A) \[\frac{4}{\sqrt{3}}\]
B) \[\frac{6}{\sqrt{2}}\]
C) \[\frac{2}{\sqrt{6}}\]
D) \[2\sqrt{6}\]
Correct Answer: A
Solution :
From the formula of maximum energy during simple harmonic motion \[{{E}_{\max }}=\frac{1}{2}m{{\omega }^{2}}{{A}^{2}}\] ?(1) Again at\[x=4,\] the potential energy \[\frac{{{E}_{\max }}}{3}=\frac{1}{2}m{{\omega }^{2}}{{x}^{2}}\] ?(2) Now from Eqs. (1) and (2), we obtain \[\frac{1}{6}m{{\omega }^{2}}{{A}^{2}}=\frac{1}{2}m{{\omega }^{2}}{{x}^{2}}\] \[{{x}^{2}}=\frac{{{A}^{2}}}{3}\]or \[A=\sqrt{3}x\] So, at \[x=4\,cm,\,A=4\sqrt{3}\]
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