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Q.

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

2ab

b

b2a

c

ab

d

2ba

answer is B.

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Detailed Solution

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 F(x) = 0 , therefore
4ax5−2bx3= 0 ⇒  x= ±2ab

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