First slide

Simple harmonic motion

Question

A particle of mass m moves in one dimension, whose potential energy varies as $U (x) = - a x^{2} + b x^{4}$ , where a and b are positive constants. The angular frequency of small oscillations about the minima of the potential energy is equal to

Difficult

A

$π \sqrt{\frac{a}{2 b}}$
B

$2 \sqrt{\frac{a}{m}}$
C

$\sqrt{\frac{2 a}{m}}$
D

$\sqrt{\frac{a}{2 m}}$

Solution

Potential energy
$U (x) = - a x^{2} + b x^{4}$
Force $F = - \frac{dU}{d x} = 2 a x - 4 b x^{3}$
At mean position $F = 0$
$\Rightarrow 2 a x = 4 b x^{3}$
$∴ x = \sqrt{\frac{a}{2 b}}$
$\frac{d^{2} U}{d x^{2}} = - 2 a + 12 b x^{2} = - 2 a + 6 a = 4 a i.e + v e$
$\Rightarrow U$ is minima at $x = \sqrt{\frac{a}{2 b}}$
$∴$ Force constant $K_{eff} = \frac{F}{x} o r \frac{d F}{d x}$
$\Rightarrow K_{eff} = \frac{d^{2} u}{d x^{2}} = 2 a - 12 b x^{2}$
$\Rightarrow K = 2 a - 12 b \times \frac{a}{2 b} = 4 a$
Now,
$F = m a' = - k x$
$\Rightarrow a' = - \frac{4 a}{m} . x$
This represents equation of SHM.
$\Rightarrow ω = \sqrt{\frac{4 a}{m}} = 2 \sqrt{\frac{a}{m}}$

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