What is the time of small oscillation in this question? Question :

*

*A particle rests in equilibrium under the attraction of two centres of force which attract directly as the distance, their attractions per unit of mass at unit distance being u and u'; the particle is slightly displaced towards one of them;

*Show that the time of small oscillation is
$$
{2\pi \over \left(\,{u + u'}\,\right)^{1/2}}
$$
Solution
Force diagram
-In this picture F1 is the force on the particle due to B and
F2 is the force on the particle due to A.

*

*$$F1 = m.u.(a - x) $$ and


*$$F2 = m.u'.(a + x) $$
where m = mass of the particle 
u = force per unit mass per unit distance


*Net Resultant force $$= F2 - F1 $$
$$=m.(u'a + u'x - ua + ux) $$
$$=m.(u + u')[x - a.(u-u')/(u+u')]$$


*Accelaration $$=(u + u')[x - a.(u-u')/(u+u')] $$


*Time period = $${2\pi \over \left( {u + u'}\, \right)^{1/2} }$$
 A: Let's make sure we get all the signs right.
Your origin is at the middle of the distance between the two objects, With the positive direction towards $B$. Then $$\frac{F_A}m=-u(a+x)\\\frac{F_B}m=+u'(a-x)$$
Then the acceleration is $$\ddot x=-u(a+x)+u'(a-x)=-(u+u')x+(u'-u)a$$
This seems similar to the equation for a mass/spring system:
$$\ddot x=-\frac km x$$
The period for this is $$T=\frac{2\pi}{\sqrt{\frac km}}$$
We can see a similarity, if we write $u+u'=\frac km$. So what's the difference? Is that extra term at the end. We can rewrite that equation as $$\ddot x=(u+u')(x-x_0)$$where $$x_0=\frac{u'-u}{u'+u}a$$
Then change the variable $y=x-x_0$, with $\ddot y=\ddot x$, so the period for $y$ is then $$T=\frac{2\pi}{(u+u')^{1/2}}$$
Wondering what that $x_0$ is? It's the equilibrium position, where the two forces cancel. Going back to the first formula for acceleration, and setting it to be $0$ at $x_0$ you get
$$-(u+u')x_0+(u'-u)a=0$$
Solve for $x_0$, and you get the formula that I've put in before:$$x_0=\frac{u'-u}{u'+u}a$$
