Answer:
The position of stable equilibrium is -a
And the period of small oscillations must be: c/(ma^3)
Explanation:
Since the potential is:
[tex]V(x) = \frac{c x}{a^2+x^2}[/tex]
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.
[tex]V'(x) = \frac{c}{a^2+x^2}-\frac{2 c x^2}{\left(a^2+x^2\right)^2}[/tex]
Which vanish for
x = a ; x =-a
The second derivative of V(x) is:
[tex]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}[/tex]
And:
[tex]V''(a) = -\frac{c}{2 a^3}\\V''(-a) = \frac{c}{2 a^3}\\[/tex]
Therefore:
a)
The position of stable equilibrium is -a
And the period of small oscillations must be:
[tex]\omega = \sqrt{2 V''(-a)/m} = \sqrt{\frac{c}{a^3 m}}[/tex]
(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
