Two waves have equations x_1 = a sin(omega t | vedclass

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(B) The resultant wave is $x' = x_1 + x_2 = a[\sin(\omega t + \phi_1) + \sin(\omega t + \phi_2)]$.Using the trigonometric identity $\sin A + \sin B =

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$\frac{2\pi}{3}$

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(B) The resultant wave is $x' = x_1 + x_2 = a[\sin(\omega t + \phi_1) + \sin(\omega t + \phi_2)]$.Using the trigonometric identity $\sin A + \sin B = 2 \sin(\frac{A+B}{2}) \cos(\frac{A-B}{2})$,we get:$x' = 2a \sin(\omega t + \frac{\phi_1 + \phi_2}{2}) \cos(\frac{\phi_1 - \phi_2}{2})$.The amplitude of the resultant wave is $A_{res} = |2a \cos(\frac{\phi_1 - \phi_2}{2})|$.Given that the resultant amplitude is equal to the amplitude of the individual waves $(a)$:$|2a \cos(\frac{\phi_1 - \phi_2}{2})| = a$.$\cos(\frac{\phi_1 - \phi_2}{2}) = \frac{1}{2}$.Let $\Delta\phi = \phi_1 - \phi_2$. Then $\cos(\frac{\Delta\phi}{2}) = \frac{1}{2}$.$\frac{\Delta\phi}{2} = \frac{\pi}{3}$.$\Delta\phi = \frac{2\pi}{3}$.

Two waves have equations $x_1 = a \sin(\omega t + \phi_1)$ and $x_2 = a \sin(\omega t + \phi_2)$. If in the resultant wave the frequency and amplitude remain equal to the amplitude of the superimposing waves,the phase difference between them is ........

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