22-Mec-A2 Kinematics and Dynamics of Machines · May 2014

Question 6 of 6: Two-DOF forced vibration — steady-state response

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

National Exams, May 2014 — 07-Mec-A2, Kinematics and Dynamics of Machines. 3 hours, OPEN BOOK. Six questions of 20 marks each; candidates answer any FIVE. All SIX are solved in full below (Part A — mechanisms and machine dynamics, Q1–Q4; Part B — mechanical vibration, Q5–Q6).

Reference texts: R. L. Norton, Design of Machinery, 5th ed. (instant centres and mechanical advantage, cam design, gear-train kinematics, linkage balancing); S. S. Rao, Mechanical Vibrations, 6th ed. (impulsive loading, single- and two-degree-of-freedom free and forced response); J. E. Shigley & C. R. Mischke, Mechanical Engineering Design (planetary-train cross-check).

Question 6: Two-DOF forced vibration — steady-state response (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. Two equal masses $m=5\ \text{kg}$ in a vertical chain from a fixed ceiling, each stage a spring $k=500\ \text{N/m}$ in parallel with a damper $c=20\ \text{N}\cdot\text{s/m}$. Harmonic force $Q_1=10\sin\Omega t$ on the upper mass, $Q_2=0$, drive frequency $\Omega=20\ \text{rad/s}$.

Find. The steady-state displacements $x_1(t)$ and $x_2(t)$.

Figure 6.1 — two-DOF chain: ceiling–(c,k)–mass 1–(c,k)–mass 2, driven by Q₁ on mass 1.

Approach. Write the mass, damping and stiffness matrices, assume a complex harmonic response, and invert the dynamic-stiffness (impedance) matrix at $\Omega$.

Equations of motion. With $x_1,x_2$ measured downward from equilibrium, $$\mathbf M=\begin{bmatrix}m&0\\0&m\end{bmatrix},\ \mathbf C=\begin{bmatrix}2c&-c\\-c&c\end{bmatrix},\ \mathbf K=\begin{bmatrix}2k&-k\\-k&k\end{bmatrix},\ \mathbf Q=\begin{Bmatrix}10\\0\end{Bmatrix}\sin\Omega t.$$
Harmonic (impedance) form. For $\mathbf x=\mathrm{Im}(\mathbf X e^{i\Omega t})$ the amplitudes solve $\mathbf Z\,\mathbf X=\mathbf F$ with the dynamic stiffness $\mathbf Z=\mathbf K-\Omega^2\mathbf M+i\Omega\mathbf C$ and $\mathbf F=(10,0)^{T}$.
Evaluate $\mathbf Z$ at $\Omega=20$. With $m\Omega^2=2000$: $$Z_{11}=-1000+800i,\quad Z_{12}=Z_{21}=-500-400i,\quad Z_{22}=-1500+400i.$$ The determinant is $\det\mathbf Z=1.09\times10^{6}-2.00\times10^{6}i$.
Solve for the amplitudes. By Cramer’s rule $X_1=Z_{22}F_1/\det\mathbf Z$ and $X_2=-Z_{21}F_1/\det\mathbf Z$, giving $$X_1=-4.69-4.94i\ \text{mm},\qquad X_2=-0.49+2.77i\ \text{mm}.$$ Hence $|X_1|=6.82\ \text{mm}$ at $-133.5^\circ$, $|X_2|=2.81\ \text{mm}$ at $+100.1^\circ$.
Steady-state response. $$x_1(t)=6.82\sin(20t-133.5^\circ)\ \text{mm},\qquad x_2(t)=2.81\sin(20t+100.1^\circ)\ \text{mm}.$$ Both masses oscillate at the drive frequency 20 rad/s; the driven upper mass leads with the larger amplitude, and the phase lags reflect the 20 N·s/m damping.

Quantity	Value
$Z_{11},Z_{12},Z_{22}$	$-1000+800i$, $-500-400i$, $-1500+400i$
Upper mass $x_1$	6.82 mm, phase −133.5°
Lower mass $x_2$	2.81 mm, phase +100.1°
Response frequency	20 rad/s (drive frequency)

Check. The stiffness in the source is printed “$k=500\ \text{N/s}$”, which is dimensionally impossible for a spring; it is read as $k=500\ \text{N/m}$ (the only physically consistent value), matching $c$ in N·s/m. The two undamped natural frequencies are $\omega_{1,2}=\sqrt{(3\mp\sqrt5)k/2m}=6.18$ and $16.2\ \text{rad/s}$, so the 20 rad/s drive is above both resonances.

Topic. Steady-state harmonic response of a damped two-degree-of-freedom system by the dynamic-stiffness (impedance) method.

Key relations. $\mathbf Z(\Omega)=\mathbf K-\Omega^2\mathbf M+i\Omega\mathbf C$; amplitudes $\mathbf X=\mathbf Z^{-1}\mathbf F$; physical response $\mathbf x=\mathrm{Im}(\mathbf X e^{i\Omega t})$, i.e. $|X_j|\sin(\Omega t+\angle X_j)$.

Why this works. A linear system driven harmonically responds at the same frequency, so replacing $d/dt\to i\Omega$ turns the coupled differential equations into complex algebra. Inverting $\mathbf Z$ once yields both amplitudes and phases directly, with damping captured by the imaginary part.

Common pitfall. Assembling $\mathbf C$ or $\mathbf K$ without the coupling terms — the upper mass sees both springs/dampers ($2k,2c$) while the lower sees one ($k,c$), with $-k,-c$ off-diagonal; and dropping the phase (taking $|\mathbf Z^{-1}\mathbf F|$ of a real force) loses the lag. (Rao, Mechanical Vibrations, ch. 5.)

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