Coupled Vertical Spring-Mass Systems

[graviton_10]

Homework Statement

A vertical spring mass system is attached in series with another vertical spring mass system. Masses of both objects are different and so are the spring constants of both springs. y1 and y2 are the coordinates measured from the rest position of mass 1 and mass 2 respectively.

I have identified the forces on both masses using free body diagram, but I do not know how to proceed next.
My final goal is to set up matrix equation and then solve for normal frequencies.

Relevant Equations

Mq'' = Kq

How should I proceed?

 

[erobz]

I'm getting $\downarrow^+$:

$$m_1 \ddot y_1 = m_1 g - k_1( y_1 - l_1) + k_2 \Big[ (l_2 + y_2) - (l_1 + y_1) \Big]$$

$$m_2 \ddot y_2 = m_2 g - k_2 \Big[ (l_2 + y_2) - (l_1 + y_1) \Big]$$

 

[erobz]

As for solving the system, another option is to solve a single fourth order ODE by either solving $y_1$ in terms of $y_2$ and computing $\ddot y_1$ or visa-versa.

 

[erobz]

Actually, I blundered too.

Let $l_1$ and $l_2$ be the unloaded free lengths of the springs. The equation for static equilibrium ( LHS of the diagram)

$$ m_1 g - k_1( y_1 - l_1) + k_2( y_2 - l_2) = 0 \tag{1} $$

$$ m_2 g - k_2( y_2 - l_2 ) = 0 \tag{2}$$

Now, displace $m_1, m_2$ from equilibrium by some distance $\Delta y_1, \Delta y_2$ respectively. What follows is that:

$$ m_1 \ddot{\Delta y_1} = m_1 g - k_1( y_1 + \Delta y_1 - l_1 )+ k_2( y_2 + \Delta y_2 - \Delta y_1 - l_2) \tag{3}$$

Now, expand (3) as follows:

$$ m_1 \ddot{\Delta y_1} = \cancel{ \boxed{m_1 g - k_1( y_1 - l_1) + k_2( y_ 2 - l_2)}}^0 -k_1 \Delta y_1 + k_2( \Delta y_2 - \Delta y_1)$$

Notice the boxed terms are identically 0 by (1). It follows that:

$$ m_1 \ddot{\Delta y_1} = -k_1 \Delta y_1 + k_2( \Delta y_2 - \Delta y_1) \tag{4}$$

I'll let you determine the result for $m_2$.