12.1 Single-Degree-of-Freedom Vibrations & Resonance

12.1 Single-Degree-of-Freedom Vibrations & Resonance

Mechanical vibration represents the oscillatory motion of dynamic systems about an equilibrium position. In machine design, uncontrolled vibration causes alternating fatigue stresses, premature bearing wear, structural loosening, acoustic noise, and catastrophic resonance failures. Mastering the dynamics of Single-Degree-of-Freedom (SDOF) systems provides the core analytical foundation for diagnosing machine dynamics, predicting dynamic magnification, and designing resilient rotating machinery.

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Undamped Free Vibration of SDOF Systems

A single-degree-of-freedom model consists of a concentrated mass $m$, a massless restoring element of linear stiffness $k$, and a single generalized coordinate $x(t)$ describing translational or angular displacement from static equilibrium.

          +-------------------+
          |      Ceiling      |
          +---------+---------+
                    |
                    |  Restoring Element (k)
                    |
                    v
              +-----------+
              |  Mass (m) |  --->  x(t) [Displacement]
              +-----------+

Applying Newton's second law or d'Alembert's dynamic equilibrium principle in the direction of motion yields:

$$\sum F_x = -k x = m \ddot{x} \implies m \ddot{x} + k x = 0$$

Dividing by mass $m$ produces the canonical second-order linear homogeneous differential equation:

$$\ddot{x} + \omega_n^2 x = 0$$

where $\omega_n$ is the circular natural frequency in radians per second:

$$\omega_n = \sqrt{\frac{k}{m}}$$

The corresponding cyclic natural frequency $f_n$ (measured in Hertz, or cycles per second) and the natural period of oscillation $T_n$ (in seconds) are:

$$f_n = \frac{\omega_n}{2\pi} = \frac{1}{2\pi} \sqrt{\frac{k}{m}}$$ $$T_n = \frac{1}{f_n} = \frac{2\pi}{\omega_n}$$

Given initial conditions $x(0) = x_0$ and $\dot{x}(0) = v_0$, the exact time-domain response is:

$$x(t) = x_0 \cos(\omega_n t) + \frac{v_0}{\omega_n} \sin(\omega_n t) = X_0 \cos(\omega_n t - \phi)$$

where the peak amplitude $X_0$ and phase angle $\phi$ are:

$$X_0 = \sqrt{x_0^2 + \left(\frac{v_0}{\omega_n}\right)^2}, \qquad \tan\phi = \frac{v_0}{\omega_n x_0}$$

Static Deflection Relationship

When a mechanical component of mass $m$ is suspended vertically under gravitational acceleration $g$, it produces a static deflection $\delta_{st}$ of the spring support:

$$W = m g = k \delta_{st} \implies \frac{k}{m} = \frac{g}{\delta_{st}}$$

Substituting this relationship into the natural frequency formula links natural frequency directly to measurable static deflection:

$$\omega_n = \sqrt{\frac{g}{\delta_{st}}}$$ $$f_n = \frac{1}{2\pi} \sqrt{\frac{g}{\delta_{st}}}$$

In standard engineering units:

• SI Units ($g = 9.81\,\text{m/s}^2$, with $\delta_{st}$ in meters): $f_n = \frac{1}{2\pi} \sqrt{\frac{9.81}{\delta_{st}}} \approx \frac{0.4984}{\sqrt{\delta_{st}\,(\text{m})}}\,\text{Hz} = \frac{15.76}{\sqrt{\delta_{st}\,(\text{mm})}}\,\text{Hz}$
• USCS Units ($g = 386.1\,\text{in/s}^2$, with $\delta_{st}$ in inches): $f_n = \frac{1}{2\pi} \sqrt{\frac{386.1}{\delta_{st}}} \approx \frac{3.127}{\sqrt{\delta_{st}\,(\text{in})}}\,\text{Hz}$

Equivalent Stiffness Formulations

Complex mechanical supports can be reduced to an equivalent single spring stiffness $k_{eq}$:

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Viscously Damped Free Vibration

