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

Potential energy
$U\left(x\right)=-a{x}^2+b{x}^4$ 
Force  $F=-\frac{\text{dU}}{\text{d}x}=2ax-4b{x}^3$
At mean position  $F=0$
$\Rightarrow 2ax=4b{x}^3$ 
 $\therefore x=\sqrt{\frac{a}{2b}}$
 $\frac{{\text{d}}^2\text{U}}{\text{d}{x}^2}=-2a+12b{x}^2=-2a+6a=4a\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}i.e\hspace{0.17em}}+ve$
$\Rightarrow U$ is minima at  $x=\sqrt{\frac{a}{2b}}$
$\therefore$  Force constant $K_{\text{eff}}=\frac{F}{x}or\frac{\text{d}F}{\text{d}x}$
$\Rightarrow K_{\text{eff}}=\frac{{\text{d}}^{2}u}{\text{d}{x}^2}=2a-12b{x}^2$ 
$\Rightarrow K=2a-12b\times \frac{a}{2b}=4a$ 
Now,
$F=ma\text{'}=-kx$

$\Rightarrow a\text{'}=-\frac{4a}{m}.x$

This represents equation of SHM.

$\Rightarrow \omega =\sqrt{\frac{4a}{m}}=2\sqrt{\frac{a}{m}}$