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

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(B) The resultant amplitude $A$ of two superimposing waves with equal amplitude $a$ is given by the formula: $A^2 = a^2 + a^2 + 2a^2 \cos \phi$,where

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

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(B) The resultant amplitude $A$ of two superimposing waves with equal amplitude $a$ is given by the formula: $A^2 = a^2 + a^2 + 2a^2 \cos \phi$,where $\phi$ is the phase difference.Given that the resultant amplitude $A$ is equal to the amplitude of the individual waves,$A = a$.Substituting this into the equation: $a^2 = a^2 + a^2 + 2a^2 \cos \phi$.$a^2 = 2a^2 + 2a^2 \cos \phi$.$-a^2 = 2a^2 \cos \phi$.$\cos \phi = -\frac{1}{2}$.Since $\cos(120^\circ) = -\frac{1}{2}$,the phase difference $\phi = \frac{2\pi}{3}$ radians.

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

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