Question

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

A

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B

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C

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D

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Solution

## The correct option is B The position of equilibrium corresponds to F(x) = 0 Since F(x)=−dU(x)dx so F(x)=ddx(ax4−bx2) or F(x)=4ax5−2bx3 For equilibrium m, F(x) = 0, therefore 4ax5−2bx3=0⇒x=±√2ab −d2U(x)dx2=−20ax6+8bx4 Putting x=±√2ab gives −d2U(x)dx2 as negative So U is maximum. Hence, it is position of unstable equilibrium.

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