Question 5 of 6: Coupled two-degree-of-freedom vibration

Question 5: Coupled two-degree-of-freedom vibration (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. A vertical chain: top mass $2m$ connected to the middle mass $6m$ by two springs of stiffness $k$ each (parallel, $2k$); the middle mass connected to ground by two springs of $5k$ each (parallel, $10k$). Let $x_1$ (top, mass $2m$) and $x_2$ (middle, mass $6m$) be downward displacements from equilibrium. Numbers: $m=10\ \text{kg}$, $k=50\,000\ \text{N/m}$.

Find. parts (a)–(g) as listed.

Approach. Assemble $\mathbf{M},\mathbf{K}$, solve the eigenproblem $(\mathbf{K}-\omega^2\mathbf{M})\boldsymbol{\phi}=\mathbf 0$ for the natural frequencies and mode shapes, form modal masses and mass-normalized modes, then use modal superposition for the initial-velocity response.

1. (a) Equations of motion. Newton on each mass (upper spring $2k$, lower spring $10k$): $$2m\ddot x_1 + 2k(x_1-x_2)=0,\qquad 6m\ddot x_2 + 2k(x_2-x_1) + 10k\,x_2 = 0.$$ In matrix form, $$\begin{bmatrix}2m&0\\0&6m\end{bmatrix}\!\begin{Bmatrix}\ddot x_1\\\ddot x_2\end{Bmatrix} + \begin{bmatrix}2k&-2k\\-2k&12k\end{bmatrix}\!\begin{Bmatrix}x_1\\x_2\end{Bmatrix} = \mathbf 0.$$ Numerically $\mathbf M=\begin{bmatrix}20&0\\0&60\end{bmatrix}\text{kg}$, $\mathbf K=\begin{bmatrix}1.0&-1.0\\-1.0&6.0\end{bmatrix}\times10^{5}\ \text{N/m}$.
2. (b) Natural frequencies. Setting $\det(\mathbf K-\omega^2\mathbf M)=0$ with $\lambda=\omega^2$ gives $3m^2\lambda^2-9mk\lambda+5k^2=0$, i.e. $\lambda = \dfrac{k}{m}\dfrac{9\pm\sqrt{21}}{6}$. With $k/m=5000$, $$\lambda_1=3681\ \Rightarrow\ \boxed{\omega_1=60.67\ \text{rad/s}}\ (9.66\,\text{Hz}),\quad \lambda_2=11\,319\ \Rightarrow\ \boxed{\omega_2=106.39\ \text{rad/s}}\ (16.93\,\text{Hz}).$$
3. (c) Amplitude ratios. From the first row, $r=\dfrac{x_1}{x_2}=\dfrac{2k}{2k-2m\lambda}=\dfrac{k}{k-m\lambda}$: $$r_1 = \frac{50\,000}{50\,000-10(3681)} = +3.79,\qquad r_2 = \frac{50\,000}{50\,000-10(11\,319)} = -0.791.$$ Mode 1 is in-phase (top leads), mode 2 is out-of-phase.
4. (d) Modal vectors. Taking $x_2=1$: $$\boldsymbol{\phi}^{(1)}=\begin{Bmatrix}3.79\\1\end{Bmatrix},\qquad \boldsymbol{\phi}^{(2)}=\begin{Bmatrix}-0.791\\1\end{Bmatrix}.$$
5. (e) Modal masses. $M_r=\boldsymbol{\phi}^{(r)\mathsf T}\mathbf M\,\boldsymbol{\phi}^{(r)}$: $$M_1 = 20(3.79)^2+60(1)^2 = 347.5\ \text{kg},\qquad M_2 = 20(-0.791)^2+60 = 72.5\ \text{kg}.$$ (Check: $K_r/M_r=\omega_r^2$ recovers 3681 and 11 319.)
6. (f) Mass-normalized modes. Divide each mode by $\sqrt{M_r}$: $$\boldsymbol{\varphi}^{(1)}=\frac{1}{\sqrt{347.5}}\begin{Bmatrix}3.79\\1\end{Bmatrix}=\begin{Bmatrix}0.2034\\0.0536\end{Bmatrix},\quad \boldsymbol{\varphi}^{(2)}=\frac{1}{\sqrt{72.5}}\begin{Bmatrix}-0.791\\1\end{Bmatrix}=\begin{Bmatrix}-0.0929\\0.1174\end{Bmatrix},$$ satisfying $\boldsymbol\Phi^{\mathsf T}\mathbf M\boldsymbol\Phi=\mathbf I$.
7. (g) Free response to $\dot x_1(0)=10$ m/s. Initial state $\mathbf x(0)=\mathbf 0$, $\dot{\mathbf x}(0)=\{10,0\}^{\mathsf T}$. Modal velocities $\dot\eta_r(0)=\boldsymbol{\varphi}^{(r)\mathsf T}\mathbf M\dot{\mathbf x}(0)$ give $\dot\eta_1(0)=40.68$, $\dot\eta_2(0)=-18.58$. Each mode is $\eta_r=(\dot\eta_r(0)/\omega_r)\sin\omega_r t$, so $$\mathbf x(t)=\boldsymbol{\varphi}^{(1)}\frac{40.68}{60.67}\sin(60.67\,t)+\boldsymbol{\varphi}^{(2)}\frac{-18.58}{106.39}\sin(106.39\,t).$$ Componentwise (metres): $$x_1(t)=0.1364\sin(60.67t)+0.01623\sin(106.39t),$$ $$x_2(t)=0.03597\sin(60.67t)-0.02051\sin(106.39t).$$