InterviewSolution
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InterviewSolution
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When a particle is mass `m` moves on the `x-` axis in a potential of the from `V(x) = kx^(2)`, it performs simple harmonic motion. The corresponding thime periond is proportional to `sqrt((m)/(k))`, as can be seen easily asing dimensional analysis. However, the motion of a pariticle can be periodic even when its potential enem increases on both sides `x = 0` in a way different from `kx^(2)` and its total energy is such that the particel does not escape to infinity. consider a particle of mass `m` moving onthe `x-`axis . Its potential energy is `V(x) = omega (alpha gt 0`) for `|x|` near the origin and becomes a constant equal to `V_(0)` for `|x| ge X_(0)` (see figure) If the total energy of the particle is `E`, it will perform is periodic motion why if :A. `Asqrt((m)/(alpha))`B. `(1)/(A)sqrt((m)/(alpha))`C. `Asqrt((alpha)/(m))`D. `(1)/(A)sqrt((alpha)/(m))` |
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Answer» Correct Answer - B `V = alphaX^(4)` `T.E. = (1)/(2) momega^(2)A^(2) = alphaA^(4)` (not stricltly applicable just for dimension matching it is used) `omega^(2) = (2alphaA^(2))/(m) rArr T prop (1)/(A)sqrt((m)/(alpha))` |
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