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A particle which is constant to move alo...

A particle which is constant to move along the `x- axis` , is subjected to a force in the same direction which varies with the distance `x` of the particle from the origin as `F(x) = -Kx + ax^(3)` . Hero `K` and `a` are positive constant . For `x ge 0`, the fanctional from of the patential every `U(x) of the particle is

A

B

C

D

Text Solution

Verified by Experts

The correct Answer is:
D
Given, `F(x) = - kx + ax^(3)" "…(i)`
`rArr dU= -F dx` or `U(x) =-int_(0)^(x) (-kx + ax^(3) ) dx`
`rArr U(x) = (kx^(2))/( 2) - (ax^(4))/( 4) " "…(ii)`
From eqn. (ii) , `U(x) =0`
`rArr (kx^(2) ) /( 2) (1- (ax^2)/( 2k) )=0`
`rArr x=0 and x= pm sqrt((2k)/( a))`.
Also, `U(x) =` negative for `x gt sqrt((2k)/( a))`
Now finding points where force is zero
`x(-k + ax^(2) ) = 0`
or `x=0 and x= pm sqrt((k)/( a))`
The slope of U-x graph is zero at these points. Therefore, the most appropriate option is (d).
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