Wave Mechanics Question 565
Question: A particle of mass m executes simple harmonic motion with amplitude a and frequency v. The average kinetic energy during its motion from the position of equilibrium to the end is
Options:
A) $ 2{{\pi }^{2}}ma^{2}v^{2} $
B) $ {{\pi }^{2}}ma^{2}v^{2} $
C) $ \frac{1}{4}ma^{2}v^{2} $
D) $ 4{{\pi }^{2}}ma^{2}v^{2} $
Show Answer
Answer:
Correct Answer: B
Solution:
[b] The kinetic energy of a particle executing S.H.M. is given by $ K=\frac{1}{2}ma^{2}{{\omega }^{2}},{{\sin }^{2}}\omega t $
$ =<\frac{1}{2}m{{\omega }^{2}}a^{2}{{\sin }^{2}}\omega t>,=\frac{1}{2}m{{\omega }^{2}}a^{2}<{{\sin }^{2}}\omega t> $
$ =\frac{1}{2}m{{\omega }^{2}}a^{2}( \frac{1}{2} ) $
$ =\frac{1}{4}m{{\omega }^{2}}a^{2},( \
Therefore <{{\sin }^{2}}\theta >=\frac{1}{2} ) $
$ =\frac{1}{4}ma^{2},{{(2\pi v)}^{2}} $
$ (\
Therefore \omega =2\pi v) $ or $
BETA
