Find and interpret the mode ratios for the system shown in Figure 13.4.1, for the case m_1 = m, m_2 = 3m, k_1 = k, and k_2 = k_3 = 2k.
For this case, equation (3) in Example 13.4.1 becomes
\left(m_1s^2+k_1+k_2\right)\left(m_2s^2+k_2+k_3\right)-k_2^2=0 (3) in Example 13.4.1
3u² + 13αu + 8α² = 0
where u = s² and α = k/m. From the quadratic formula we obtain u = −0.743α and u = −3.591α. Thus the two modal frequencies are ω_1 = \sqrt{0.743α} = 0.862 \sqrt{k/m} and ω_2 = \sqrt{3.591α} = 1.89 \sqrt{k/m}. From equation (1) of Example 13.4.1 the mode ratios are computed as
\left(m_1s^2+k_1+k_2\right)A_1 – k_2 A_2=0 (1) of Example 13.4.1
\frac{A_1}{A_2}=\frac{2 \alpha}{s^2+3 \alpha}=0.886for mode 1, and
\frac{A_1}{A_2}=\frac{2 \alpha}{s^2+3 \alpha}=-3.39for mode 2.
Thus in mode 1 the masses move in the same direction with the amplitude of mass m_1 equal to 0.886 times the amplitude of mass m_2. This oscillation has a frequency of ω_1 = 0.862 \sqrt{k/m}.
In mode 2, the masses move in the opposite direction with amplitude of mass m_1 equal to 3.39 times the amplitude of mass m_2. This oscillation has a higher frequency of ω_2 = 1.89 \sqrt{k/m}.
To stimulate the first mode, displace mass m_1 0.866 times the initial displacement of mass m_2, in the same direction. To stimulate the second mode, displace mass m_1 3.39 times the initial displacement of mass m_2, but in the opposite direction.