Question

A particle of mass 10 kg is placed in a potential field given by $V = (50{x^2} + 100)\;erg/g.$ What will be the frequency of oscillation of a particle ?

Answer 1

Frequency is the number of oscillations per second. The particle is oscillating due to the presence of the field. Here we find the force in two different ways and equate it to one another to get the second order differential equation. From there we get the $\omega$ value. From this $\omega$ we can find the frequency.
Potential field and force is related by the equation, $\;F = - \dfrac{{dU}}{{dx}}$
The second order equation of oscillatory motion is $\dfrac{{{d^2}x}}{{{d^2}t}} + {\omega ^2}x = 0$

Complete answer:
The oscillation of the particle is due to the effect of the potential.
Given , $V = (50{x^2} + 100)$
Also the potential is calculated by $U = M*V$ , where M is the mass and v is the potential.
$U = 10(50{x^2} + 100)$
We differentiate this to get F
$F = - \dfrac{{dU}}{{dx}} = - 1000x - - - - - - - - - - \left( 1 \right)$
Also force is given by $F = m*a$ , where m is the mass and a is acceleration.
$F = 10\dfrac{{{d^2}x}}{{{d^2}t}} - - - - - - - - - \left( 2 \right)$
From equations $\left( 1 \right)$ and $\left( 2 \right)$
$- 1000x = 10\dfrac{{{d^2}x}}{{{d^2}t}}$
$\dfrac{{{d^2}x}}{{{d^2}t}} + 100x = 0$
But we know that $\dfrac{{{d^2}x}}{{{d^2}t}} + {\omega ^2}x = 0.$
Therefore ${\omega ^2} = 100$
$\Rightarrow \omega = 10$
Frequency of oscillation $f = \dfrac{\omega }{{2\pi }}$
$f = \dfrac{{10}}{{6.28}}$
$f = 1.59Hertz$
Thus the frequency of oscillation of the particle is $1.59Hertz$ .
A potential is a scalar field that represents the potential energy per unit of a quantity as a result of a vector field. It has a close connection with potential energy.

Note:
Oscillation is described as the process of repeated variations of any quantity or measure in time about its equilibrium value. Oscillation may also be described as a periodic variation in a matter's value between two values or near its centre value. Hertz is the unit of frequency.
Potential and field (E) have a differential relationship: electric field is the gradient of potential (V) in the x direction. This may be written as ${E_x} = \dfrac{{dV}}{{dx}}$ E .As a result, as the test charge moves in the x direction, the rate of change in potential equals the value of the electric field.