2+2A_1{A}_2\cos \varphi }$

Substituting $A_1=A_2=A$ and $\varphi =120^{\circ }$:

$A_R=\sqrt{A^2+A^2+2A^2\cos 120^{\circ }}$

Since $\cos 120^{\circ }=-0.5$, we have:

$A_R=\sqrt{2A^2+2A^2(-0.5)}$

$A_R=\sqrt{2A^2-A^2}=\sqrt{A^2}=A$

Answer: A