Real physical systems dissipate energy through internal material friction, fluid resistance, or joint slippage. The standard linear model incorporates a viscous dashpot exhibiting a damping force proportional to velocity: $F_d = -c \dot{x}$, where $c$ is the viscous damping coefficient (units of $\text{N}\cdot\text{s/m}$ or $\text{lbf}\cdot\text{s/in}$).

$$m \ddot{x} + c \dot{x} + k x = 0$$

Dividing by $m$:

$$\ddot{x} + \frac{c}{m} \dot{x} + \omega_n^2 x = 0$$

Assuming a solution form $x(t) = C e^{s t}$ yields the characteristic algebraic equation:

$$s^2 + \frac{c}{m} s + \omega_n^2 = 0 \implies s_{1,2} = -\frac{c}{2m} \pm \sqrt{\left(\frac{c}{2m}\right)^2 - \omega_n^2}$$

Critical Damping and Damping Ratio

Critical damping $c_c$ is defined as the damping coefficient that sets the discriminant to zero, separating oscillatory behavior from non-oscillatory decay:

$$\left(\frac{c_c}{2m}\right)^2 - \omega_n^2 = 0 \implies c_c = 2 m \omega_n = 2 \sqrt{k m}$$

The dimensionless damping ratio $\zeta$ (zeta) is defined as:

$$\zeta = \frac{c}{c_c} = \frac{c}{2m\omega_n} = \frac{c}{2\sqrt{km}}$$

The characteristic equation expressed in terms of $\zeta$ and $\omega_n$ becomes:

$$s^2 + 2 \zeta \omega_n s + \omega_n^2 = 0 \implies s_{1,2} = -\zeta \omega_n \pm \omega_n \sqrt{\zeta^2 - 1}$$

Damping Regimes

                               Damping Regimes Comparison
               +---------------------------------------------------------+
               | Damping Ratio (zeta) | Nature of Roots  | Response Type |
               |----------------------+------------------+---------------|
               | Underdamped (zeta<1) | Complex Conj.    | Oscillatory   |
               | Critical    (zeta=1) | Real & Repeated  | No Oscillation|
               | Overdamped  (zeta>1) | Real & Distinct  | Sluggish Decay|
               +---------------------------------------------------------+

1. Underdamped ($\,0 \le \zeta < 1\,$): The roots are complex conjugates: $s_{1,2} = -\zeta \omega_n \pm i \omega_d$, where the damped natural frequency $\omega_d$ is:

   $$\omega_d = \omega_n \sqrt{1 - \zeta^2}$$

   The transient response oscillates with an exponentially decaying envelope:

   $$x(t) = X_0 e^{-\zeta \omega_n t} \cos(\omega_d t - \phi)$$

2. Critically Damped ($\,\zeta = 1\,$): Roots are repeated real values: $s_1 = s_2 = -\omega_n$. The system returns to equilibrium in the shortest possible time without overshoot:

   $$x(t) = (A + B t) e^{-\omega_n t}$$

3. Overdamped ($\,\zeta > 1\,$): Roots are unequal, distinct negative real numbers. The motion is non-oscillatory and sluggish, returning slowly toward equilibrium:

   $$x(t) = A e^{s_1 t} + B e^{s_2 t}, \qquad s_{1,2} = \left(-\zeta \pm \sqrt{\zeta^2 - 1}\right) \omega_n$$

Logarithmic Decrement

In underdamped systems, the rate of amplitude decay between successive peaks separated by one period $T_d = 2\pi / \omega_d$ quantifies system damping. The logarithmic decrement $\delta$ is:

$$\delta = \ln\left(\frac{x_1}{x_2}\right) = \ln\left(\frac{X_0 e^{-\zeta \omega_n t_1}}{X_0 e^{-\zeta \omega_n (t_1 + T_d)}}\right) = \zeta \omega_n T_d = \zeta \omega_n \left(\frac{2\pi}{\omega_n \sqrt{1 - \zeta^2}}\right) = \frac{2\pi \zeta}{\sqrt{1 - \zeta^2}}$$

