The path difference between two waves Y_1 = a | vedclass

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(B) Given waves are $Y_1 = a_1 \sin \left(\omega t - \frac{2 \pi x}{\lambda}\right)$ and $Y_2 = a_2 \cos \left(\omega t - \frac{2 \pi x}{\lambda} + \p

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

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(B) Given waves are $Y_1 = a_1 \sin \left(\omega t - \frac{2 \pi x}{\lambda}ight)$ and $Y_2 = a_2 \cos \left(\omega t - \frac{2 \pi x}{\lambda} + \phiight)$.To compare the phases,we convert the cosine function to sine: $\cos(	heta) = \sin(	heta + \frac{\pi}{2})$.Thus,$Y_2 = a_2 \sin \left(\omega t - \frac{2 \pi x}{\lambda} + \phi + \frac{\pi}{2}ight)$.The phase difference $\delta$ is the difference between the arguments of the sine functions: $\delta = \left(\omega t - \frac{2 \pi x}{\lambda} + \phi + \frac{\pi}{2}ight) - \left(\omega t - \frac{2 \pi x}{\lambda}ight) = \phi + \frac{\pi}{2}$.The relationship between path difference $\Delta x$ and phase difference $\delta$ is $\Delta x = \frac{\lambda}{2 \pi} \delta$.Substituting the value of $\delta$,we get $\Delta x = \frac{\lambda}{2 \pi} \left(\phi + \frac{\pi}{2}ight)$.

The path difference between 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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