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3.5 Single-DOF and 2-DOF Vibrations, Damping & Dynamic Balancing

Key Takeaways

  • The fundamental natural frequency of an SDOF system is $\omega_n = \sqrt{k/m} = \sqrt{g/\delta_{st}}$, where $\delta_{st}$ is the static deflection under gravity, providing a rapid calculation method ($f_n \approx \frac{15.76}{\sqrt{\delta_{st}\text{ (mm)}}}$ in SI, $f_n \approx \frac{3.127}{\sqrt{\delta_{st}\text{ (in)}}}$ in US Customary).
  • The viscous damping ratio $\zeta = c / c_{cr} = c / (2\sqrt{km})$ governs free decay: underdamped systems ($\zeta < 1$) oscillate at damped natural frequency $\omega_d = \omega_n \sqrt{1 - \zeta^2}$, with decay measured by logarithmic decrement $\delta = \frac{1}{n} \ln(x_0/x_n) \approx 2\pi \zeta$.
  • Steady-state forced vibration amplitude is governed by the Dynamic Magnification Factor $DMF = 1 / \sqrt{(1 - r^2)^2 + (2\zeta r)^2}$; at resonance ($r = \omega/\omega_n = 1.0$), response is strictly damping-limited ($DMF_{\text{peak}} = 1/(2\zeta)$) with phase lag $\phi = 90^\circ$.
  • Vibration isolation occurs exclusively in the supercritical zone $r > \sqrt{2}$ where Transmissibility $TR = \sqrt{\frac{1 + (2\zeta r)^2}{(1 - r^2)^2 + (2\zeta r)^2}} < 1.0$; all transmissibility curves intersect at $TR = 1.0$ when $r = \sqrt{2}$.
  • In 2-DOF systems, a Frahm Tuned Mass Damper tuned to $\omega_a = \sqrt{k_2/m_2} = \omega$ completely nullifies steady-state vibration of the primary mass ($X_1 = 0$), splitting the original resonance into two safe peak frequencies away from the excitation.
Last updated: August 2026

Single-DOF and 2-DOF Vibrations, Damping & Dynamic Balancing

Mechanical vibration is the oscillatory motion of dynamic systems about an equilibrium configuration. On the NCEES PE Mechanical exam, vibration and dynamics questions assess five essential topic areas: single-degree-of-freedom (SDOF) undamped and damped natural frequencies, viscous damping regimes and logarithmic decrement, harmonically excited steady-state forced response (Dynamic Magnification Factor), vibration isolation and force transmissibility, and two-degree-of-freedom (2-DOF) tuned vibration absorbers and dynamic balancing.

+---------------------------------------------------------------------------------------------------+
|                         MECHANICAL VIBRATIONS & DYNAMICS TAXONOMY                                 |
|                                                                                                   |
|   [FREE SDOF VIBRATION]    [DAMPING REGIMES]        [FORCED RESONANCE]       [ISOLATION & TMD]    |
|   - ODE: m*x'' + k*x = 0   - c_cr = 2*sqrt(k*m)     - F(t) = F_0*sin(w*t)    - r = w / w_n        |
|   - w_n = sqrt(k/m)        - zeta = c / c_cr        - DMF = 1/sqrt(...)      - TR = sqrt(...)     |
|   - w_n = sqrt(g/delta_st) - Underdamped: zeta < 1  - Resonance at r = 1     - Isolation: r>sqrt2 |
|   - Springs Ser/Parallel   - Log Dec: delta = 2pi*z - Phase Lag: phi=90 deg  - 2-DOF Tuned Damper |
+---------------------------------------------------------------------------------------------------+

1. SDOF Free Undamped Vibration & Equivalent Stiffness

The fundamental mass-spring model represents an elastic structure of equivalent stiffness $k$ supporting an inertial mass $m$.

+---------------------------------------------------------------------------------------------------+
|                                 SDOF MASS-SPRING-DAMPER MODEL                                     |
|                                                                                                   |
|                                    +-----------------------+                                      |
|                                    |    FIXED CEILING      |                                      |
|                                    +---+---------------+---+                                      |
|                                        |               |                                          |
|                         Spring (k)    [k]             [c]    Dashpot / Damper (c)                 |
|                                        |               |                                          |
|                                    +---+---------------+---+                                      |
|                                    |       MASS (m)        | ---> Displacement x(t)              |
|                                    +-----------------------+                                      |
|                                                |                                                  |
|                                                v External Excitation: F(t) = F_0 * sin(w*t)       |
+---------------------------------------------------------------------------------------------------+

Equation of Motion & Natural Frequency

Applying Newton's second law ($m \ddot{x} + k x = 0$):

