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    Question

    A particle of mass m is placed in a potential V(x) =

    ax² + bx⁴. For small oscillations around equilibrium, its time period is:
    A 2π√(m/2a) Correct Answer Incorrect Answer
    B 2π√(m/a) Correct Answer Incorrect Answer
    C 2π√(m/4a) Correct Answer Incorrect Answer
    D 2π√(m/3a) Correct Answer Incorrect Answer

    Solution

    Potential energy: V(x) = a·x² + b·x⁴ Mass = m We are to find time period for small oscillations around equilibrium. Find equilibrium point Equilibrium occurs where dV/dx = 0 dV/dx = 2a·x + 4b·x³ = x(2a + 4b·x²) ⇒ x = 0 is the equilibrium point

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