Two equations of two S.H.M. are y = asin(omeg | vedclass

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(C) The first equation is $y_1 = a\sin(\omega t - \alpha)$.Using the trigonometric identity $\sin(	heta) = \cos(	heta - 90^\circ)$,we can rewrite th

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$90$

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(C) The first equation is $y_1 = a\sin(\omega t - \alpha)$.Using the trigonometric identity $\sin(	heta) = \cos(	heta - 90^\circ)$,we can rewrite this as:$y_1 = a\cos(\omega t - \alpha - 90^\circ)$.The second equation is $y_2 = b\cos(\omega t - \alpha)$.Comparing the arguments of the cosine functions,the phase of the first equation is $\phi_1 = \omega t - \alpha - 90^\circ$ and the phase of the second equation is $\phi_2 = \omega t - \alpha$.The phase difference is $\Delta\phi = |\phi_2 - \phi_1| = |(\omega t - \alpha) - (\omega t - \alpha - 90^\circ)| = 90^\circ$.

Two equations of two $S.H.M.$ are $y = a\sin(\omega t - \alpha)$ and $y = b\cos(\omega t - \alpha)$. The phase difference between the two is .... $^\circ$.

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