Across $n$ successive full cycles between peak amplitude $x_0$ and peak amplitude $x_n$:

$$\delta = \frac{1}{n} \ln\left(\frac{x_0}{x_n}\right)$$

Solving for the damping ratio $\zeta$:

$$\zeta = \frac{\delta}{\sqrt{(2\pi)^2 + \delta^2}}$$

For lightly damped structures ($\zeta < 0.2$), $\sqrt{1 - \zeta^2} \approx 1$, yielding the widely applied engineering approximation:

$$\delta \approx 2\pi \zeta \implies \zeta \approx \frac{\delta}{2\pi}$$

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Harmonically Excited SDOF Vibrations

When a dynamic system experiences an external periodic harmonic excitation force $F(t) = F_0 \cos(\omega t)$, the equation of motion is:

$$m \ddot{x} + c \dot{x} + k x = F_0 \cos(\omega t)$$

Dividing by $m$ and expressing in standard parameters:

$$\ddot{x} + 2\zeta\omega_n \dot{x} + \omega_n^2 x = \frac{F_0}{m} \cos(\omega t)$$

The total solution consists of a transient complementary solution $x_h(t)$ (which decays to zero) and a steady-state particular solution $x_p(t)$ at the driving frequency $\omega$:

$$x_p(t) = X \cos(\omega t - \phi)$$

Substituting $x_p(t)$ into the governing equation and employing trigonometric identities or complex exponential phasors yields the steady-state displacement amplitude $X$ and phase lag $\phi$:

$$X = \frac{F_0}{\sqrt{(k - m\omega^2)^2 + (c\omega)^2}} = \frac{F_0 / k}{\sqrt{(1 - r^2)^2 + (2\zeta r)^2}}$$ $$\tan\phi = \frac{c\omega}{k - m\omega^2} = \frac{2\zeta r}{1 - r^2} \implies \phi = \arctan\left(\frac{2\zeta r}{1 - r^2}\right)$$

where:

• $r = \frac{\omega}{\omega_n}$ is the dimensionless frequency ratio.
• $\delta_{st} = \frac{F_0}{k}$ is the zero-frequency static displacement.

Dynamic Magnification Factor ($M$)

The ratio of steady-state dynamic amplitude to static deflection is the Dynamic Magnification Factor (also called amplification factor):

$$M = \frac{X}{F_0 / k} = \frac{1}{\sqrt{(1 - r^2)^2 + (2\zeta r)^2}}$$

       Dynamic Magnification Factor M vs Frequency Ratio r
  M ^
    |           /|\  zeta = 0.05 (Sharp high peak)
    |          / | \
 10 +         /  |  \
    |        /   |   \    zeta = 0.1
  5 +       /    |    \
    |      /     |     \____ zeta = 0.3
  1 +----+-------+----------------------> Asymptote = 0
    0           r=1.0                  r
         Stiffness   Resonance     Inertia
         Dominated   (Damping)     Dominated

Frequency Response Regimes

1. Low Frequency ($r \ll 1$, Stiffness-Dominated): As $r \to 0$, $M \to 1.0$ and $\phi \to 0^\circ$. The response displacement is in-phase with the excitation force, and amplitude is governed strictly by spring stiffness: $X \approx F_0 / k$.

2. Resonant Region ($r \approx 1.0$, Damping-Dominated): At $r = 1.0$, the inertia and spring restoring forces cancel identically ($k - m\omega^2 = 0$). The motion is opposed solely by viscous damping:

   $$M_{r=1} = \frac{1}{2\zeta} = Q \quad (\text{Quality Factor})$$ $$\phi_{r=1} = 90^\circ = \frac{\pi}{2}\;\text{rad}$$

   The maximum dynamic magnification occurs slightly below $r = 1.0$ at the peak frequency ratio:

   $$r_{\text{peak}} = \sqrt{1 - 2\zeta^2} \qquad (\text{for } \zeta < 1/\sqrt{2} \approx 0.707)$$ $$M_{\max} = \frac{1}{2\zeta \sqrt{1 - \zeta^2}} \approx \frac{1}{2\zeta}$$

3. High Frequency ($r \gg 1$, Inertia/Mass-Dominated): As $r \to \infty$, $M \to 0$ and $\phi \to 180^\circ$ ($\pi\;\text{rad}$). The displacement vector opposes the force vector, and response amplitude decays as $1/r^2$.

