A) \[\frac{U}{F}+x=0\]
B) \[\frac{2U}{F}+x=0\]
C) \[\frac{F}{U}+x=0\]
D) \[\frac{F}{2U}+x=0\]
Correct Answer: B
Solution :
Potential energy of an oscillating particle \[U=\frac{1}{2}k{{x}^{2}}\] \[2U=k{{x}^{2}}\] \[2U=-kx\] \[\frac{2U}{F}=-x\] or \[\frac{2U}{F}+x=0\]You need to login to perform this action.
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