[tex]x_{0} = v \sqrt{\frac{m}{2k}\\}[/tex] is the velocity of the object.
It states that energy can neither be created nor destroyed only converted from one form of energy to another.
According to law of conservation of energy:
T.E = Constant.
Also , P.E. = [tex]\frac{1}{2} kx^{2}[/tex] , K.E. = [tex]\frac{1}{2} mv^{2}[/tex]
T.E = P.E + K.E = [tex]\frac{1}{2} mv^{2}[/tex]
According to the question ,
2P.E. = [tex]\frac{1}{2} mv^{2}[/tex]
P.E. = [tex]\frac{mv^{2} }{4} = \frac{1}{2} k x_{0} ^{2}[/tex]
[tex]x_{0} ^2= v^{2} \frac{m}{2k}\\[/tex]
[tex]x_{0} = v \sqrt{\frac{m}{2k}\\}[/tex]
Therefore, the velocity of the object when the spring returns to the equilibrium position is best represented by [tex]x_{0} = v \sqrt{\frac{m}{2k}\\}[/tex]
Learn more about Law of conservation of energy here:
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