A) \[\pi \]
B) 0.707\[\pi \]
C) zero
D) 0.5\[\pi \]
Correct Answer: D
Solution :
[d] Let\[y=A\,\sin \omega t\] \[{{v}_{inst}}=\frac{dy}{dt}=A\omega \,\cos \,\omega t=A\omega \,\sin (\omega t+\pi /2)\] Acceleration=\[=-A{{\omega }^{2}}\sin \,\omega t=A{{\omega }^{2}}\sin (\pi +\omega t)\] \[\therefore \,\phi =\frac{\pi }{2}=0.5\pi \]
You need to login to perform this action.
You will be redirected in
3 sec