ωn=km(rad/s),fn=ωn2π=12πkm(Hz),Tn=1fn=2πωn(seconds)\omega_n = \sqrt{\frac{k}{m}} \quad (\text{rad/s}), \quad f_n = \frac{\omega_n}{2\pi} = \frac{1}{2\pi}\sqrt{\frac{k}{m}} \quad (\text{Hz}), \quad T_n = \frac{1}{f_n} = \frac{2\pi}{\omega_n} \quad (\text{seconds})

Static Deflection Equivalence ($\delta_{st}$)

Under gravity, static deflection is $\delta_{st} = \frac{m g}{k}$. Substituting $\frac{k}{m} = \frac{g}{\delta_{st}}$ provides a rapid shortcut to estimate natural frequency directly from measured static sag:

ωn=gδst\omega_n = \sqrt{\frac{g}{\delta_{st}}}

  • In SI Units ($g = 9.81\text{ m/s}^2$, $\delta_{st}$ in mm): fn=12π9810δst (mm)≈15.76δst (mm)(Hz)f_n = \frac{1}{2\pi}\sqrt{\frac{9810}{\delta_{st}\text{ (mm)}}} \approx \frac{15.76}{\sqrt{\delta_{st}\text{ (mm)}}} \quad (\text{Hz})
  • In US Customary Units ($g = 386.4\text{ in/s}^2$, $\delta_{st}$ in inches): fn=12π386.4δst (in)≈3.127δst (in)(Hz)f_n = \frac{1}{2\pi}\sqrt{\frac{386.4}{\delta_{st}\text{ (in)}}} \approx \frac{3.127}{\sqrt{\delta_{st}\text{ (in)}}} \quad (\text{Hz})

Equivalent Spring Stiffness Combinations ($k_{eq}$)

Configuration / ElementEquivalent Stiffness ($k_{eq}$)Physical Boundary Condition
Springs in Parallelkeq=k1+k2+⋯+knk_{eq} = k_1 + k_2 + \dots + k_nIdentical deflection ($x_1 = x_2 = x$)
Springs in Series1keq=1k1+1k2  ⟹  keq=k1k2k1+k2\frac{1}{k_{eq}} = \frac{1}{k_1} + \frac{1}{k_2} \implies k_{eq} = \frac{k_1 k_2}{k_1 + k_2}Shared load ($F_1 = F_2 = F$)
Cantilever Beam (End Load)k=3EIL3k = \frac{3 E I}{L^3}Shaft or bracket tip deflection
Simply Supported Beam (Center Load)k=48EIL3k = \frac{48 E I}{L^3}Shaft mid-span between bearings
Fixed-Fixed Beam (Center Load)k=192EIL3k = \frac{192 E I}{L^3}Clamped shaft mid-span
Torsional Shaftkt=GJLk_t = \frac{G J}{L}Rotational shaft stiffness ($\text{N}\cdot\text{m/rad}$)

2. Viscous Damping & Response Regimes

Adding a linear viscous dashpot (damping force $F_d = -c \dot{x}$) yields the governing differential equation:

mx¨+cx˙+kx=0  ⟹  x¨+2ζωnx˙+ωn2x=0m \ddot{x} + c \dot{x} + k x = 0 \implies \ddot{x} + 2 \zeta \omega_n \dot{x} + \omega_n^2 x = 0

Where:

  • Critical Damping Coefficient ($c_{cr}$): The minimum damping required to prevent oscillatory motion. ccr=2mωn=2kmc_{cr} = 2 m \omega_n = 2 \sqrt{k m}
  • Damping Ratio ($\zeta$): Dimensionless measure of damping severity. ζ=cccr=c2mωn=c2km\zeta = \frac{c}{c_{cr}} = \frac{c}{2 m \omega_n} = \frac{c}{2 \sqrt{k m}}
+---------------------------------------------------------------------------------------------------+
|                                DAMPING RESPONSE CLASSIFICATIONS                                   |
|                                                                                                   |
|   UNDERDAMPED (zeta < 1):      CRITICALLY DAMPED (zeta = 1):    OVERDAMPED (zeta > 1):            |
|   - Oscillatory decaying sine  - Fastest return to equilibrium  - Sluggish non-oscillatory decay  |
|   - w_d = w_n * sqrt(1-zeta^2) - No overshoot                   - Two real negative roots         |
|   - x(t) = X0*e^(-z*w_n*t)*cos - x(t) = (C1 + C2*t)*e^(-w_n*t)  - x(t) = C1*e^(s1*t)+C2*e^(s2*t) |
+---------------------------------------------------------------------------------------------------+

