Question

Two sinusoidal waves given below are superposed
$y_{1}=A \sin \left(k x-\omega t+\frac{\pi}{6}\right), y_{2}=A \sin \left(k x-\omega t-\frac{\pi}{6}\right)$
The equation of resultant wave is

• A. $y=\frac{A}{\sqrt{3}} \sin (k x-\omega t)$
• B. $y=A \sqrt{3} \sin (k x-\omega t)$
• C. $y=A \sqrt{3} \sin \left(k x-\omega t-\frac{\pi}{3}\right)$
• D. $y=\frac{A}{\sqrt{3}} \sin \left(k x-\omega t-\frac{\pi}{3}\right)$

Answer 1

Answer: (B)

$y^{\prime}=y_{1}+y_{2}$
or $y'=A \sin \left(k x-\omega t+\frac{\pi}{6}\right)+A \sin \left(k x-\omega t-\frac{\pi}{6}\right)$
or $y'=2 A \sin (k x-\omega t) \cdot \cos \left(\frac{\left(\frac{\pi}{6}+\frac{\pi}{6}\right)}{2}\right)$
or $y'=2 A \frac{\sqrt{3}}{2} \sin (k x-\omega t)$
$\therefore y'=A \sqrt{3} \sin (k x-\omega t)$