Wave Mechanics Question 558
Question: A particle executing harmonic motion is having velocities $ v _{1} $ and $ v _{2} $ at distances is x, and x, from the equilibrium position. The amplitude of the motion is
Options:
A) $ \sqrt{\frac{v _{1}^{2}x _{2}-v _{2}^{2}x _{1}}{v _{1}^{2}+v _{2}^{2}}} $
B) $ \sqrt{\frac{v _{1}^{2}x _{1}^{2}-v _{2}^{2}x _{2}^{2}}{v _{1}^{2}+v _{2}^{2}}} $
C) $ \sqrt{\frac{v _{1}^{2}x _{2}^{2}-v _{2}^{2}x _{1}^{2}}{v _{1}^{2}-v _{2}^{2}}} $
D) $ \sqrt{\frac{v _{1}^{2}x _{2}^{2}+v _{2}^{2}x _{1}^{2}}{v _{1}^{2}+v _{2}^{2}}} $
Show Answer
Answer:
Correct Answer: C
Solution:
[c] $ v _{1}=\omega \sqrt{a^{2}-x _{1}^{2}},v _{2}=\omega \sqrt{a^{2}-x _{2}^{2}} $ We get, $ a=\sqrt{\frac{v _{1}^{2}x _{2}^{2}-v _{2}^{2}x _{1}^{2}}{v _{1}^{2}-v _{2}^{2}}} $