Question 6.1: For the two degree of freedom system shown in Fig. 6.2, let ...
For the two degree of freedom system shown in Fig. 6.2, let m_1 = m_2 = 5 \text{kg and} k_{1} = k_{2} = 2000 \text{N/m and let} x_{10} = x_{20} = \dot{x}_{10}= 0 and \dot{x}_{20} = 0.3m/s.
Determine the system response as a function of time.
The natural frequencies ω_{1} and ω_{2} can be calculated by using Eq. 6.12, where the constants in this equation are given by
a = m_{1}m_{2} = (5)(5) = 25\\[0.5 cm]b = −[m_{1}k_{2} + m_{2}(k_{1} + k_{2})] = −[5(2000) + 5(2000 + 2000)]\\[0.5 cm]= −30,000\\[0.5 cm]c = k_{1}k_{2} = (2000)(2000) = 4 × 10^{6}Thus
ω^{2}_{1} = \frac{30,000 + \sqrt{(30,000)^{2} − 4(25)(4 × 10^{6})}}{2(25)} = 1047.2\\[0.5cm]ω^{2}_{2} = \frac{30,000 −\sqrt{(30,000)2 − 4(25)(4 × 10^{6})}}{2(25)} = 152.786
that is,
ω_{1} = 32.361 rad/s, ω_{2} = 12.361 rad/s
The amplitude ratios β1 and β2 can be determined from Eqs. 6.14 and 6.17 as follows:
β_{1} = \frac{k_{2}}{k_{1} + k_{2} − ω^{2}_{1}m_{1}} = \frac{2000}{4000 − (1047.21)(5)} = −1.6181\\[0.5cm]β_{2} = \frac{k_{2}}{k_{1} + k_{2} − ω^{2}_{2}m_{1}} = \frac{2000}{4000 − (152.786)(5)} = 0.618
Applying the initial conditions, Eq. 6.22 yields
0 = β_{1}X_{21} \sin \phi_{1} + β_{2}X_{22} \sin \phi_{2}\\[0.5cm]0 = X_{21} \sin \phi_{1} + X_{22} \sin \phi_{2}\\[0.5cm]0 = ω_{1}β_{1}X_{21} \cos \phi_{1} + ω_{2}β_{2}X_{22} \cos \phi_{2}\\[0.5cm]0.3 = ω_{1}X_{21} \cos \phi_{1} + ω_{2}X_{22} \cos \phi_{2}
Given the values of ω_{1}, ω_{2}, β_{1}, and β_{2}, these four equations can be written as
−1.6181X_{21} \sin \phi_{1} + 0.618X_{22} \sin \phi_{2} = 0\\[0.5cm]X_{21} \sin \phi_{1} + X_{22} \sin \phi_{2} = 0\\[0.5cm]−52.363X_{21} \cos \phi_{1}+ 7.639 X_{22} \cos \phi_{2} = 0\\[0.5cm]32.361X_{21} \cos \phi_{1}+ 12.361X_{22} \cos \phi_{2} = 0
These algebraic equations can be solved for the four unknowns X_{21}, X_{22}, \phi_{1}, and \phi_{2}. It is clear from the first two equations that X_{21} \sin \phi_{1} = X_{22} \sin \phi_{2} = 0. The third and fourth equations can be written in the following matrix form
\begin{bmatrix} −52.363 & 7.639 \\ 32.361 & 12.361 \end{bmatrix} \begin{bmatrix} X_{21} & \cos\phi_{1} \\ X_{22} & \cos\phi_{2} \end{bmatrix} = \begin{bmatrix} 0 \\ 0.3 \end{bmatrix}The solution of this matrix equation can be obtained by using Cramer’s rule or matrix methods as
\begin{bmatrix} X_{21} & \cos\phi_{1} \\ X_{22} & \cos\phi_{2} \end{bmatrix} = \begin{bmatrix} 2.562 × 10^{−3} \\ 1.756 × 10^{−2} \end{bmatrix}One, therefore, has X_{21} = 2.562 × 10^{−3}, X_{22} = 1.756 × 10^{−2}, and \phi_{1} =\phi_{2} = 0.