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(D) The resultant wave is given by $y = y_1 + y_2 = A \sin(kx - \omega t + \phi)$.Using the phasor addition method for two waves with amplitudes $A_1

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$[2\sqrt{3}, \frac{\pi}{6}]$

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(D) The resultant wave is given by $y = y_1 + y_2 = A \sin(kx - \omega t + \phi)$.Using the phasor addition method for two waves with amplitudes $A_1 = 4$ and $A_2 = 2$ and a phase difference $\Delta\phi = \frac{2\pi}{3} = 120^{\circ}$,the resultant amplitude $A$ is:$A = \sqrt{A_1^2 + A_2^2 + 2A_1A_2 \cos(\Delta\phi)}$$A = \sqrt{4^2 + 2^2 + 2(4)(2) \cos(120^{\circ})}$$A = \sqrt{16 + 4 + 16(-0.5)} = \sqrt{20 - 8} = \sqrt{12} = 2\sqrt{3}$.The phase angle $\phi$ of the resultant wave is given by:$	an \phi = \frac{A_2 \sin(\Delta\phi)}{A_1 + A_2 \cos(\Delta\phi)}$$	an \phi = \frac{2 \sin(120^{\circ})}{4 + 2 \cos(120^{\circ})} = \frac{2(\frac{\sqrt{3}}{2})}{4 + 2(-0.5)} = \frac{\sqrt{3}}{4 - 1} = \frac{\sqrt{3}}{3} = \frac{1}{\sqrt{3}}$.Thus,$\phi = \frac{\pi}{6}$.

The amplitude and phase of a wave that is formed by the superposition of two harmonic travelling waves,$y_1(x, t) = 4 \sin(kx - \omega t)$ and $y_2(x, t) = 2 \sin(kx - \omega t + \frac{2\pi}{3})$,are (Take the angular frequency of initial waves same as $\omega$):

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