Question 6.4: The bar A B shown in Fig. 6.5, which represents a simplified...

Since the displacement of the bar in the x direction is neglected, the system configuration can be identified using the two variables y and θ, where y is the vertical displacement of the center of mass and θ is the angular rotation of the bar. It is left to the reader as an exercise to show that the weight of the bar cancels with the deflection of the springs at the static equilibrium position. Therefore, the differential equations of motion are given by

m\ddot{y} = −k_{1}\left( y − \frac{l}{2}θ\right) − k_2\left(y + \frac{l}{2}θ\right)

I \ddot{θ} = k_1 (y − \frac{l}{2}θ) \frac{l}{2}\ cos θ − k_2 (y + \frac{l}{2}θ) \frac{l}{2} cos θ

For small angular oscillations, cos θ ≈ 1, and the differential equations reduce to

m\ddot{y} + (k_{1} + k_{2})y + (k_{2} − k_{1}) \frac{l}{2} θ = 0

I \ddot{θ} + (k_{2} + k_{1}) \frac{l^2}{4} θ + (k_{2} − k_{1}) \frac{l}{2}y = 0

which can be written in matrix form as

\begin{bmatrix} m & 0 \\ 0 & I \end{bmatrix} \begin{bmatrix} \ddot{y} \\ \ddot{θ} \end{bmatrix}+ \begin{bmatrix} k_1 + k_2 & (k_2 − k_1) \frac{l}{2} \\  (k_2 − k_1) \frac{l}{2} & (k_2 + k_1) \frac{l^2}{4} \end{bmatrix} \begin{bmatrix} y \\ θ \end{bmatrix}= \begin{bmatrix} 0 \\  0 \end{bmatrix}

which can be written in a more compact form as
M\ddot{x} + Kx = 0

where M and K are the symmetric mass and stiffness matrices given by

M = \begin{bmatrix} m & 0 \\ 0 & I \end{bmatrix},         K = \begin{bmatrix} k_1 + k_2 & (k_2 − k_1) \frac{l}{2} \\  (k_2 − k_1) \frac{l}{2} & (k_2 + k_1) \frac{l^2}{4} \end{bmatrix}

and the vectors x and \ddot{x} are

One can then assume a solution in the form x = Xsin(ωt + Φ), where X is the vector of amplitudes defined as

X= \begin{bmatrix} Y\\\Theta \end{bmatrix}

The acceleration vector \ddot{x}is given by \ddot{x}= −ω^2X\sin(ωt + \phi) . Substituting x and \ddot{x} into the differential equation one obtains
[K − ω²M]X = 0
For a nontrivial solution, the determinant of the coefficient matrix must be equal to zero, and hence
|K − ω²M| = 0
which can be written in a more explicit form as

\left|\begin{matrix}k_{11} − ω^{2}m&k_{12}\\k_{21}&k_{22} − ω^{2}I\end{matrix}\right|  = 0

where k_{11} = k_{1} +k_{2}, k_{12} = k_{21} = (k_{2} −k_{1})(l/2), and k_{22} = (k_{1} +k_{2})(l²/4).
The characteristic equation is then given by
(k_{11} − ω^{2}m)(k_{22} − ω^{2}I) − k^{2}_{12} = 0
or
ω^{4}Im − ω^{2}[mk_{22} + Ik_{11}] + k_{11}k_{22} − k^{2}_{12} = 0

which yields the following equation:

The roots of this equation are given by

ω^{2}_{1}= \frac{ −b +\sqrt{b^2 − 4ac}}{2a}            ω^{2}_{2}=\frac{ −b -\sqrt{b^2 − 4ac}}{2a}

where
a = ml, b = −(k_1 + k_2) \left(m\frac{l^2}{4}+ I\right) , c = 4k_1k_2 \frac{l^2}{4}

The amplitude ratios β_1 and β_2 are given by

The solution is then given by
x_1 = β_1X_{21} \sin(ω_1t + \phi_1) + β_2X_{22} \sin(ω_2t + \phi_2)\\[0.5 cm] x_2 = X_{21} \sin(ω_1t + \phi_1) + X_{22} \sin(ω_2t + \phi_2)