Trapped in the potential well
A particle can execute one dimensional motion in a potential given by
Assume that the mechanical energy of the particle was 0 at t=0 Further assume, that the particle is subject to No Other influence Then what is the region of space (in terms of range of 'x') where the particle performs oscillatory motion if released, let it be
Now , what is the value of x beyond (to the right of ) which, if the particle is released, it will depart to infinity, Let it be c
Then find, a+b+c
HINT- plotting a graph will allow you to answer in less than 5 seconds,
SOURCE- INPHO 2011
The answer is 3.
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1 solution
This is the graph of its motion, (note you dont really need a plotter, you can easily predict is using calculus, finding extrema, and checking its 0, also its nature of growth at different places, a rough plot will suffice,)
Now P+K=E
as K>0 os P<E
So particle can only occupy portions of the curve which are below E,
Now energy is conserved, and initially particle had 0 energy, so
only portions of curve below y-axis are feasible,
so if it is to perform oscilatory motion, if must lie in a "valley" , so that whenever it tries to come out of the valley, it will be stopped because it reaches P=E point, in this case its when it strikes the x-axis, at that point, its K=0
So it will fall back ,, Such a valley occurs clearly from
x=0 til the next place it crosses the x- axis,
And it will depart to infinity, if clearly placed beyond the other root,
their sum is simple 0+ sum of roots = 0-b/a= 3
(U is potential, K is kinetic, E is mechanical )
Note that you could have simply seen the graph as a land scape with hills and valleys and checked where you should drop it so the ball is trapped forever, or gets lost to infinity (rolling down the hill)