Wave Mechanics Question 537
Question: If a is the amplitude of SHM, then K.E. is equal to the P.E. at………… distance from the mean position.
Options:
A) $ \frac{a}{\sqrt{2}} $
B) $ \frac{a}{2} $
C) $ \frac{a}{4} $
D) $ a $
Show Answer
Answer:
Correct Answer: A
Solution:
[a] If displacement of particle is y, then $ KE=\frac{1}{2}m{{\omega }^{2}}(a^{2}-y^{2})\And P.E.=\frac{1}{2}m{{\omega }^{2}}y^{2} $ If $ KE=PE $ then $ \frac{1}{2}m{{\omega }^{2}}y^{2} $
$ =\frac{1}{2}m{{\omega }^{2}}a^{2}-\frac{1}{2}m{{\omega }^{2}}y^{2} $
$ \Rightarrow 2y^{2}=a^{2},\
Therefore y=\frac{a}{\sqrt{2}} $
BETA
