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. propotional to `V_(0)`B. propotional to `V_(0)/(mX_(0))`C. propotional to `sqrt((V_(0))/(mX_(0)))`D. zero |
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Answer» Correct Answer - D `F = -(dU)/(dx)` as for `|x| gt x_(0) V = V_(0) =` constant `rArr (dU)/(dx) = 0 rArr F = 0`. |
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