Question

A particle of mass m is allowed to oscillate on a smooth parabola $x^2=4ay,a>0$, about the origin O (see fig). For small oscillations, find the angular frequency $\left(\omega \right)$
.

• A. $\sqrt{\frac{g}{2a}}$
• B. $\sqrt{\frac{g}{a}}$
• C. $\sqrt{\frac{2g}{a}}$
• D. $2\sqrt{\frac{g}{a}}$

Answer 1

The correct option is A $\sqrt{\frac{g}{2a}}$

During its oscillations, at a particular instant let the coorinates of the particle are (x, y).
The total energy of the particle is
$E_r=\frac{1}{2}m{\left(\frac{dx}{dt}\right)}^2+\frac{1}{2}m{\left(\frac{dy}{dt}\right)}^2+mgy$
Equation of the curve is $x^2=4ay$
$\Rightarrow 2x\frac{dx}{dt}=4a\frac{dy}{dt}$
$\Rightarrow \frac{dy}{dt}=\frac{x}{2a}\frac{dx}{dt}$
$\therefore E_r=\frac{1}{2}m{\left(\frac{dx}{dt}\right)}^2+\frac{1}{2}m{\left(\frac{x}{2a}\right)}^2{\left(\frac{dx}{dt}\right)}^2+mg\frac{x^2}{4a}$
$[\because y=\frac{x^2}{4a}]$
As oscillations are very small, we can ignore the middle term.
$E_r=\frac{1}{2}m{\left(\frac{dx}{dt}\right)}^2+\frac{mg}{4a}{x}^2$
$\frac{d{E}_T}{dt}=\frac{1}{2}m.2\left(\frac{dx}{dt}\right)\left(\frac{d^{2}x}{d{t}^2}\right)+\frac{mg}{4a}.2x\frac{dx}{dt}=0$
$m.\frac{d^{2}x}{d{t}^2}=-\frac{mg}{4a}.2x$
$\frac{d^{2}x}{d{t}^2}=-\frac{g}{2a}x$
$\Rightarrow \omega =\sqrt{\frac{g}{2a}}$