MCQ
Two sinusoidal waves given below are superposed

$y_1=A \sin \left(k x-\omega t+\frac{\pi}{6}\right), \quad y_2=A \sin \left(k x-\omega t-\frac{\pi}{6}\right)$

The equation of resultant wave is

  • A
    $y=\frac{A}{\sqrt{3}} \sin (k x-\omega t)$
  • $y=A \sqrt{3} \sin (k x-\omega t)$
  • 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

Correct option: B.
$y=A \sqrt{3} \sin (k x-\omega t)$
b
(b)

$y=y_1+y_2$

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 $y^{\prime}=2 A \frac{\sqrt{3}}{2} \sin (k x-\omega t)$

$\therefore y^{\prime}=A \sqrt{3} \sin (k x-\omega t)$

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