Wave Mechanics Question 539
Question: Starting from the origin a body oscillates simple harmonically with a period of 2 s. After what time will its kinetic energy be 75% of the total energy?
Options:
A) $ \frac{1}{6}s $
B) $ \frac{1}{4}s $
C) $ \frac{1}{3}s $
D) $ \frac{1}{12}s $
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Answer:
Correct Answer: A
Solution:
[a] $ K.E.=\frac{1}{2}ma^{2}{{\omega }^{2}}{{\cos }^{2}}\omega t $ and $ T.E.=\frac{1}{2}ma^{2}{{\omega }^{2}} $ Given $ K.E.=0.75T.E. $
$ \Rightarrow ,0.75={{\cos }^{2}}\omega t\Rightarrow \omega t=\frac{\pi }{6} $
$ \Rightarrow ,t=\frac{\pi }{6\times \omega }\Rightarrow t=\frac{\pi \times 2}{6\times 2\pi }\Rightarrow t=\frac{1}{6}s $
BETA
