The phase difference between the following tw | vedclass

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(B) Given equations are:$y_1 = a \sin(\omega t - kx)$$y_2 = b \cos(\omega t - kx + \frac{\pi}{3})$To find the phase difference,we convert the cosine f

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$\frac{5\pi}{6}$

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(B) Given equations are:$y_1 = a \sin(\omega t - kx)$$y_2 = b \cos(\omega t - kx + \frac{\pi}{3})$To find the phase difference,we convert the cosine function into a sine function using the identity $\cos(	heta) = \sin(	heta + \frac{\pi}{2})$.$y_2 = b \sin(\omega t - kx + \frac{\pi}{3} + \frac{\pi}{2})$$y_2 = b \sin(\omega t - kx + \frac{2\pi + 3\pi}{6})$$y_2 = b \sin(\omega t - kx + \frac{5\pi}{6})$The phase of $y_1$ is $\phi_1 = \omega t - kx$.The phase of $y_2$ is $\phi_2 = \omega t - kx + \frac{5\pi}{6}$.The phase difference $\Delta\phi = \phi_2 - \phi_1 = \frac{5\pi}{6}$.

The phase difference between the following two waves $y_1$ and $y_2$ is:
$y_1 = a \sin(\omega t - kx)$
$y_2 = b \cos(\omega t - kx + \frac{\pi}{3})$

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The phase difference between the following two waves $y_1$ and $y_2$ is:$y_1 = a \sin(\omega t - kx)$$y_2 = b \cos(\omega t - kx + \frac{\pi}{3})$