InterviewSolution
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InterviewSolution
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A particle located in a one-dimensional potential field has its potential energy function as `U(x)(a)/(x^4)-(b)/(x^2)`, where a and b are positive constants. The position of equilibrium x corresponds toA. (a) `(b)/(2a)`B. (b) `sqrt((2a)/(b))`C. (c) `sqrt((2b)/(a))`D. (d) `(a)/(2a)` |
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Answer» Correct Answer - B The position of equilibrium corresponds to `F(x)=0` Since `F(x)=(-dU(x))/(dx)` so `F(x)=-(d)/(dx)(a/x^4-b/x^2)` or `F(x)=(4a)/(x^5)-(2b)/(x^3)` For equilibrium, `F(x)=0`, therefore `(4a)/(x^5)-(2b)/(x^3)=0impliesx=+-sqrt((2a)/(b))` `(d^2U(x))/(dx^2)=-(20a)/(x^6)+(8b)/(x^4)` Putting `x=+-sqrt((2a)/(6))` gives `(d^2U(x))/(dx^2)` as negative So U is maximum. Hence, it is position of unstable equilibrium |
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