The path difference between the two waves y_1 | vedclass

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(C) Given the wave equations:$y_1 = a_1 \sin \left( \omega t - \frac{2\pi x}{\lambda} \right)$$y_2 = a_2 \cos \left( \omega t - \frac{2\pi x}{\lambda}

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$\frac{\lambda}{2\pi} \left( \phi + \frac{\pi}{2} \right)$

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(C) Given the wave equations:$y_1 = a_1 \sin \left( \omega t - \frac{2\pi x}{\lambda} \right)$$y_2 = a_2 \cos \left( \omega t - \frac{2\pi x}{\lambda} + \phi \right)$To compare the phases,convert the cosine function to sine:$y_2 = a_2 \sin \left( \omega t - \frac{2\pi x}{\lambda} + \phi + \frac{\pi}{2} \right)$The phase difference $\Delta \phi$ is the difference between the arguments of the sine functions:$\Delta \phi = \left( \omega t - \frac{2\pi x}{\lambda} + \phi + \frac{\pi}{2} \right) - \left( \omega t - \frac{2\pi x}{\lambda} \right) = \phi + \frac{\pi}{2}$The relationship between path difference $\Delta x$ and phase difference $\Delta \phi$ is given by $\Delta \phi = \frac{2\pi}{\lambda} \Delta x$.Therefore,$\Delta x = \frac{\lambda}{2\pi} \Delta \phi = \frac{\lambda}{2\pi} \left( \phi + \frac{\pi}{2} \right)$.

The path difference between the two waves $y_1 = a_1 \sin \left( \omega t - \frac{2\pi x}{\lambda} \right)$ and $y_2 = a_2 \cos \left( \omega t - \frac{2\pi x}{\lambda} + \phi \right)$ is

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