Question

A particle of mass m moves under a conservative force with potential energy V ( x )= cx/(x2+a2),where c and a are positive constants. Find the position of stable equilibrium and the period of small oscillations about it. (b) If the particle starts from this point with velocity v, find the r

Answer 1

Answer:

The position of stable equilibrium is -a

And the period of small oscillations  must be: c/(ma^3)

Explanation:

Since the potential is:

$V(x) = \frac{c x}{a^2+x^2}$

We first look for a position of stable equilibrium. This posiiton must satisfy two considtions, that the first derivative of the potential must vanish at this point and the second derivative must be positive.

$V'(x) = \frac{c}{a^2+x^2}-\frac{2 c x^2}{\left(a^2+x^2\right)^2}$

Which vanish for

x = a   ; x =-a

The second derivative of V(x) is:

$V''(x) = \frac{8 c x^3}{\left(a^2+x^2\right)^3}-\frac{6 c x}{\left(a^2+x^2\right)^2}$

And:

$V''(a) = -\frac{c}{2 a^3}\\V''(-a) = \frac{c}{2 a^3}$

Therefore:

a)

The position of stable equilibrium is -a

And the period of small oscillations  must be:

$\omega = \sqrt{2 V''(-a)/m} = \sqrt{\frac{c}{a^3 m}}$

(c/(ma^3))^1/2

b)

Let's find the maximum amplitude if the particle starts at this point with velocity v

If this is the case, the total energy will be:

(mv^2)/2

And the maximum amplitude will be

x = a^3/c mv^2 = (m v^2 a^3)/ c