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(A) Given the displacement equations:$y_1 = a_1 \sin \left(\frac{2 \pi x}{\lambda} - \omega t\right) \quad \dots(i)$$y_2 = a_2 \cos \left(\frac{2 \pi

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

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(A) Given the displacement equations:$y_1 = a_1 \sin \left(\frac{2 \pi x}{\lambda} - \omega tight) \quad \dots(i)$$y_2 = a_2 \cos \left(\frac{2 \pi x}{\lambda} - \omega t + \phiight) \quad \dots(ii)$To compare the phases,we convert the cosine function into a sine function using the identity $\cos(	heta) = \sin(	heta + \frac{\pi}{2})$:$y_2 = a_2 \sin \left(\frac{2 \pi x}{\lambda} - \omega t + \phi + \frac{\pi}{2}ight) \quad \dots(iii)$Comparing the phase of $y_1$ (which is $\frac{2 \pi x}{\lambda} - \omega t$) with the phase of $y_2$ (which is $\frac{2 \pi x}{\lambda} - \omega t + \phi + \frac{\pi}{2}$),the phase difference is:$\Delta \phi = \left(\frac{2 \pi x}{\lambda} - \omega t + \phi + \frac{\pi}{2}ight) - \left(\frac{2 \pi x}{\lambda} - \omega tight)$$\Delta \phi = \phi + \frac{\pi}{2}$

Two simple harmonic progressive waves have displacements $y_1 = a_1 \sin \left(\frac{2 \pi x}{\lambda} - \omega t\right)$ and $y_2 = a_2 \cos \left(\frac{2 \pi x}{\lambda} - \omega t + \phi\right)$. What is the phase difference between the two waves?

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