The path difference between two waves,represe | vedclass

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Question

(B) Given equations are:$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} + \ph

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) Given equations are:$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)$We know that $\cos 	heta = \sin \left(	heta + \frac{\pi}{2}ight)$.Therefore,$y_2$ can be rewritten as:$y_2 = a_2 \sin \left(\omega t - \frac{2 \pi x}{\lambda} + \phi + \frac{\pi}{2}ight)$The phase difference $\Delta \Phi$ between the two waves is the difference in their arguments:$\Delta \Phi = \left(\omega t - \frac{2 \pi x}{\lambda} + \phi + \frac{\pi}{2}ight) - \left(\omega t - \frac{2 \pi x}{\lambda}ight) = \phi + \frac{\pi}{2}$The relationship between path difference $\Delta x$ and phase difference $\Delta \Phi$ is given by:$\Delta x = \frac{\lambda}{2 \pi} 	imes \Delta \Phi$Substituting the value of $\Delta \Phi$:$\Delta x = \frac{\lambda}{2 \pi} \left(\phi + \frac{\pi}{2}ight)$

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

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