Question 13.4.1: Find and interpret the mode ratios for the system shown in F......

The equations of motions for the system are

m_1 \ddot{x}_1=-k_1 x_1-k_2\left(x_1-x_2\right)

m_2 \ddot{x}_2=k_2\left(x_1-x_2\right)-k_3 x_2

Substitute x_1(t) = A_1e^{st}  and  x_2(t) = A_2e^{st}  into the preceding differential equations, cancel the e^{st} terms, and collect the coefficients of A_1  and  A_2 to obtain

\left(m_1 s^2+k_1+k_2\right) A_1-k_2 A_2=0        (1)

-k_2 A_1+\left(m_2 s^2+k_2+k_3\right) A_2=0          (2)

To have nonzero solutions for A_1  and  A_2, the determinant of the above equations must be zero. Thus,

\left|\begin{array}{cc}m_1 s^2+k_1+k_2 & -k_2 \\-k_2 & m_2 s^2+k_2+k_3\end{array}\right|=0

Expanding this determinant gives

\left(m_1s^2+k_1+k_2\right)\left(m_2s^2+k_2+k_3\right)-k_2^2=0          (3)

Using m_1 = m_2 = m, k_1 = k_3 = k,  and  k_2 = 2k, we obtain

\left(m s^2+3 k\right)^2-4 k^2=0

or

m^2 s^4+6 k m s^2+5 k^2=0

This can be simplified to

s^4+6 \alpha s^2+5 \alpha^2=0

where α = k/m.
This polynomial has four roots because it is fourth order. We can solve it for s² using the quadratic formula because it is quadratic in s² (there is no s term or s³ term). To see why this is true, let u = s². Then the preceding equation becomes

u² + 6αu + 5α² = 0 (4)

which has the solutions u = −α and u = −5α. Thus s = ±j \sqrt{α} and s = ±j \sqrt{5α}. The two
modal frequencies are thus ω_1 = \sqrt{α}  and  ω_2 = \sqrt{5α}.
The mode ratio can be found from either equation (1) or (2). Choosing the former, we obtain

\frac{A_1}{A_2}=\frac{2 k}{m s^2+3 k}=\frac{2 \alpha}{s^2+3 \alpha}

The mode ratio A_1/A_2 can be thought of as the ratio of the amplitudes of x_1  and  x_2 in that mode. For the first mode, s² = −α and A_1/A_2 = 1. For the second mode, s² = −5α and A_1/A_2 = −1.
Thus, in mode 1 the masses move in the same direction with the same amplitude. This oscillation has a frequency of ω_1 = \sqrt{k/m}. In mode 2, the masses move in the opposite direction but with the same amplitude. This oscillation has a higher frequency of ω_2 = \sqrt{5k/m}.
The specific motion depends on the initial conditions, and in general is a combination of both modes. If the masses are initially displaced an equal distance in the same direction and then released, only the first mode will be stimulated. Only the second mode will be stimulated if the masses are initially displaced an equal distance but in opposite directions.