Question

Two waves are represented $y_1=A\sin (kx-\omega t)$ and $y_2=A\cos (kx-\omega t)$. The amplitude of resultant wave is?

• A. $4A$
• B. $2A$
• C. $\sqrt{2}A$
• D. A

Answer 1

The correct option is C $\sqrt{2}A$

Given, $y_1=A\sin (kx-\omega t)$
$y_2=A\cos (kx-\omega t)$
or $y_2=A\sin \left(kx-\omega t+\frac{\pi }{2}\right)$
Phase difference of two waves $=\frac{\pi }{2}$
$\because$ Resultant amplitude
$R=\sqrt{A^2+A^2+2AA\cos \varphi }$
$=\sqrt{A^2+A^2+2A^2\cos \frac{\pi }{2}}$
$=\sqrt{2A^2}$ $(\because \cos \frac{\pi }{2}=0)$
$R=\sqrt{2}A$.y1=Asin⁡(kx−ωt)