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Standard XII

Physics

SHM Formulae

Potential ene...

Question

Potential energy of a particle along $x -$ axis, varies as, $U = - 20 + (x - 2)^{2},$ where $U$ is in joule and $x$ in meter. The equilibrium position is :

x = 2 m

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x = 1.5 m

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x = 1 m

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x = 2.5 m

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Solution

The correct option is B $x = 2 m$
At equilibrium $F = - \frac{d U}{d x} = 0$

$\Rightarrow \frac{d (- 20 + (x - 2)^{2})}{d x} = 0$

$\Rightarrow 2 (x - 2) = 0$

$\Rightarrow x = 2$

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Potential energy of a particle along x− axis, varies as, U=−20+(x−2)2, where U is in joule and x in meter. The equilibrium position is :

The correct option is B x=2mAt equilibrium F=−dUdx=0⇒d(−20+(x−2)2)dx=0⇒2(x−2)=0⇒x=2

Potential energy of a particle along $x -$ axis, varies as, $U = - 20 + (x - 2)^{2},$ where $U$ is in joule and $x$ in meter. The equilibrium position is :

The correct option is B $x = 2 m$
At equilibrium $F = - \frac{d U}{d x} = 0$

$\Rightarrow \frac{d (- 20 + (x - 2)^{2})}{d x} = 0$

$\Rightarrow 2 (x - 2) = 0$

$\Rightarrow x = 2$