If $x = a\sin \left( {\omega t + \frac{\pi }{6}} \right)$ and $x' = a\cos \omega t$, then what is the phase difference between the two waves
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(a) $x = a\sin \left( {\omega t + \frac{\pi }{6}} \right)$ and $x' = a\cos \omega t = a\sin \,\left( {\omega \,t + \frac{\pi }{2}} \right)$
$\therefore \Delta \phi = \left( {\omega t + \frac{\pi }{2}} \right) - \left( {\omega t + \frac{\pi }{6}} \right) = \frac{\pi }{3}$
$\therefore \Delta \phi = \left( {\omega t + \frac{\pi }{2}} \right) - \left( {\omega t + \frac{\pi }{6}} \right) = \frac{\pi }{3}$
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