Question

A particle of mass M is placed at the lowest point of smooth parabola x squared is equal to a y what is the time period of small oscillation

Answer 1

Answer:

Time period $=2\pi\sqrt{\frac{2a}{g}}$

Explanation:

Given equation of the parabola

$x^2=4ay$

If the particle of mass M is displaced by a small distance y vertically

The potential energy

$U=-Mgy$

$\implies U=-Mg\frac{x^2}{4a}$

Restoring force

$F=\frac{dU}{dx}$

$\implies F=-\frac{d}{dx}(Mg\frac{x^2}{4a})$

$\implies F=-\frac{Mg}{4a}\frac{dx^2}{dx}$

$\implies F=-\frac{Mg}{4a}\times 2x$

$\implies F=-\frac{Mg}{2a}x$

or, $F=-kx$

where $k=\frac{Mg}{2a}$

Therefore, the angular frequency of small oscillations

$\omega=\sqrt{\frac{k}{M}}$

$\omega=\sqrt{\frac{g}{2a}}$

Thus, the time period of small oscillations

$T=\frac{2\pi}{\omega}$

$\implies T=\frac{2\pi}{\sqrt{g/2a}}$

$\implies T=2\pi\sqrt{\frac{2a}{g}}$

Hope this helps.

Answer 2

Answer:

its answer is given in attachment

Explanation:

may be it will help you