InterviewSolution
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InterviewSolution
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If a partical moves in a potential energy held `U = U_(0) - ax + bx^(2)`, where are a and b partical constents obtian an expression for the force acting on if as a function of position. At what point does the force vanish? Is this a point of stable equilibriun ? Calculate the force constant and friquency of the partical. |
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Answer» `F = - (dU)/(dx) = a - 2b x` `F = 0, at x = a//2b` If stable equilibrum is present at this position, `(d^(2(U)/(dx^(2))) gt 0)` Hence, `(d^(2(U)/(dx^(2))) = 2b gt 0)` i.e., `x = (a)/(2 b)` is a point of minimum potential energy Hence, the equilibrium is stable. The partical will oscillate about `x = (a)/(2 b)` The effective force constant of oscillation `K_(eff) = ((d^(2)U)/(dx^(2))|_(x= (a)/(2b))) = 2b` As `K_(eff) = m omeg^(2) = 2b = m omeg^(2)` or `omega^(2) = (2 b)/(m) implies omega = 2 pi f = sqrt((2b)/(m)) implies f = (1)/(2pi) sqrt((k)/(m))` |
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