The potential energy of a particle varies with X coordinate according to the relation U x - x 2-4 x- The point x -2 is a point ofA- Stable equilibriumB- None of the aboveC- Unstable equilibriumD- Neutral equilibrium

The correct option is A Stable equilibriumGiven U=x^2 4x For equilibrium dUdx=0 ⇒ 2x 4=0 ⇒ x=2m Now double differentiation d^2Udx^2= 2 gt;0→Stable equilibriu ...

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Standard XII

Physics

Conservative and Non Conservative Forces

The potential...

Question

The potential energy of a particle varies with $x$ coordinate according to the relation $U (x) = x^{2} - 4 x$ . The point $x = 2$ is a point of

Stable equilibrium

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Unstable equilibrium

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Neutral equilibrium

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None of the above

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Solution

The correct option is A Stable equilibrium
Given $U = x^{2} - 4 x$
For equilibrium $\frac{d U}{d x} = 0$
$\Rightarrow 2 x - 4 = 0 \Rightarrow x = 2 m$
Now double differentiation
$\frac{d^{2} U}{d x^{2}} = 2 > 0 \to$ Stable equilibrium

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The potential energy of a particle varies with x coordinate according to the relation U(x)=x2−4x. The point x=2 is a point of

The correct option is A Stable equilibriumGiven U=x2−4x For equilibrium dUdx=0 ⇒2x−4=0⇒x=2m Now double differentiation d2Udx2=2>0→Stable equilibrium

The potential energy of a particle varies with $x$ coordinate according to the relation $U (x) = x^{2} - 4 x$ . The point $x = 2$ is a point of

The correct option is A Stable equilibrium
Given $U = x^{2} - 4 x$
For equilibrium $\frac{d U}{d x} = 0$
$\Rightarrow 2 x - 4 = 0 \Rightarrow x = 2 m$
Now double differentiation
$\frac{d^{2} U}{d x^{2}} = 2 > 0 \to$ Stable equilibrium