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

Q: 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?

0.066. The logarithmic decrement over 5 cycles is delta = (1/n) * ln(x_0 / x_n) = (1/5) * ln(12.0 / 1.5) = (1/5) * ln(8.0) = (1/5) * 2.0794 = 0.4159. For light damping: zeta = delta / sqrt((2*pi)^2 + delta^2) = 0.4159 / sqrt(39.478 + 0.173) = 0.4159 / 6.297 = 0.0661 (approx 0.066). 0.033 divides by 10 cycles; 0.416 is the raw logarithmic decrement delta; and 0.125 is the simple amplitude ratio.

Q: 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?

2.449. For an undamped isolator in the isolation zone (r > sqrt(2)), transmissibility is TR = 1 / (r^2 - 1). Setting TR = 0.20: 0.20 = 1 / (r^2 - 1) implies r^2 - 1 = 1 / 0.20 = 5. Therefore, r^2 = 6, which gives r = sqrt(6) = 2.449. 1.414 is the isolation crossover threshold (TR = 1.0); 2.000 yields TR = 0.333 (67% isolation); and 2.236 is sqrt(5).

Q: 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?

7.88 Hz. Using the static deflection shortcut: f_n = 15.76 / sqrt(delta_st in mm) = 15.76 / sqrt(4.0) = 15.76 / 2.0 = 7.88 Hz. 49.5 rad/s represents angular natural frequency (omega_n) rather than frequency in Hz (f_n); 12.4 Hz results from an incorrect gravity constant; and 2.47 Hz is an arithmetic error.

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 ( $m \ddot{x} + k x = 0$ ):

\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:

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

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

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

Configuration / Element	Equivalent Stiffness ( $k_{eq}$ )	Physical Boundary Condition
Springs in Parallel	$k_{eq} = k_1 + k_2 + \dots + k_n$	Identical deflection ( $x_1 = x_2 = x$ )
Springs in Series	$\frac{1}{k_{eq}} = \frac{1}{k_1} + \frac{1}{k_2} \implies k_{eq} = \frac{k_1 k_2}{k_1 + k_2}$	Shared load ( $F_1 = F_2 = F$ )
Cantilever Beam (End Load)	$k = \frac{3 E I}{L^3}$	Shaft or bracket tip deflection
Simply Supported Beam (Center Load)	$k = \frac{48 E I}{L^3}$	Shaft mid-span between bearings
Fixed-Fixed Beam (Center Load)	$k = \frac{192 E I}{L^3}$	Clamped shaft mid-span
Torsional Shaft	$k_t = \frac{G J}{L}$	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:

m \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. $c_{cr} = 2 m \omega_n = 2 \sqrt{k m}$
Damping Ratio ( $\zeta$ ): Dimensionless measure of damping severity. $\zeta = \frac{c}{c_{cr}} = \frac{c}{2 m \omega_n} = \frac{c}{2 \sqrt{k m}}$

+---------------------------------------------------------------------------------------------------+
|                                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:

\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 $\delta$ :

\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 = \frac{F_0 / k}{\sqrt{(1 - r^2)^2 + (2\zeta r)^2}} = X_{st} \times DMF

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

\text{Phase Angle (}\phi\text{): } \tan\phi = \frac{2 \zeta r}{1 - r^2}

Three Dynamic Operating Regimes

Low Frequency ( $r \ll 1$ ): In phase ( $\phi \approx 0^\circ$ ), governed strictly by stiffness ( $X \approx F_0 / k$ ).
Resonant Peak ( $r \approx 1$ ): Lags by $\phi = 90^\circ$ , governed strictly by damping. At exact resonance ( $r = 1$ ): $DMF_{\text{resonance}} = \frac{1}{2\zeta} = Q \quad (\text{Quality Factor})$
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 = \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 \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 ( $\eta_{\text{iso}}$ )

\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:

\omega_a = \sqrt{\frac{k_2}{m_2}} = \omega

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

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

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:

Operating excitation frequency $f$ and $\omega$ .
Required frequency ratio $r$ .
System natural frequency $f_n$ in Hz.
Required stiffness of each of the 4 spring mounts.
Static deflection $\delta_{st}$ of the mounts.
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?

0.033

0.066

0.125

0.416

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?

1.414

2.000

2.236

2.449

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?

7.88 Hz

12.4 Hz

49.5 Hz

2.47 Hz

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))?

Damping reduces peak vibration amplitude at resonance (r = 1.0) and also improves isolation efficiency (reduces TR) when r > sqrt(2).

Damping has no effect at resonance (r = 1.0), but dramatically improves isolation efficiency when r > sqrt(2).

Damping substantially reduces peak vibration amplitude at resonance (r = 1.0), but degrades isolation performance (increases TR) when r > sqrt(2).

Damping increases vibration amplitude at all operating frequencies.

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4.1 Fluid Statics, Manometry, Submerged Surfaces & Buoyancy

Chapter 4: Thermal & Fluid Systems

PE Mechanical

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

Key Takeaways

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

1. SDOF Free Undamped Vibration & Equivalent Stiffness

Equation of Motion & Natural Frequency

Static Deflection Equivalence (δst\delta_{st}δst​)

Equivalent Spring Stiffness Combinations (keqk_{eq}keq​)

2. Viscous Damping & Response Regimes

Logarithmic Decrement (δ\deltaδ)

3. Harmonically Forced Vibration & Dynamic Magnification

Steady-State Amplitude (XXX) and Dynamic Magnification Factor (DMFDMFDMF)

Three Dynamic Operating Regimes

4. Vibration Isolation & Force Transmissibility

The Transmissibility Equation (TRTRTR)

The Fundamental Laws of Vibration Isolation

Isolation Efficiency (ηiso\eta_{\text{iso}}ηiso​)

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

Absorber Sizing & Response Equations

6. Dynamic Balancing of Rotating Machinery

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

8. Exam Tips & Common Traps

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

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

Logarithmic Decrement ( $\delta$ )

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

The Transmissibility Equation ( $TR$ )

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