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. `E lt 0`B. `E gt 0`C. `V_(0) gt E gt 0`D. `E gt V_(0)`

When a particle is mass `m` moves on the `x-` axis in a potential of t

1.	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. `E lt 0`B. `E gt 0`C. `V_(0) gt E gt 0`D. `E gt V_(0)`
Answer» Correct Answer - C When `0 lt E lt V_(0)` there will be acting a restoring force to perform oscillation because in this case paticle will be in the region `\|x\|lex_(0)`.

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