A) 0°
B) a°
C) 90°
D) 180°
Correct Answer: C
Solution :
\[y=a\sin (\omega \,t-\alpha )=a\cos \left( \omega \,t-\alpha -\frac{\pi }{2} \right)\] Another equation is given \[y=\cos (\omega \,t-\alpha )\] So, there exists a phase difference of \[\frac{\pi }{2}=90{}^\circ \]You need to login to perform this action.
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