Logarithmic Decrement ($\delta$)

The rate of decay of underdamped free vibrations is measured by the natural log of the ratio of successive peak amplitudes separated by $n$ cycles:

δ=ln⁡(x1x2)=1nln⁡(x0xn)=2πζ1−ζ2\delta = \ln\left( \frac{x_1}{x_2} \right) = \frac{1}{n} \ln\left( \frac{x_0}{x_n} \right) = \frac{2 \pi \zeta}{\sqrt{1 - \zeta^2}}

Solving for damping ratio $\zeta$ from measured log decrement $\delta$:

ζ=δ(2π)2+δ2≈δ2π(for light damping ζ<0.20)\zeta = \frac{\delta}{\sqrt{(2\pi)^2 + \delta^2}} \approx \frac{\delta}{2\pi} \quad (\text{for light damping } \zeta < 0.20)

3. Harmonically Forced Vibration & Dynamic Magnification

Subjecting an SDOF system to harmonic excitation $F(t) = F_0 \sin(\omega t)$ produces steady-state response $x_p(t) = X \sin(\omega t - \phi)$.

+---------------------------------------------------------------------------------------------------+
|                         DYNAMIC MAGNIFICATION FACTOR (DMF) CURVES                                 |
|                                                                                                   |
|   DMF (X / X_st)                                                                                  |
|     ^                                                                                             |
|   5 |             |  zeta = 0.1 (Peak DMF = 1/(2*zeta) = 5.0)                                     |
|   4 |            / \                                                                              |
|   3 |           /   \                                                                             |
|   2 |          /     \   zeta = 0.25 (Peak = 2.0)                                                 |
|   1 +---------+-------+---------------------------> Isolation Zone (r > sqrt(2))                  |
|     0        0.5     1.0     1.414 (sqrt(2))  2.0     3.0  Frequency Ratio r = w / w_n            |
+---------------------------------------------------------------------------------------------------+

Steady-State Amplitude ($X$) and Dynamic Magnification Factor ($DMF$)

Let $X_{st} = \frac{F_0}{k}$ be zero-frequency static deflection, and $r = \frac{\omega}{\omega_n}$ be frequency ratio:

X=F0/k(1−r2)2+(2ζr)2=Xst×DMFX = \frac{F_0 / k}{\sqrt{(1 - r^2)^2 + (2\zeta r)^2}} = X_{st} \times DMF

DMF=1(1−r2)2+(2ζr)2DMF = \frac{1}{\sqrt{(1 - r^2)^2 + (2\zeta r)^2}}

Phase Angle (ϕ): tan⁡ϕ=2ζr1−r2\text{Phase Angle (}\phi\text{): } \tan\phi = \frac{2 \zeta r}{1 - r^2}

Three Dynamic Operating Regimes

  1. Low Frequency ($r \ll 1$): In phase ($\phi \approx 0^\circ$), governed strictly by stiffness ($X \approx F_0 / k$).
  2. Resonant Peak ($r \approx 1$): Lags by $\phi = 90^\circ$, governed strictly by damping. At exact resonance ($r = 1$): DMFresonance=12ζ=Q(Quality Factor)DMF_{\text{resonance}} = \frac{1}{2\zeta} = Q \quad (\text{Quality Factor})
  3. High Frequency ($r \gg 1$): Out of phase ($\phi \approx 180^\circ$), governed strictly by mass inertia ($X \to 0$).

4. Vibration Isolation & Force Transmissibility

Vibration isolation prevents machine dynamic forces from transmitting into the support structure.

+---------------------------------------------------------------------------------------------------+
|                         FORCE TRANSMISSIBILITY (TR) MASTER CURVE                                  |
|                                                                                                   |
|   Transmissibility TR                                                                             |
|     ^                                                                                             |
|   4 |        RESONANCE AMPLIFICATION ZONE            VIBRATION ISOLATION ZONE                     |
|     |        (r < sqrt(2), TR > 1.0)                 (r > sqrt(2), TR < 1.0)                      |
|   3 |                 /\                                                                          |
|     |                /  \                                                                         |
|   2 |               /    \                                                                        |
|   1 +--------------+------+-------+------------------------------------------ (TR = 1.0)           |
|     |             /        \       \                                                              |
|   0 +------------+----------+-------+------------------------------> r = w / w_n                  |
|     0           0.5        1.0    1.414 (sqrt(2))   2.0      3.0                                  |
+---------------------------------------------------------------------------------------------------+

The Transmissibility Equation ($TR$)

The ratio of transmitted force $F_T$ to excitation force $F_0$ is the Transmissibility ($TR$):

