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 .... $^o$
A$0$
B$\alpha$
C$90$
D$180$
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C$90$
c (c) $y = a\sin (\omega \,t - \alpha ) = a\cos \left( {\omega \,t - \alpha - \frac{\pi }{2}} \right)$
Another equation is given $y = \cos (\omega \,t - \alpha )$
So, there exists a phase difference of $\frac{\pi }{2} = 90^\circ $
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