potential enrgy of a particle along x-axis varies as, U=-20 + (x-2)^(2), where U is in joule and x in meter. Find the equilibrium position and state whether it is stable or unstable equilibrium.
Answers
particle is in stable equilibrium at x = 2
given, potential energy of a particle along x - axis varies as U = -20 + (x - 2)²
= -20 + x² + 4 - 4x
= x² - 4x - 16
To find equilibrium position,
differentiating U with respect to x,
dU/dx = 2x - 4
at dU/dx = 0, x = 2
now again differentiating with respect to x,
d²U/dx² = 2 > 0
so, at x = 2 particle is in stable equilibrium.
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