TR=FTF0=1+(2ζr)2(1−r2)2+(2ζr)2TR = \frac{F_T}{F_0} = \sqrt{\frac{1 + (2\zeta r)^2}{(1 - r^2)^2 + (2\zeta r)^2}}

For undamped or lightly damped isolators ($\zeta \approx 0$):

TR≈1∣1−r2∣=1r2−1(for r>2)TR \approx \frac{1}{|1 - r^2|} = \frac{1}{r^2 - 1} \quad (\text{for } r > \sqrt{2})

The Fundamental Laws of Vibration Isolation

  • All transmissibility curves pass through $TR = 1.0$ at exact frequency ratio $r = \sqrt{2} \approx 1.414$, regardless of damping $\zeta$.
  • Vibration isolation occurs ONLY when $r > \sqrt{2}$ ($TR < 1.0$). If $r < \sqrt{2}$, mounting the machine on springs amplifies transmitted forces.
  • The Damping Paradox: In the isolation zone ($r > \sqrt{2}$), increasing damping $\zeta$ increases transmissibility $TR$ (reducing isolation efficiency). However, damping is mandatory to suppress destructive resonance peaks when accelerating through $r = 1$ during startup and shut-down coast-down.

Isolation Efficiency ($\eta_{\text{iso}}$)

ηiso=1−TR=1−1r2−1  ⟹  r=1TR+1\eta_{\text{iso}} = 1 - TR = 1 - \frac{1}{r^2 - 1} \implies r = \sqrt{\frac{1}{TR} + 1}

5. Two-Degree-of-Freedom Systems & Tuned Mass Dampers

When a primary structure $(m_1, k_1)$ experiences severe resonant vibration at excitation frequency $\omega = \sqrt{k_1/m_1}$, an auxiliary Tuned Mass Damper (Frahm Dynamic Vibration Absorber) $(m_2, k_2)$ can be attached.

+---------------------------------------------------------------------------------------------------+
|                         FRAHM TUNED DYNAMIC VIBRATION ABSORBER (2-DOF)                            |
|                                                                                                   |
|                                   +-----------------------+                                       |
|                                   |    FIXED CEILING      |                                       |
|                                   +-----------+-----------+                                       |
|                                               |                                                   |
|                                           Spring [k_1]                                            |
|                                               |                                                   |
|                                   +-----------+-----------+                                       |
|                                   |    PRIMARY MASS (m_1) | <--- F(t) = F_0 * sin(w*t)            |
|                                   +-----------+-----------+                                       |
|                                               |                                                   |
|                                           Spring [k_2]  (Absorber)                                |
|                                               |                                                   |
|                                   +-----------+-----------+                                       |
|                                   |   ABSORBER MASS (m_2) |                                       |
|                                   +-----------------------+                                       |
+---------------------------------------------------------------------------------------------------+

Absorber Sizing & Response Equations

Setting the absorber natural frequency equal to the excitation frequency:

ωa=k2m2=ω\omega_a = \sqrt{\frac{k_2}{m_2}} = \omega

At this tuned condition, the steady-state amplitudes are:

X1=0(Primary mass comes to a COMPLETE STANDSTILL!)X_1 = 0 \quad (\text{Primary mass comes to a COMPLETE STANDSTILL!})

X2=−F0k2(Absorber spring force exactly cancels F0)X_2 = -\frac{F_0}{k_2} \quad (\text{Absorber spring force exactly cancels } F_0)

[!IMPORTANT] Frequency Splitting: Adding the absorber eliminates vibration at $\omega$, but splits the original single resonance peak into two new resonant peaks ($\omega_1 < \omega < \omega_2$). The system must operate at constant speed $\omega$ without drifting into $\omega_1$ or $\omega_2$.

6. Dynamic Balancing of Rotating Machinery

  • Static Unbalance (Single-Plane): The center of mass is offset from the rotational axis ($e \ne 0$), producing a single rotating centrifugal force vector $\vec{F}_c = m e \omega^2$. Corrected by adding or removing mass in one correction plane.
  • Dynamic Unbalance (Two-Plane): The principal inertia axis is both offset and skewed relative to the shaft rotational axis, producing both a net centrifugal force and a rotating dynamic couple (moment). Requires correction in two distinct axial planes using 4-run or vibration analyzer phase measurements.

