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(C) Given equations are $x = a \sin(\omega t - \alpha)$ and $y = b \cos(\omega t - \alpha)$.We know that $\cos(	heta) = \sin(	heta + \pi/2)$.Therefo

Vedclass · Accepted Answer

$90^o$

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(C) Given equations are $x = a \sin(\omega t - \alpha)$ and $y = b \cos(\omega t - \alpha)$.We know that $\cos(	heta) = \sin(	heta + \pi/2)$.Therefore,we can rewrite the equation for $y$ as $y = b \sin(\omega t - \alpha + \pi/2)$.The phase of $x$ is $\phi_1 = \omega t - \alpha$.The phase of $y$ is $\phi_2 = \omega t - \alpha + \pi/2$.The phase difference $\Delta \phi = \phi_2 - \phi_1 = (\omega t - \alpha + \pi/2) - (\omega t - \alpha) = \pi/2$.Since $\pi/2$ radians is equal to $90^o$,the phase difference is $90^o$.

What is the phase difference between the simple harmonic motions $x = a \sin(\omega t - \alpha)$ and $y = b \cos(\omega t - \alpha)$?

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