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Rotating Unbalance Excitation

Common industrial machinery—such as centrifugal blowers, turbines, pumps, and motors—contains residual rotor mass unbalance. Consider a total machine structure of mass $M_m$ supporting an eccentric unbalance mass $m_e$ situated at distance $e$ from the axis of rotation, spinning with angular velocity $\omega$.

                    Total Machine Mass M_m
            +------------------------------------+
            |           Rotating Shaft           |
            |              /\                    |
            |             /  \ e (Eccentricity)  |
            |            O----(m_e)              |
            +------------------------------------+
               |                              |
              [k] (Spring)                   [c] (Dashpot)
               |                              |
            ===+==============================+===

The rotating unbalance produces a rotating centrifugal force vector whose vertical harmonic component is:

$$F(t) = m_e e \omega^2 \sin(\omega t)$$

The equation of motion governing vertical displacement $x(t)$ is:

$$M_m \ddot{x} + c \dot{x} + k x = m_e e \omega^2 \sin(\omega t)$$

Solving for the steady-state amplitude $X$:

$$X = \frac{m_e e \omega^2}{\sqrt{(k - M_m \omega^2)^2 + (c\omega)^2}} = \frac{m_e e}{M_m} \left[ \frac{r^2}{\sqrt{(1 - r^2)^2 + (2\zeta r)^2}} \right]$$

where $\omega_n = \sqrt{k / M_m}$, $\zeta = c / (2\sqrt{k M_m})$, and $r = \omega / \omega_n$.

Asymptotic Behavior of Rotating Unbalance

• At low operating speeds ($r \to 0$): $X \to 0$.
• At resonance ($r = 1.0$): $X_{r=1} = \frac{m_e e}{M_m} \left(\frac{1}{2\zeta}\right)$.
• At supercritical operating speeds ($r \to \infty$):

$$\lim_{r \to \infty} X = \frac{m_e e}{M_m}$$

In supercritical operation, the physical center of mass of the combined system becomes stationary in space, and the geometric shaft axis precesses around it by displacement amplitude $\frac{m_e e}{M_m}$.

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Step-by-Step Worked Vibration Engineering Scenario

Problem Statement

An industrial electric drive motor has a total mass $m = 250\,\text{kg}$. It is mounted on four identical spring-damper isolator feet operating in parallel. Each isolator has a vertical linear spring stiffness $k_i = 60\,\text{kN/m}$. A fluid viscous dashpot provides a total lumped damping coefficient $c = 1,\!200\,\text{N}\cdot\text{s/m}$.

The motor operates at a steady-state running speed $N = 1,\!750\,\text{rpm}$. An internal rotating unbalance exists within the rotor, consisting of an effective unbalance mass $m_e = 0.50\,\text{kg}$ located at radius $e = 80\,\text{mm}$ from the shaft centerline.

Calculate:

1. Total equivalent stiffness $k_{\text{tot}}$, circular natural frequency $\omega_n$, cyclic natural frequency $f_n$, and static deflection $\delta_{st}$.
2. Critical damping coefficient $c_c$, damping ratio $\zeta$, damped frequency $\omega_d$, and logarithmic decrement $\delta$.
3. Operating angular frequency $\omega$, frequency ratio $r$, rotating unbalance centrifugal excitation force $F_0$, dynamic magnification factor $M$, and steady-state vibration displacement amplitude $X$.

Step-by-Step Solution

Step 1: System Natural Frequencies & Static Deflection

The four isolator mounts operate in parallel, so their stiffnesses sum directly:

$$k_{\text{tot}} = 4 \times k_i = 4 \times 60,\!000\,\text{N/m} = 240,\!000\,\text{N/m} = 240\,\text{kN/m}$$

The undamped circular natural frequency is:

$$\omega_n = \sqrt{\frac{k_{\text{tot}}}{m}} = \sqrt{\frac{240,\!000\,\text{N/m}}{250\,\text{kg}}} = \sqrt{960} \approx 30.984\,\text{rad/s}$$

The cyclic natural frequency is:

$$f_n = \frac{\omega_n}{2\pi} = \frac{30.984}{2\pi} \approx 4.931\,\text{Hz}$$

The gravitational static deflection under motor weight ($g = 9.81\,\text{m/s}^2$) is:

$$\delta_{st} = \frac{m g}{k_{\text{tot}}} = \frac{250 \times 9.81}{240,\!000} = \frac{2,\!452.5}{240,\!000} = 0.01022\,\text{m} = 10.22\,\text{mm}$$

Verification: $\omega_n = \sqrt{g / \delta_{st}} = \sqrt{9.81 / 0.01021875} = \sqrt{960} = 30.984\,\text{rad/s}$.

