KVPY SX SB Exam 2016 Question Paper

Question : 27 of 120

Marks: +1, -0

A particle with total mechanical energy, which is small and negative, is under the influence of a one dimensional potential

U(x) = x^4/4-x^2/2

J Where x is in meters. At time t = 0s, it is at x = – 0.5 m. Then at a later time it can be found

Solution:

at t = 0, x = 0.5

u=x^4/4-x^2/2

⇒ 1/4×1/16-1/4×1/2⇒|1/4|

{du}/{dx}={4x^3}/4-{2x}/2

=x^3-x

=x(x^2-1)

{du}/{dx}=0

at point of maxima & minima

x = 0 ; x = ± 1

({d^2u}/{dx^2})_{x=0}=-1

point of maxima

({d^2u}/{dx^2})_{x=±1}2

point of minima

particle will found between (–1, 0)

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