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(B) Given the two wave equations: $y_1 = a \sin(\omega t)$ $y_2 = b \cos(\omega t)$ We can rewrite the second equation in terms of the sine function u

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$\frac{\pi}{2}$

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(B) Given the two wave equations: $y_1 = a \sin(\omega t)$ $y_2 = b \cos(\omega t)$ We can rewrite the second equation in terms of the sine function using the trigonometric identity $\cos(	heta) = \sin(	heta + \frac{\pi}{2})$: $y_2 = b \sin(\omega t + \frac{\pi}{2})$ Comparing both equations with the general form $y = A \sin(\omega t + \phi)$,we find the phase of the first wave is $\phi_1 = 0$ and the phase of the second wave is $\phi_2 = \frac{\pi}{2}$. The phase difference is $\Delta \phi = \phi_2 - \phi_1 = \frac{\pi}{2} - 0 = \frac{\pi}{2}$.

The two waves represented by $y_1 = a \sin(\omega t)$ and $y_2 = b \cos(\omega t)$ have a phase difference of

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