Wave Mechanics Question 536
Question: Part of a simple harmonic motion is graphed in the figure, where y is the displacement from the mean position. The correct equation describing this S.H. M. is
Options:
A) $ y=4\cos (0.6t) $
B) $ y=2\sin ( \frac{10}{3}t+\frac{\pi }{2} ) $
C) $ y=4\sin ( \frac{10}{3}t+\frac{\pi }{2} ) $
D) $ y=2\cos ( \frac{10}{3}t+\frac{\pi }{2} ) $
Show Answer
Answer:
Correct Answer: B
Solution:
[b] From the figure, $ \frac{T}{2}=0.3\pi , $
$ \
Therefore T=0.6\pi ,and,{\phi _{0}}=-\frac{\pi }{2} $ Thus $ y=A,\sin (\omega t+{\phi _{0}})=2\sin ( \frac{2\pi }{0.6\pi }t+-\frac{\pi }{2} ) $
$ =2\sin ( \frac{10t}{3}-\frac{\pi }{2} ) $ .
BETA
