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22-Mec-A2 Kinematics and Dynamics of Machines · December 2013

Question 6 of 6: Steady-state response of a 2-DOF system

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

Notes on this paper

Paper: National Examinations — 07-Mec-A2 Kinematics and Dynamics of Machines, December 2013. Open-book, 3-hour paper; answer FIVE of six questions, all of equal value (20 marks). Part A (mechanisms & machine dynamics, Q1–Q4) and Part B (mechanical vibration, Q5–Q6). All six questions are solved here.

Reference texts: R. L. Norton, Design of Machinery (6th ed., McGraw-Hill) — velocity/acceleration analysis, cams, gear trains, engine balancing; C. E. Wilson & J. P. Sadler, Kinematics and Dynamics of Machinery (3rd ed.); S. S. Rao, Mechanical Vibrations (6th ed., Pearson) — Part B; J. E. Shigley & C. R. Mischke, Mechanical Engineering Design.

Note on the figure-based questions. Q1 (six-bar) is a scaled graphical problem (“Scale 1:5”) with no printed dimensions; its link geometry is read from the drawing, so angular results carry a graphical tolerance of a few percent and the linear accelerations carry the 1:5 length calibration (±5 %). Q3 (planetary box) is solved from the kinematic topology inferred from the sectioned schematic. Both interpretations are stated explicitly with each answer.

Question 6: Steady-state response of a 2-DOF system (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 $m=5$ kg hung in series from a fixed support; a spring $k=500$ N/m and damper $c=15$ N·s/m in parallel above each mass. Harmonic force $Q_1=10\sin\Omega t$ N on the upper mass, $Q_2=0$, at $\Omega=20$ rad/s. Coordinates $x_1,x_2$ from static equilibrium.

Find. The steady-state amplitudes and phases of $x_1$ and $x_2$.

m (x₁)kcm (x₂)kcQ₁=10 sinΩtQ₂=0Two equal masses in series; harmonic force on the upper mass only.
Two-degree-of-freedom chain: each mass hangs on a parallel spring–damper; only the upper mass is forced.

Approach. Write the coupled equations of motion, assume a complex harmonic response $x_j=\mathrm{Re}(X_j e^{i\Omega t})$, and solve the $2\times2$ complex impedance system for the amplitudes and phase lags.

  1. Equations of motion. $$m\ddot x_1+2c\dot x_1+2k x_1-c\dot x_2-k x_2=Q_1,$$ $$m\ddot x_2+c\dot x_2+k x_2-c\dot x_1-k x_1=0.$$
  2. Complex impedance. With $x_j=X_j e^{i\Omega t}$ the system is $\mathbf Z\,\mathbf X=\mathbf Q$, $$\mathbf Z=\begin{bmatrix}2k-m\Omega^{2}+2ic\Omega & -(k+ic\Omega)\\ -(k+ic\Omega) & k-m\Omega^{2}+ic\Omega\end{bmatrix}.$$
  3. Substitute numbers. $m\Omega^{2}=2000$, so $$\mathbf Z=\begin{bmatrix}-1000+600i & -500-300i\\ -500-300i & -1500+300i\end{bmatrix},\quad \mathbf Q=\begin{bmatrix}10\\0\end{bmatrix}.$$
  4. Solve. Inverting the $2\times2$ complex system gives $X_1=(-6.09-5.29i)\times10^{-3}$ and $X_2=(0.365+3.05i)\times10^{-3}$ m, hence $|X_1|=8.07$ mm lagging the force by $139^\circ$, and $|X_2|=3.08$ mm leading by $83^\circ$.
  5. Steady-state motions. $$x_1(t)=8.07\sin(20t-139^\circ)\ \text{mm},\qquad x_2(t)=3.08\sin(20t+83^\circ)\ \text{mm}.$$
QuantityValue
Upper mass $x_1$8.07 mm, phase −139°
Lower mass $x_2$3.08 mm, phase +83°
Check: The source prints the stiffness as “k = 500 N/s”; a stiffness must have units N/m, so it is read as $k=500$ N/m. All numbers use that reading.
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