Step 2: Damping Parameters and Transient Decay

The critical damping coefficient is:

$$c_c = 2 m \omega_n = 2 \times 250\,\text{kg} \times 30.984\,\text{rad/s} = 15,\!492\,\text{N}\cdot\text{s/m}$$

The damping ratio is:

$$\zeta = \frac{c}{c_c} = \frac{1,\!200\,\text{N}\cdot\text{s/m}}{15,\!492\,\text{N}\cdot\text{s/m}} = 0.07746 \quad (\approx 7.75\%)$$

Since $\zeta < 1.0$, the system is underdamped and oscillates with damped natural frequency:

$$\omega_d = \omega_n \sqrt{1 - \zeta^2} = 30.984 \times \sqrt{1 - (0.07746)^2} = 30.984 \times \sqrt{0.9940} = 30.891\,\text{rad/s}$$ $$f_d = \frac{\omega_d}{2\pi} = \frac{30.891}{2\pi} \approx 4.916\,\text{Hz}$$

The logarithmic decrement between consecutive free vibration peaks is:

$$\delta = \frac{2\pi \zeta}{\sqrt{1 - \zeta^2}} = \frac{2\pi \times 0.07746}{\sqrt{1 - (0.07746)^2}} = \frac{0.48669}{0.996995} \approx 0.4882$$

The amplitude ratio between consecutive free oscillation cycles is:

$$\frac{x_1}{x_2} = e^\delta = e^{0.4882} \approx 1.629$$

Step 3: Forced Response Under Rotating Unbalance

Convert motor running speed to operating angular frequency:

$$\omega = \frac{2\pi N}{60} = \frac{2\pi \times 1,\!750}{60} = 183.26\,\text{rad/s} \quad (f_{\text{drive}} = 29.17\,\text{Hz})$$

The operating frequency ratio is:

$$r = \frac{\omega}{\omega_n} = \frac{183.26}{30.984} \approx 5.915$$

The magnitude of the rotating centrifugal excitation force is:

$$F_0 = m_e e \omega^2 = (0.50\,\text{kg}) \times (0.080\,\text{m}) \times (183.26\,\text{rad/s})^2 = 0.040 \times 33,\!584.2 = 1,\!343.4\,\text{N}$$

Compute the dynamic magnification components:

$$1 - r^2 = 1 - (5.915)^2 = 1 - 34.987 = -33.987 \implies (1 - r^2)^2 = 1,\!155.1$$ $$2\zeta r = 2 \times 0.07746 \times 5.915 = 0.9164 \implies (2\zeta r)^2 = 0.8398$$ $$M = \frac{1}{\sqrt{1,\!155.1 + 0.8398}} = \frac{1}{\sqrt{1,\!155.94}} = \frac{1}{34.00} \approx 0.02941$$

The steady-state vibration displacement amplitude is:

$$X = \left(\frac{F_0}{k_{\text{tot}}}\right) M = \left(\frac{1,\!343.4\,\text{N}}{240,\!000\,\text{N/m}}\right) \times 0.02941 = (0.005598\,\text{m}) \times 0.02941 = 1.646 \times 10^{-4}\,\text{m} = 0.165\,\text{mm}$$

Verification via the direct rotating unbalance formula:

$$X = \frac{m_e e}{m} \left[\frac{r^2}{\sqrt{(1 - r^2)^2 + (2\zeta r)^2}}\right] = \left(\frac{0.040}{250}\right) \left[\frac{34.987}{34.00}\right] = (0.000160\,\text{m}) \times 1.029 = 0.165\,\text{mm}$$

Because $r = 5.915 \gg 1.0$, the machine operates deep within the supercritical isolation zone, where vibration amplitude closely tracks the theoretical mass unbalance offset $\frac{m_e e}{m} = 0.160\,\text{mm}$.