Wave Mechanics Question 555
Question: Two simple harmonic motions are represented by the equations $ y _{1}=0.1\sin ( 100\pi t+\frac{\pi }{3} ) $ and $ y _{2}=0.1\cos \pi t $ . The phase difference of the velocity of particle 1 with respect to the velocity of particle 2 is
Options:
A) $ \frac{\pi }{3} $
B) $ \frac{-\pi }{6} $
C) $ \frac{\pi }{6} $
D) $ \frac{-\pi }{3} $
Show Answer
Answer:
Correct Answer: B
Solution:
[b] $ v _{1}=\frac{dy _{1}}{dt}=0.1\times 100\pi \cos ( 100\pi t+\frac{\pi }{3} ) $
$ v _{2}=\frac{dy _{2}}{dt}=-0.1\pi \sin \pi t $
$ =0.1\pi \cos ( \pi t+\frac{\pi }{2} ) $
$ \
Therefore Phase,diff.,={\phi _{1}}-{\phi _{2}}=\frac{\pi }{3}-\frac{\pi }{2}=\frac{2\pi -3\pi }{6}=\frac{\pi }{6} $
BETA
