The path difference between the two waves y_1 | vedclass

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(B) Given the two 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}{\lam

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

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

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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