A) \[4\sqrt{3}\]
B) \[6\sqrt{3}\]
C) \[2\sqrt{6}\]
D) \[2/\sqrt{6}\]
Correct Answer: A
Solution :
The maximum energy during S.H.M. \[{{E}_{\max }}=\frac{1}{2}m{{\omega }^{2}}{{A}^{2}}\] ?(i) where A = amplitude of the S.H.M. now at x = 4 \[P.E.=\frac{{{E}_{\max }}}{3}=\frac{1}{2}m{{\omega }^{2}}{{x}^{2}}\] ?(ii) Then from equation (i) and (ii) we get \[\frac{1}{6}m{{\omega }^{2}}{{A}^{2}}=\frac{1}{2}m{{\omega }^{2}}{{x}^{2}}\] \[{{x}^{2}}=\frac{{{A}^{2}}}{3},\] so\[A=\sqrt{3}x=4\sqrt{3}\]
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