Question

A particle of mass $5\times {10}^{-5}$kg is placed at lowest point of smooth parabola $x^2=40y$ (x and y in m) (in rad/s). If its is displaced slightly such that it is constrained to move along parabola, angular frequency of oscillation will be approximately ($g=10m/s^2$.

Answer 1

$F=mg\sin \theta$
$\approx mg\tan \theta$
$=mg\frac{dy}{dx}=-mg\times \frac{2x}{40}$
$a=-\frac{x}{2}$

$a=-\frac{-x}{2}$

we know that $a=-{\omega }^{2}x$

$\omega =\frac{1}{\sqrt{2}}$