7. Step-by-Step Worked Problem: Complete Vibration Isolation Design

Problem: A $350\text{ kg}$ motor-driven exhaust fan operates at $1750\text{ rpm}$ and produces a vertical unbalance excitation force $F_0 = 800\text{ N}$. An engineer must design 4 identical corner spring mounts to achieve $85%$ force isolation ($\eta_{\text{iso}} = 0.85$, $TR = 0.15$) with negligible damping ($\zeta \approx 0$). Calculate:

  1. Operating excitation frequency $f$ and $\omega$.
  2. Required frequency ratio $r$.
  3. System natural frequency $f_n$ in Hz.
  4. Required stiffness of each of the 4 spring mounts.
  5. Static deflection $\delta_{st}$ of the mounts.
  6. Dynamic force $F_T$ transmitted to the supporting floor.
+---------------------------------------------------------------------------------------------------+
|                                 STEP-BY-STEP SOLUTION WORKFLOW                                    |
|                                                                                                   |
|   STEP 1: Excitation Frequencies                                                                  |
|           f = 1750 rpm / 60 = 29.167 Hz                                                           |
|           omega = 2 * pi * 29.167 = 183.26 rad/s                                                  |
|                                                                                                   |
|   STEP 2: Frequency Ratio (r) for TR = 0.15 in Isolation Zone (r > sqrt(2))                       |
|           TR = 1 / (r^2 - 1)  ===>  0.15 = 1 / (r^2 - 1)                                          |
|           r^2 - 1 = 1 / 0.15 = 6.667  ===>  r^2 = 7.667                                           |
|           r = sqrt(7.667) = 2.7689                                                                |
|                                                                                                   |
|   STEP 3: Required Natural Frequency (f_n)                                                        |
|           f_n = f / r = 29.167 Hz / 2.7689 = 10.534 Hz                                            |
|           omega_n = omega / r = 183.26 / 2.7689 = 66.185 rad/s                                    |
|                                                                                                   |
|   STEP 4: Total Spring Stiffness & Stiffness per Mount                                            |
|           k_total = m * (omega_n)^2 = 350 kg * (66.185 rad/s)^2 = 350 * 4380.5 = 1,533,160 N/m   |
|           k_total = 1.533 MN/m                                                                    |
|           For 4 parallel corner mounts: k_each = k_total / 4 = 1,533,160 / 4 = 383,290 N/m        |
|           k_each = 383.3 kN/m                                                                     |
|                                                                                                   |
|   STEP 5: Static Deflection Check                                                                 |
|           delta_st = m * g / k_total = (350 * 9.81) / 1,533,160 = 3433.5 / 1,533,160              |
|           delta_st = 0.002239 m = 2.24 mm                                                         |
|           (Check: f_n = 15.76 / sqrt(2.24 mm) = 15.76 / 1.497 = 10.53 Hz [Matches!])              |
|                                                                                                   |
|   STEP 6: Transmitted Dynamic Force (F_T)                                                         |
|           F_T = TR * F_0 = 0.15 * 800 N = 120.0 N                                                 |
+---------------------------------------------------------------------------------------------------+

8. Exam Tips & Common Traps

[!TIP] Rapid Isolation Formulas:

  • For $90%$ isolation ($TR = 0.10$): $r = \sqrt{10 + 1} = \sqrt{11} \approx 3.32$.
  • For $85%$ isolation ($TR = 0.15$): $r = \sqrt{6.67 + 1} = \sqrt{7.67} \approx 2.77$.
  • For $80%$ isolation ($TR = 0.20$): $r = \sqrt{5 + 1} = \sqrt{6} \approx 2.45$.

[!WARNING] Common Pitfalls:

  • The Isolation Boundary: Placing mounts such that $r < \sqrt{2}$ causes force amplification, transmitting higher vibration forces into the floor than if the machine were bolted down rigidly!
  • Series vs. Parallel Stiffness: Placing identical springs in parallel doubles total stiffness ($2k$), which raises natural frequency and can accidentally push $r$ below $\sqrt{2}$ into the amplification zone.
Test Your Knowledge

An underdamped SDOF mechanical system exhibits free vibration decay where the amplitude of the first peak is x_0 = 12.0 mm and the amplitude after 5 complete cycles is x_5 = 1.5 mm. What is the damping ratio (zeta) of the system?

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Test Your Knowledge

An electric motor rotating at 1750 rpm is mounted on spring isolators. To achieve 80% force isolation (transmissibility TR = 0.20), what frequency ratio r = w / w_n is required assuming negligible damping?

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Test Your Knowledge

A 250 kg machine is placed onto a set of elastic mounting pads, causing an observed static gravity deflection of 4.0 mm. What is the fundamental natural frequency of the mounted system in Hertz?

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Test Your Knowledge

How does adding viscous damping (increasing zeta) affect a harmonically excited SDOF system when operating at resonance (r = 1.0) versus operating deep in the isolation zone (r > sqrt(2))?

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