Question

Two waves are passing through a region in the same direction at the same time. If the equation of these waves are
$y_1=\alpha \sin \frac{2\pi }{\lambda }(vt-x)$
and $y_2=\beta \sin \frac{2\pi }{\lambda }\left[\right(vt-x)+\varphi ]$,
then the amplitude of the resultant wave for $\varphi =\frac{\lambda }{2}$ is

• A. $|\alpha -\beta |$
• B. $\alpha +\beta$
• C. $\sqrt{{\alpha }^2+{\beta }^2}$
• D. $\sqrt{{\alpha }^2+{\beta }^2+2\alpha \beta cosx}$

Answer 1

The correct option is A $|\alpha -\beta |$

Let ${\theta }_{1}and{\theta }_2$ represent the phase angles of the first and second waves. Then
${\theta }_1=\frac{2\pi }{\lambda }(vt-x)$
and ${\theta }_2=\frac{2\pi }{\lambda }\left[\right(vt-x)+\varphi ]$
But $\varphi =\frac{\lambda }{2}$
So, ${\theta }_2-{\theta }_1=\pi$

Hence, phase difference $\theta =\pi ,$ so amplitude of resultant wave
$A_{res}=\sqrt{{\alpha }^2+{\beta }^2+2\alpha \beta \cos \theta }$
$=\sqrt{{\alpha }^2+{\beta }^2+2\alpha \beta \cos \pi }=\sqrt{(\alpha -\beta {)}^2}$
Or , Amplitude of Resultant = $|\propto -\beta |$