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}}} $