Wave Mechanics Question 582
Question: A particle of mass m oscillates with a potential energy $ U=U _{0}+\alpha x^{2} $ , where $ U _{0} $ and a are constants and x is the displacement of particle from equilibrium position. The time period of oscillation is
Options:
A) $ 2\pi \sqrt{\frac{m}{\alpha }} $
B) $ 2\pi \sqrt{\frac{m}{2\alpha }} $
C) $ \pi \sqrt{\frac{2m}{\alpha }} $
D) $ 2\pi \sqrt{\frac{m}{{{\alpha }^{2}}}} $
Show Answer
Answer:
Correct Answer: B
Solution:
[b] $ U=U _{0}+\alpha x^{2}\Rightarrow F=-\frac{DU}{dx}=-2\alpha x $
$ a=\frac{F}{m}=-\frac{2\alpha }{m}x $
$ \Rightarrow {{\omega }^{2}}=\frac{2\alpha }{m}\Rightarrow T=\frac{2\pi }{\omega }\Rightarrow T=\frac{2\pi }{\omega }=2\pi \sqrt{\frac{m}{2\alpha }} $
BETA
