9.4 Free, Damped & Forced Single-Degree-of-Freedom Vibrations

Free, Damped & Forced Single-Degree-of-Freedom Vibrations

Quick Reference: Natural circular frequency is $\omega_n = \sqrt{k/m}$. Damping ratio $\zeta = c / (2\sqrt{km})$. Damped frequency $\omega_d = \omega_n \sqrt{1 - \zeta^2}$. Logarithmic decrement $\delta = 2\pi\zeta / \sqrt{1 - \zeta^2} \approx 2\pi\zeta$. Transmissibility $\text{TR} < 1$ (effective isolation) occurs strictly when frequency ratio $r = \omega / \omega_n > \sqrt{2}$.

Mechanical vibrations are ubiquitous in Coal India mining operations: vibrating sizing screens, coal crushing ball mills, heavy mine drainage slurry pumps, and longwall shearer drives. Controlling structural vibrations and designing isolation mounts prevents structural fatigue, foundation failure, and noise pollution.

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1. Undamped Free SDOF Vibrations

Equation of Motion & Natural Frequency

For a lumped mass $m$ attached to a linear spring of stiffness $k$, applying Newton's second law or energy conservation yields:

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

$\text{Natural Circular Frequency: } \omega_n = \sqrt{\frac{k}{m}} \text{ (rad/s)}$

$\text{Natural Frequency (Hz): } f_n = \frac{\omega_n}{2\pi} = \frac{1}{2\pi} \sqrt{\frac{k}{m}} = \frac{1}{2\pi} \sqrt{\frac{g}{\Delta_{\text{st}}}}$

$\text{Time Period: } T_n = \frac{1}{f_n} = 2\pi \sqrt{\frac{m}{k}} = 2\pi \sqrt{\frac{\Delta_{\text{st}}}{g}}$

Where $\Delta_{\text{st}} = \frac{m g}{k}$ is the static deflection of the spring under mass weight.

+-------------------------------------------------------------------------+
|                   EQUIVALENT STIFFNESS FORMULATIONS                    |
+-------------------------------------------------------------------------+
| Configuration    | Equivalent Stiffness (k_eq)   | Physical Condition   |
+------------------+-------------------------------+----------------------+
| Parallel Springs | k_eq = k1 + k2 + ... + kn     | Same deflection,     |
|                  |                               | forces add           |
+------------------+-------------------------------+----------------------+
| Series Springs   | 1/k_eq = 1/k1 + 1/k2 + ...    | Same force,          |
|                  | k_eq = (k1 * k2) / (k1 + k2)  | deflections add      |
+------------------+-------------------------------+----------------------+
| Cantilever Beam  | k_eq = (3 * E * I) / L^3      | Point load at free tip|
+------------------+-------------------------------+----------------------+
| Simply Supported | k_eq = (48 * E * I) / L^3     | Point load at midspan|
+------------------+-------------------------------+----------------------+
| Fixed-Fixed Beam | k_eq = (192 * E * I) / L^3    | Point load at midspan|
+------------------+-------------------------------+----------------------+
| Torsional Shaft  | k_t = (G * J) / L             | Angle of twist theta |
+------------------+-------------------------------+----------------------+

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2. Viscously Damped Free SDOF Vibrations

Governing Differential Equation

When a viscous dashpot producing damping force $F_d = -c \dot{x}$ is added:

$m \ddot{x} + c \dot{x} + k x = 0 \implies \ddot{x} + 2\zeta \omega_n \dot{x} + \omega_n^2 x = 0$

• Critical Damping Coefficient ($c_c$): The minimum damping coefficient for which motion becomes non-oscillatory: $c_c = 2\sqrt{k m} = 2 m \omega_n$
• Damping Ratio / Damping Factor ($\zeta$): $\zeta = \frac{c}{c_c} = \frac{c}{2\sqrt{k m}} = \frac{c}{2 m \omega_n}$

  DAMPED FREE RESPONSE REGIMES:
  
  Displacement x(t)
     ^
     |  Overdamped (zeta > 1) ---------\
     |  Critically Damped (zeta = 1) ---\
     |                                   \
     |    Underdamped (zeta < 1)          \
     |      /---\                          \
  ---+-----+-----+---\----------------------+------------------> Time t
     |    /       \   \---/\
     |   /         \        \---
     v

The Three Damping Regimes:

