Two sinusoidal waves given below are superpos | vedclass

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(B) The resultant wave is given by the principle of superposition: $y = y_1 + y_2$.Substituting the given equations:$y = A \sin \left(kx - \omega t +

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$y = A \sqrt{3} \sin (kx - \omega t)$

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(B) The resultant wave is given by the principle of superposition: $y = y_1 + y_2$.Substituting the given equations:$y = A \sin \left(kx - \omega t + \frac{\pi}{6}\right) + A \sin \left(kx - \omega t - \frac{\pi}{6}\right)$.Using the trigonometric identity $\sin C + \sin D = 2 \sin \left(\frac{C+D}{2}\right) \cos \left(\frac{C-D}{2}\right)$:Let $C = kx - \omega t + \frac{\pi}{6}$ and $D = kx - \omega t - \frac{\pi}{6}$.Then $\frac{C+D}{2} = kx - \omega t$ and $\frac{C-D}{2} = \frac{\pi}{6}$.Therefore,$y = 2A \sin(kx - \omega t) \cos\left(\frac{\pi}{6}\right)$.Since $\cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}$,we get:$y = 2A \sin(kx - \omega t) \cdot \frac{\sqrt{3}}{2} = A \sqrt{3} \sin(kx - \omega t)$.

Two sinusoidal waves given below are superposed:
$y_1 = A \sin \left(kx - \omega t + \frac{\pi}{6}\right), \quad y_2 = A \sin \left(kx - \omega t - \frac{\pi}{6}\right)$
The equation of the resultant wave is:

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Two sinusoidal waves given below are superposed:$y_1 = A \sin \left(kx - \omega t + \frac{\pi}{6}\right), \quad y_2 = A \sin \left(kx - \omega t - \frac{\pi}{6}\right)$The equation of the resultant wave is: