Question 13.4.1: Modes of Two Masses in Translation EXAMPLE 13.4.1 Find and i...

The equations of motions for the system are
m_{1} \ddot{x}_{1} = −k_{1} x_{1}  −  k_{2}(x_{1}  −  x_{2})
m_{2}\ddot{x}_{2} = k_{2}(x_{1}  −  x_{2})  −  k_{3} x_{2}
Substitute x_{1}(t) = A_{1} e^{st} and x_{2}(t) = A_{2} e^{st} into the preceding differential equations, cancel the est terms, and collect the coefficients of A_{1} and A_{2} to obtain

(m_{1} s^{2} + k_{1} + k_{2}) A_{1}  −  k_{2} A_{2} = 0          (1)
−k_{2}A_{1} + (m_{2}s^{2} + k_{2} + k_{3}) A_{2} = 0        (2)
To have nonzero solutions for A_{1} and A_{2}, the determinant of the above equations must be zero.
Thus,

Expanding this determinant gives

(m_{1}s^{2} + k_{1} + k_{2})(m_{2} s^{2} + k_{2} + k_{3})  −  k^{2}_{2} = 0                    (3)
Using m_{1} = m_{2} = m, k_{1} = k_{3} = k, and k_{2} = 2k , we obtain

or
m^{2} s^{4} + 6 kms^{2} + 5k^{2} = 0
This can be simplified to
s^{4} + 6αs^{2} + 5α^{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} = 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{2k}{ms^{2}  +  3k} = \frac{2α}{s^{2}  +  3α}
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.