Wave Mechanics Question 579
Question: A particle is executing SHM along a straight line. Its velocities at distances $ x _{1} $ and $ x _{2} $ from the mean position are $ V _{1} $ and $ V _{2} $ , respectively. Its time period is
Options:
A) $ 2\pi \sqrt{\frac{x _{2}^{2}-x _{1}^{2}}{V _{1}^{2}-V _{2}^{2}}} $
B) $ 2\pi \sqrt{\frac{V _{1}^{2}+V _{2}^{2}}{x _{1}^{2}+x _{2}^{2}}} $
C) $ 2\pi \sqrt{\frac{V _{1}^{2}-V _{2}^{2}}{x _{1}^{2}-x _{2}^{2}}} $
D) $ 2\pi \sqrt{\frac{x _{1}^{2}-x _{2}^{2}}{V _{1}^{2}-V _{2}^{2}}} $
Show Answer
Answer:
Correct Answer: A
Solution:
[a] As we know, for particle undergoing SHM, $ V=\sqrt[\omega ]{A^{2}-X^{2}}; $
$ V _{1}^{2}={{\omega }^{2}}(A^{2}-x _{1}^{2}) $
$ V _{2}^{2}={{\omega }^{2}}(A^{2}-x _{2}^{2}) $ Substructing we get, $ \frac{V _{1}^{2}}{{{\omega }^{2}}}+x _{1}^{2}=\frac{V _{2}^{2}}{{{\omega }^{2}}}+x _{2}^{2} $
$ w=\sqrt{\frac{V _{1}^{2}-V _{2}^{2}}{x _{2}^{2}-x _{1}^{2}}}\Rightarrow T=2\pi \sqrt{\frac{x _{2}^{2}-x _{1}^{2}}{V _{1}^{2}-V _{2}^{2}}} $
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