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(B) Given the equations for the two waves are $x_1 = A \sin \left(\omega t + \frac{\pi}{6}\right)$ and $x_2 = A \cos \omega t$.To find the phase diffe

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$\frac{\pi}{3}$

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(B) Given the equations for the two waves are $x_1 = A \sin \left(\omega t + \frac{\pi}{6}ight)$ and $x_2 = A \cos \omega t$.To find the phase difference,we express both waves in terms of the same trigonometric function (sine or cosine).We know that $\cos 	heta = \sin \left(	heta + \frac{\pi}{2}ight)$.Therefore,$x_2 = A \sin \left(\omega t + \frac{\pi}{2}ight)$.Now,the phase of $x_1$ is $\phi_1 = \omega t + \frac{\pi}{6}$ and the phase of $x_2$ is $\phi_2 = \omega t + \frac{\pi}{2}$.The phase difference $\Delta \phi = \phi_2 - \phi_1 = \left(\omega t + \frac{\pi}{2}ight) - \left(\omega t + \frac{\pi}{6}ight)$.$\Delta \phi = \frac{\pi}{2} - \frac{\pi}{6} = \frac{3\pi - \pi}{6} = \frac{2\pi}{6} = \frac{\pi}{3}$.

Two waves are represented by: $x_1 = A \sin \left(\omega t + \frac{\pi}{6}\right)$ and $x_2 = A \cos \omega t$. Then the phase difference between them is

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