A particle of mass m is allowed to oscillate on a smooth parabola: $x^2=4\text{a}y,\text{a}>1$ , as shown in the figure. The angular frequency $\left(\omega \right)$ of small oscillations is

Given$x^2=4\text{a}y;y=\frac{x^2}{4\text{a}}$

Slope$\left(\text{m}\right)=\tan \theta =\frac{\text{dy}}{\text{d}x}$

$\Rightarrow \tan \theta =\frac{\text{d}}{\text{d}x}\left(\frac{x^2}{4\text{a}}\right)=\frac{1}{4\text{a}}\left(2x\right)=\frac{x}{2\text{a}}$

We know that

${\text{F}}_{\text{Restring}}=\text{mg}\tan \theta =\text{mg}\left(\frac{x}{2\text{a}}\right)=\left(\frac{\text{mg}}{\text{2a}}\right)x\cdots \left(1\right)$

$i.e.,{\text{F}}_{\text{Restring}}=\text{k}x\cdots \left(2\right)$

From (1) & (2)

$\text{k}=\frac{\text{mg}}{\text{2a}};\omega =\sqrt{\frac{\text{k}}{\text{m}}}=\sqrt{\frac{\text{mg}}{\text{2am}}}=\sqrt{\frac{\text{g}}{\text{2a}}}$