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Question

A particle located in a one - dimentional potential field has its potential energy functions as U(x) = ax4bx2, 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(ax4bx2) or F(x)=4ax52bx3

For equilibrium m, F(x) = 0, therefore

4ax52bx3=0x=±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.


Physics

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