Two hanging masses connected by springs I had this problem for a candidacy exam, but wasn't able to get the complete answer. Their spring constants and masses are not the same, find the equilibrium position and frequencies of the system. Apparently once you find their equilibrium position, i.e. when gravity and the spring forces cancel out, you can transform to about that position and ignore gravity. Why do you get to ignore gravity? This is a normal modes problem, but I couldn't justify why gravity just disappears, but it does. Anyone have a simple reason why?
 A: With a downwardly $x$-axis, you have the equations of movement: 
$m_1 \ddot x_1 = m_1g - k_1x_1 +  k_2x_2 \tag{1}$
$m_2 \ddot x_2 = m_2g - k_2x_2 \tag{2}$
Equilibrium position means that $\ddot x_1 = \ddot x_2 = 0$, so you have, naming $X_1$ and $X_2$ the equilibrium positions :
$0 = m_1g - k_1X_1 +  k_2X_2 \tag{3}$
$0 = m_2g - k_2X_2 \tag{4}$
The second equation gives $X_2 = \frac{m_2g}{k_2}$, and replacing $X_2$ in the first equation gives $X_1 = \frac{(m_1 + m_2)g}{k_1}$
Now, we are interested on oscillations around the equilibrium positions, so we make a change of variables : 
$ y_1 = x_1 - X_1, \quad  y_2 = x_2 - X_2\tag{5}$
Now, we rewrite equations $(1)$ and $(2)$ using $y_1, y_2$ :
$m_1 \ddot y_1 =  - k_1y_1 +  k_2y_2 \tag{6}$
$m_2 \ddot y_2 = - k_2y_2 \tag{7}$
We see, that the equilibrium positions correspond to $y_1=y_2=0$, as wished, and the terms in "$g$" (gravity) have disappeared.
Now, the equation $(7)$ has the solution : 
$y_2 = Y_2 ~\cos(\omega_2 t + \phi_2)\tag{8}$, with $ \omega_2 = \large \sqrt{\frac{k_2}{m_2}}$
Knowing $y_2$, we may search a solution of the equation $(6)$, as :
$y_1 = Y_1 ~\cos(\omega_1 t + \phi_1) + Z ~\cos(\omega_2 t + \phi_2)\tag{9}$
One find : $\omega_1 = \large \sqrt{\frac{k_1}{m_1}}$ and $Z =  \large \frac{k_2 Y_2}{k_1 - m_1 \omega_2^2}$.
