Wave Mechanics Question 542
Question: The phase difference between the instantaneous velocity and acceleration of a particle executing simple harmonic motion is
Options:
A) $ \pi $
B) 0.707 $ \pi $
C) zero
D) 0.5 $ \pi $
Show Answer
Answer:
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 $
BETA
