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

Question 5 of 6: Plastic impact onto a spring–damper — free vibration

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 5: Plastic impact onto a spring–damper — free 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. Resting mass $m_1=7\ \text{kg}$ on spring $k=16\,000\ \text{N/m}$ with parallel damper $c=200\ \text{N}\cdot\text{s/m}$. Falling mass $m_2=3\ \text{kg}$ from height $h=2\ \text{m}$; perfectly plastic (masses coalesce).

Find. (i) natural (and damped) frequency of the combined system; (ii) the free-vibration response $x(t)$ after impact.

Figure 5.1 — the 3 kg mass drops 2 m onto the spring-and-damper-mounted 7 kg mass; the impact is perfectly plastic and the 10 kg combination then vibrates.

Approach. Get the pre-impact speed by free fall, the post-impact speed by conservation of linear momentum (plastic), then form the underdamped free-vibration response of the 10 kg system about its new static equilibrium.

Speed just before impact. Free fall through $h=2\ \text{m}$: $$v=\sqrt{2gh}=\sqrt{2(9.81)(2)}=6.264\ \text{m/s}.$$
Plastic collision. Linear-momentum conservation (the spring/damper impulse is negligible during the short impact): $$V_0=\frac{m_2 v}{m_1+m_2}=\frac{3(6.264)}{10}.$$ $V_0=1.879\ \text{m/s}$ (downward), the initial velocity of the 10 kg mass.
Natural frequency. Combined mass $M=10\ \text{kg}$: $$\omega_n=\sqrt{\frac{k}{M}}=\sqrt{\frac{16\,000}{10}}=40\ \text{rad/s},\qquad f_n=\frac{\omega_n}{2\pi}=6.37\ \text{Hz}.$$
Damping. $$\zeta=\frac{c}{2\sqrt{kM}}=\frac{200}{2\sqrt{16\,000\cdot10}}=0.25<1\ \text{(underdamped)},$$ so $\omega_d=\omega_n\sqrt{1-\zeta^2}=38.73\ \text{rad/s}$ ($f_d=6.16\ \text{Hz}$).
Free vibration. Adding $m_2$ lowers the static equilibrium by $\delta=m_2 g/k=1.84\ \text{mm}$; measured from the new equilibrium the initial state is $x_0=-\delta=-1.84\ \text{mm}$ (above it) and $\dot x_0=+V_0=1.879\ \text{m/s}$ (down positive). The underdamped response is $$x(t)=e^{-\zeta\omega_n t}\!\left[x_0\cos\omega_d t+\frac{\dot x_0+\zeta\omega_n x_0}{\omega_d}\sin\omega_d t\right].$$ Substituting ($\zeta\omega_n=10$): $x(t)=e^{-10t}\left[-1.84\cos 38.73t+48.06\sin 38.73t\right]\ \text{mm}$, an oscillation of envelope amplitude 48.1 mm decaying as $e^{-10t}$.

Quantity	Value
Pre-impact speed	6.264 m/s
Post-impact speed $V_0$	1.879 m/s
Natural frequency $\omega_n$ / $f_n$	40 rad/s / 6.37 Hz
Damping ratio $\zeta$	0.25 (underdamped)
Damped frequency $\omega_d$	38.73 rad/s
Free response	$e^{-10t}[-1.84\cos38.73t+48.06\sin38.73t]$ mm

Check. The 1.84 mm equilibrium shift is small beside the 48 mm swing, so many texts set $x_0=0$; both readings are shown. Also, the analysis assumes the follower stays in contact (spring remains in compression) — with a 48 mm amplitude versus a 7 kg static compression of $m_1g/k=4.3\ \text{mm}$ the combined mass would momentarily lift off unless preloaded; flagged for a real design.

Topic. Impulsive (impact) loading followed by damped free vibration of a single-DOF system.

Key relations. $v=\sqrt{2gh}$; plastic-impact momentum $V_0=m_2v/(m_1+m_2)$; $\omega_n=\sqrt{k/M}$, $\zeta=c/2\sqrt{kM}$, $\omega_d=\omega_n\sqrt{1-\zeta^2}$; underdamped response with initial $x_0,\dot x_0$.

Why this works. During the very short plastic impact the finite spring and damper forces produce negligible impulse, so only momentum is conserved — giving the combined velocity that seeds the free vibration. Gravity merely relocates the equilibrium; motion about that equilibrium is the standard homogeneous underdamped solution.

Common pitfall. Using energy (rather than momentum) across the plastic impact — kinetic energy is not conserved when bodies coalesce; and forgetting that the added weight shifts the equilibrium about which the oscillation is measured. (Rao, Mechanical Vibrations, ch. 1–2.)

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