RRB Technician 2014 Ques 25
Question: The kinetic energy and potential energy of a particle executing simple harmonic motion will be equal when (amplitude = a) displacement is
Options:
A) $ 2,a $
B) a
C) $ a\sqrt{2} $
D) $ \frac{a}{\sqrt{2}} $
Show Answer
Answer:
Correct Answer: D
Solution:
- In SHM, KE= $ =\frac{1}{2}m,(a^{2}-y^{2}){{\omega }^{2}} $ $ PE=\frac{1}{2}m{{\omega }^{2}}y^{2} $ They are equal if $ \frac{1}{2}m{{\omega }^{2}}(a^{2}-y^{2})=\frac{1}{2}m{{\omega }^{2}}y^{2} $
$ \Rightarrow $ $ a^{2}-y^{2}=y^{2} $
$ \Rightarrow $ $ 2y^{2}=a^{2} $
$ \therefore $ $ y=\frac{a}{\sqrt{2}} $
BETA
