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Question: Two equations of two S.H.M. are \(y = a\sin(\omega t - \alpha)\) and\(y = b\cos(\omega t - \alpha)\)...

Two equations of two S.H.M. are y=asin⁡(ωt−α)y = a\sin(\omega t - \alpha) andy=bcos⁡(ωt−α)y = b\cos(\omega t - \alpha). The phase difference between the two is

A

0°

B

α°

C

90°

D

180°

Answer

90°

Explanation

Solution

y=asin⁡(ωt−α)=acos⁡(ωt−α−π2)y = a\sin(\omega t - \alpha) = a\cos\left( \omega t - \alpha - \frac{\pi}{2} \right)

Another equation is given y=cos⁡(ωt−α)y = \cos(\omega t - \alpha)

So, there exists a phase difference of π2=90∘\frac{\pi}{2} = 90{^\circ}