A particle located in a $$one-dimensional$$ potential field has its potential energy function as $$U(x)=$$ $$ax4$$−$$bx2$$ ,where a and b are positive constants. The position of equilibrium x corresponds to

Question

A particle located in a $$one-dimensional$$ potential field has its potential energy function as $$U(x)=$$ $$ax4$$−$$bx2$$  ,where a and b are positive constants. The position of equilibrium x corresponds to

Infinity Learn · Accepted Answer

The position of equilibrium corresponds to $$F(x)$$ $$=$$ $$0Since$$ $$F(x)$$ $$=$$  −$$dU(x)dxSo$$ $$F(x)$$ $$=$$ −$$ddx(ax4$$−$$bx2)or$$ $$F(x)$$ $$=$$  $$4ax5$$−$$2bx3For$$ equilibrium $$F(x)$$ $$=$$ $$0$$ , $$therefore4ax5$$−$$2bx3=0$$ ⇒  $$x=$$ ±$$2abd2U(x)dx2=$$  −$$20ax6+6bx4Putting$$ x $$=$$ ±$$2ab$$ gives $$d2U(x)dx2$$  as negativeSo U is maximum. Hence, it is position of unstable equilibrium

A particle located in a one-dimensional potential field has its potential energy function as U(x)= $\frac{a}{x^{4}} - \frac{b}{x^{2}}$ ,where a and b are positive constants. The position of equilibrium x corresponds to

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A particle located in a one-dimensional potential field has its potential energy function as U(x)= ax4−bx2 ,where a and b are positive constants. The position of equilibrium x corresponds to

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A particle located in a one-dimensional potential field has its potential energy function as U(x)= $\frac{a}{x^{4}} - \frac{b}{x^{2}}$ ,where a and b are positive constants. The position of equilibrium x corresponds to