A) 0
B) \[\pi /2\]
C) \[\pi \]
D) None of these
Correct Answer: B
Solution :
Let \[\operatorname{x} - a sin\,\,\omega t\] \[v=\frac{dx}{dt}=a\omega \,cos\,\,\omega t\,\,=\,\,a\omega \,\sin \,\left( \omega t+\frac{\pi }{2} \right)\] \[a=\frac{{{d}^{2}}x}{d{{t}^{2}}}\] \[= \,\,- a{{\omega }^{2}}\,sin\,\omega t\] \[\therefore \] Phase difference between velocity and \[acceleration=\pi /2.\]You need to login to perform this action.
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