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(D) Let the equations of the two waves be $y_1 = a \sin(\omega t - kx)$ and $y_2 = a \sin(\omega t - kx + \phi)$.When they superpose,the resultant wav

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$2\pi /3$

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(D) Let the equations of the two waves be $y_1 = a \sin(\omega t - kx)$ and $y_2 = a \sin(\omega t - kx + \phi)$.When they superpose,the resultant wave is $y = y_1 + y_2$.Using the trigonometric identity $\sin A + \sin B = 2 \sin(\frac{A+B}{2}) \cos(\frac{A-B}{2})$,we get:$y = 2a \cos(\frac{\phi}{2}) \sin(\omega t - kx + \frac{\phi}{2})$.The resultant amplitude is $A_R = |2a \cos(\frac{\phi}{2})|$.Given that the resultant amplitude is equal to the individual amplitude $a$,we have:$a = |2a \cos(\frac{\phi}{2})| \implies \cos(\frac{\phi}{2}) = \pm \frac{1}{2}$.Considering the magnitude,$\cos(\frac{\phi}{2}) = \frac{1}{2}$.Since $\cos(\frac{\pi}{3}) = \frac{1}{2}$,we have $\frac{\phi}{2} = \frac{\pi}{3}$.Therefore,the phase difference is $\phi = \frac{2\pi}{3}$.

If two waves of the same frequency and amplitude on superposition produce a resultant disturbance of the same amplitude,the waves differ in phase by

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