InterviewSolution
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InterviewSolution
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The potential energy of a `1 kg` particle free to move along the x- axis is given by `V(x) = ((x^(4))/(4) - x^(2)/(2)) J` The total mechainical energy of the particle is `2 J` . Then , the maximum speed (in m//s) isA. `sqrt2`B. `(1)//(sqrt2)`C. `2`D. `(3)//(sqrt2)` |
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Answer» Correct Answer - D `V(x)=(x^4)/(4)-(x^2)/(2)` `F=-(dV_(x))/(dx)=-[x^3-x]=0impliesx(x^2-1)=0` `x=0`,`x=+-1` `(d^2V_(x))/(dx)=3x^2-1` At `x=+-1`,`(d^2V_(x))/(dx)=+ve`, i.e., at `x=+-1`, P.E. is minimum `V_(min)=-(1)/(4)` `E=K_(max)+V_(min)` `implies2=(1)/(2)xx1xxv_(max)^2-(1)/(4)impliesv_(max)=(3)/(sqrt2)(m)//(s)` |
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