Find position of unstable equilibrium.

X = 2

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X = - 2

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X = 1

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X = - 1

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Solution

The correct option is C $X = - 2$
at unstable equilibrium position. energy is maximum hence $U^{'} (x) = 0 & U^{''} (x) < 0$
$U^{'} (x) = x^{2} - 4 = 0 x = \pm 2 a t x = - 2 U^{''} (x) = 2 x = - 4 < 0$
$\Rightarrow x = - 2$ is an unstable equilibrium position.

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