1. Overdamped ($\zeta > 1$):
   • Roots are real, negative, and unequal: $s_{1,2} = (-\zeta \pm \sqrt{\zeta^2 - 1})\omega_n$.
   • Aperiodic, non-oscillatory sluggish return to equilibrium.
2. Critically Damped ($\zeta = 1$):
   • Roots are real, negative, and equal: $s_{1,2} = -\omega_n$.
   • Fastest non-oscillatory return to equilibrium without overshoot. Widely used in car suspension shock absorbers, artillery recoil mechanisms, and dial indicators.
3. Underdamped ($\zeta < 1$):
   • Roots are complex conjugates: $s_{1,2} = -\zeta \omega_n \pm i \omega_d$.
   • Damped Natural Frequency ($\omega_d$): $\omega_d = \omega_n \sqrt{1 - \zeta^2}$
   • Displacement response: $x(t) = X_0 e^{-\zeta \omega_n t} \sin(\omega_d t + \phi)$

Logarithmic Decrement ($\delta$)

The logarithmic decrement is defined as the natural logarithm of the ratio of any two successive peak amplitudes on the same side:

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

For lightly damped systems ($\zeta \le 0.2$):

$\delta \approx 2\pi \zeta$

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3. Harmonically Forced SDOF Vibrations

Steady-State Response & Magnification Factor

For a system excited by harmonic force $F(t) = F_0 \sin(\omega t)$:

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

$\text{Steady-State Displacement: } x(t) = X \sin(\omega t - \phi)$

Where:

• Static Deflection: $x_{\text{st}} = \frac{F_0}{k}$
• Frequency Ratio: $r = \frac{\omega}{\omega_n}$
• Steady-State Amplitude ($X$): $X = \frac{x_{\text{st}}}{\sqrt{(1 - r^2)^2 + (2\zeta r)^2}}$
• Magnification Factor (Dynamic Magnifier, $\text{MF}$): $\text{MF} = \frac{X}{x_{\text{st}}} = \frac{1}{\sqrt{(1 - r^2)^2 + (2\zeta r)^2}}$
• Phase Angle ($\phi$): $\tan\phi = \frac{2\zeta r}{1 - r^2}$

  MAGNIFICATION FACTOR vs. FREQUENCY RATIO (r):
  
  MF
  ^
  |         /|\  Resonance (r = 1)
  |        / | \   Peak MF = 1 / (2*zeta)
  |       /  |  \
  |      /   |   \
 1+----+    |    \------\  (zeta = 0.1)
  |    \     |            \-------\
  |     \----+---------------------\  (zeta = 0.5)
  +-----+----+-------------------------------------> Frequency Ratio r = w / w_n
  0          1                     sqrt(2)
    Stiffness   Damping           Inertia / Mass
    Controlled  Controlled        Controlled

Frequency Domain Characteristics:

1. Low Frequency ($r \ll 1$): $\text{MF} \approx 1, \phi \approx 0^{\circ}$. Dynamic amplitude equals static deflection; governed by spring stiffness.
2. Resonance ($r = 1$): $\text{MF}_{\text{res}} = \frac{1}{2\zeta}, \phi = 90^{\circ}$. Displacement lags force by $90^{\circ}$; amplitude is restrained entirely by damping.
3. High Frequency ($r \gg 1$): $\text{MF} \to 0, \phi \to 180^{\circ}$. Mass inertia dominates response; force is in direct opposition to motion.

4. Vibration Isolation & Transmissibility

Transmissibility Ratio ($\text{TR}$)

Vibration isolation mounts attenuate the dynamic force transmitted to the foundation ($F_T$). Transmissibility is the ratio of transmitted force to excitation force:

$\text{TR} = \frac{F_T}{F_0} = \frac{\sqrt{1 + (2\zeta r)^2}}{\sqrt{(1 - r^2)^2 + (2\zeta r)^2}}$

Fundamental Rules of Vibration Isolation:

1. Amplification Region ($r < \sqrt{2}$): $\text{TR} > 1$. Force transmitted to the foundation is larger than the excitation force. Damping reduces resonance peak, which is helpful.
2. Crossover Point ($r = \sqrt{2}$): $\text{TR} = 1$ for all values of damping $\zeta$. All transmissibility curves pass through $(r = \sqrt{2}, \text{TR} = 1)$.
3. Active Isolation Region ($r > \sqrt{2}$): $\text{TR} < 1$. Vibration isolation is effective only when $r > \sqrt{2}$ (i.e., operating frequency $\omega > \sqrt{2}\omega_n$).
4. Damping Trade-off in Isolation: When $r > \sqrt{2}$, increasing damping $\zeta$ increases $\text{TR}$, thereby degrading isolation efficiency. However, a small amount of damping is required to prevent catastrophic vibration while passing through resonance during engine startup and shutdown.

