Question 6.2: Derive the differential equations of motion of the two degre...

Since the spring forces balance the weights at the static equilibrium position, one can show that the effect of the weights is equal to the spring forces due to the static deflection. In the static equilibrium position, one has, from Fig. 6.3(b),

m_{1}g − k_{1}\Delta _{1} + k_{2}(\Delta _{2} − \Delta _{1}) = 0
m_{2}g − k_{2}(\Delta _{2} − \Delta _{1}) − k_{3}\Delta _{3} = 0

where m_{1} and m_{2} are the masses, g is the gravitational constant, k_{1}, k_{2}, and k_{3} are the spring constants, and \Delta_{1}, \Delta_{2}, and \Delta_{3} are the static deflections of the springs in the static equilibrium position. Without loss of generality, we assume that x_{2} > x_{1}. From the free body diagram shown in Fig. 6.3(c), the dynamic equations of the vibratory motion of the two degree of freedom system are given by

m_{1}\ddot{x_{1}} = m_{1}g − k_{1}(x_{1} + \Delta _{1}) + k_{2}(x_{2} + \Delta _{2} − x_{1} − \Delta _{1})
m_{2}\ddot{x_{2}} = m_{2}g − k_{2}(x_{2} + \Delta _{2} − x_{1} − \Delta _{1}) − k_{3} (x_{2} − \Delta _{3})
Using the static equilibrium conditions, the above differential equations
reduce to

m_{1} \ddot{x_{1}} + (k_{1} + k_{2})x_{1} − k_{2}x_{2} = 0
m_{2}\ddot{x_{2}} + (k_{2} + k_{3})x_{2} − k_{2}x_{1} = 0

which can be written in a matrix form as

\begin{bmatrix} m_{1} & 0 \\ 0 & m_{2} \end{bmatrix} \begin{bmatrix} \ddot{x_{1}}  \\  \ddot{x_{2}} \end{bmatrix} + \begin{bmatrix} k_{1} + k_{2} & − k_{2} \\ − k_{2} & k_{2} + k_{3}\end{bmatrix} \begin{bmatrix} x_{1}  \\  x_{2} \end{bmatrix} = \begin{bmatrix} 0  \\  0 \end{bmatrix}

or
M\ddot{x} + Kx = 0
where M and K are the symmetric mass and stiffness matrices defined by

M = \begin{bmatrix} m_{1} & 0 \\ 0 & m_{2} \end{bmatrix},           K = \begin{bmatrix} k_{1} + k_{2} & − k_{2} \\ − k_{2} & k_{2} + k_{3}\end{bmatrix}