Wave Mechanics Question 567
Question: A particle of mass 1 kg is placed in a potential field. Its potential energy is given by $ U=10x^{2}+5 $ .The frequency of oscillations of the particle is given by
Options:
A) $ (\sqrt{10}) $
B) $ (\sqrt{5}) $
C) $ ( \sqrt{\frac{10}{\pi }} ) $
D) $ ( \frac{\sqrt{5}}{\pi } ) $
Show Answer
Answer:
Correct Answer: D
Solution:
[d] $ F=-\frac{dU}{dx}=-20x $
$ a=\frac{F}{M}=\frac{-20x}{1}=-20x $
$ -20x=-{{\omega }^{2}}(x) $
$ \
Therefore ,\omega =\sqrt{20}=2\sqrt{5} $
$ f=\frac{\omega }{2\pi }=\frac{2\sqrt{5}}{2\pi }=( \frac{\sqrt{5}}{\pi } ) $
BETA