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5. Whirling / Critical Speed of Shafts

The critical speed (whirling speed) of a rotating shaft occurs when the shaft rotational speed $\omega$ matches its natural lateral vibration frequency $\omega_n$:

$\omega_c = \omega_n = \sqrt{\frac{k}{m}} = \sqrt{\frac{g}{\Delta_{\text{st}}}}$

For a rotor of mass $m$ with center of mass eccentric by distance $e$ from geometric center:

$r_d = \frac{e \cdot r^2}{1 - r^2} = \frac{e \cdot (\omega/\omega_c)^2}{1 - (\omega/\omega_c)^2}$

• At $\omega \ll \omega_c$: Deflection $r_d \to 0$; heavy spot rotates outside.
• At $\omega = \omega_c$: Resonant whirling ($r_d \to \infty$).
• At $\omega \gg \omega_c$: $r_d \to -e$; the center of gravity aligns itself with the rotation axis (self-centering behavior).

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6. Worked Numerical Examples

Example 1: Logarithmic Decrement & Damping Calculation

Problem: A vibrating coal screen of mass $m = 25\text{ kg}$ is supported by springs of equivalent stiffness $k = 10000\text{ N/m}$. A viscous damper is attached. An initial displacement is given, and the amplitude drops from $x_0 = 16\text{ mm}$ to $x_5 = 2.0\text{ mm}$ after 5 complete cycles. Calculate:

1. The logarithmic decrement $\delta$.
2. The damping ratio $\zeta$ and damping coefficient $c$.
3. The damped natural frequency $\omega_d$ and damped time period $T_d$.

Solution:

1. Logarithmic Decrement ($\delta$): $\delta = \frac{1}{n} \ln\left(\frac{x_0}{x_n}\right) = \frac{1}{5} \ln\left(\frac{16}{2}\right) = \frac{1}{5} \ln(8) = \frac{1}{5}(2.0794) = 0.4159$

2. Damping Ratio ($\zeta$): $\delta = \frac{2\pi\zeta}{\sqrt{1 - \zeta^2}} \implies \frac{\delta^2}{4\pi^2} = \frac{\zeta^2}{1 - \zeta^2}$ $\frac{(0.4159)^2}{4\pi^2} = \frac{0.1730}{39.478} = 0.004382 = \frac{\zeta^2}{1 - \zeta^2}$ $\zeta^2 = 0.004382 (1 - \zeta^2) \implies 1.004382 \zeta^2 = 0.004382 \implies \zeta = \sqrt{0.004363} = 0.06605$

3. Critical Damping ($c_c$) and Actual Damping ($c$): $\omega_n = \sqrt{\frac{k}{m}} = \sqrt{\frac{10000}{25}} = \sqrt{400} = 20.0\text{ rad/s}$ $c_c = 2 m \omega_n = 2 \times 25 \times 20.0 = 1000\text{ N}\cdot\text{s/m}$ $c = \zeta c_c = 0.06605 \times 1000 = 66.05\text{ N}\cdot\text{s/m}$

4. Damped Frequency ($\omega_d$) and Time Period ($T_d$): $\omega_d = \omega_n \sqrt{1 - \zeta^2} = 20.0 \times \sqrt{1 - (0.06605)^2} = 20.0 \times \sqrt{0.9956} = 19.956\text{ rad/s}$ $T_d = \frac{2\pi}{\omega_d} = \frac{2\pi}{19.956} = 0.3148\text{ seconds}$

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Example 2: Vibration Isolation Mount Design

Problem: A heavy coal pulverizer of mass $M = 500\text{ kg}$ operates at an engine speed of $N = 1200\text{ rpm}$ ($125.66\text{ rad/s}$). It is mounted on rubber isolation pads with negligible damping ($\zeta \approx 0$). Design the spring stiffness $k$ so that only $10%$ of the unbalanced dynamic force is transmitted to the supporting concrete foundation (i.e., $\text{TR} = 0.10$).

Solution:

1. Determine Required Frequency Ratio ($r$): For undamped isolation ($\zeta = 0$) in the active isolation zone ($r > \sqrt{2}$): $\text{TR} = \frac{1}{r^2 - 1} = 0.10$ $r^2 - 1 = \frac{1}{0.10} = 10 \implies r^2 = 11 \implies r = \sqrt{11} \approx 3.3166$

2. Determine Required Natural Frequency ($\omega_n$): $r = \frac{\omega}{\omega_n} \implies \omega_n = \frac{\omega}{r} = \frac{125.66\text{ rad/s}}{3.3166} = 37.888\text{ rad/s}$

3. Calculate Required Total Foundation Stiffness ($k$): $\omega_n = \sqrt{\frac{k}{M}} \implies k = M \omega_n^2$ $k = 500\text{ kg} \times (37.888)^2 = 500 \times 1435.5 = 717750\text{ N/m} = 717.75\text{ kN/m}$

4. Static Deflection Verification: $\Delta_{\text{st}} = \frac{M g}{k} = \frac{500 \times 9.81}{717750} = \frac{4905}{717750} = 0.00683\text{ m} = 6.83\text{ mm}$ (A static deflection of $6.83\text{ mm}$ is well within acceptable standard limits for industrial vibration mount pads).