A) \[\frac{3\pi }{2}\text{rad}\]
B) \[\frac{\pi }{2}\text{rad}\]
C) zero
D) \[\pi \text{ rad}\]
Correct Answer: D
Solution :
[d] If \[y=A\,\,\sin \,\omega t\] |
then \[v=\frac{dy}{dt}\] |
\[v=A\,\omega \,\,\cos \,\omega t\] |
\[a=\frac{dv}{dt}\] |
\[a=-A\,{{\omega }^{2}}\sin \,(\omega t)\] |
\[a=A\,{{\omega }^{2}}\sin \,(\omega t+\pi )\] |
So phase difference between displacement and acceleration is \[\pi \]. |
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