$y_1=A \sin \left(k x-\omega t+\frac{\pi}{6}\right), \quad y_2=A \sin \left(k x-\omega t-\frac{\pi}{6}\right)$
પરિણામી તરંગનું સમીકરણ ક્યું છે.
\(y=y_1+y_2\)
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 \(y^{\prime}=2 A \frac{\sqrt{3}}{2} \sin (k x-\omega t)\)
\(\therefore y^{\prime}=A \sqrt{3} \sin (k x-\omega t)\)
(હવામાં ધ્વનિની ઝડપ $340\, {ms}^{-1}$)