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Question

# Force (in N) varies with position as F=2x2−3x−2, Choose correct option :-

A
x=1/2 is position of stable equilibrium
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B
x=2 is position of stable equilibrium
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C
x=1/2 is position of unstable equilibrium
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D
x=2 is position of neutral equilibrium
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Solution

## The correct option is A x=−1/2 is position of stable equilibriumF=2x2−3x−2=−dUdx dUdx=−(2x2−3x−2) For equilibrium dUdx=0⇒2x2−3x−2=0 2x2−4x+x−2=0 2x(x−2)+1(x−2)=0 x=−12 or x=2 Now, d2Udx2=−4x+3 At x=−12; d2Udx2=−4(−12)+3=5=+ve ∴ stable equilibrium. At x=2, d2Udx2=−4(2)+3=−5=−ve ∴ unstable equilibrium.

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