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Chapter 8 โ€ข Theory & Derivations

Dynamical Modeling with Second-Order Systems: Oscillations, Resonance & Rocket Dynamics

Comprehensive modeling of second-order mechanical, electrical, and aerospace systems: free undamped and damped vibrations, logarithmic decrement, forced oscillations, the phenomenon of beats, resonance and Q-factors, series RLC electrical networks, and variable-mass Tsiolkovsky rocket dynamics under gravity and drag.

ยง8.1 Mechanical Vibrations: Free Undamped & Damped Oscillators

1. Newton's Second Law for the Spring-Mass-Damper System

Consider a mass $m$ attached to a linear spring of stiffness constant $k$ and a viscous damper with damping coefficient $c$. By Newton's second law, $m x'' = F_{\text{spring}} + F_{\text{damping}} + F_{\text{ext}}(t) = -kx - cx' + F(t)$:

$$m \frac{d^2x}{dt^2} + c \frac{dx}{dt} + k x = F(t)$$

Dividing by $m$, we write the canonical form: $x'' + 2\zeta \omega_0 x' + \omega_0^2 x = \frac{F(t)}{m}$, where $\omega_0 = \sqrt{\frac{k}{m}}$ is the undamped natural angular frequency and $\zeta = \frac{c}{2\sqrt{km}}$ is the dimensionless damping ratio.

2. Free Motion Classification ($F(t) \equiv 0$)

  • Undamped Motion ($\zeta = 0, c = 0$): Simple Harmonic Motion (SHM): $$x(t) = c_1 \cos(\omega_0 t) + c_2 \sin(\omega_0 t) = A \cos(\omega_0 t - \phi)$$ where amplitude $A = \sqrt{c_1^2 + c_2^2}$ and period $T = \frac{2\pi}{\omega_0}$.
  • Overdamped Motion ($\zeta > 1, c^2 > 4km$): Characteristic roots are real, negative, and distinct: $r_{1,2} = -\zeta\omega_0 \pm \omega_0\sqrt{\zeta^2 - 1}$. The system returns to equilibrium non-oscillatingly.
  • Critically Damped Motion ($\zeta = 1, c = 2\sqrt{km}$): Repeated real root $r = -\omega_0$. Solution: $$x(t) = (c_1 + c_2 t) e^{-\omega_0 t}$$ This provides the fastest possible return to equilibrium without oscillation (crucial in automobile shock absorbers and gun recoil mechanisms).
  • Underdamped Motion ($\zeta < 1, c^2 < 4km$): Complex roots $r = -\alpha \pm i \omega_d$, where $\alpha = \zeta \omega_0 = \frac{c}{2m}$ and $\omega_d = \omega_0 \sqrt{1 - \zeta^2}$ is the quasi-frequency. $$x(t) = A e^{-\alpha t} \cos(\omega_d t - \phi)$$ The amplitude decays exponentially. The rate of decay is quantified by the logarithmic decrement $\delta$: $$\delta = \ln\left(\frac{x(t)}{x(t + T_d)}\right) = \alpha T_d = \frac{2\pi \zeta}{\sqrt{1 - \zeta^2}}$$

ยง8.2 Forced Vibrations, Secular Resonance & The Quality Factor

1. Forced Oscillations with Harmonic Excitation

When subjected to a periodic external force $F(t) = F_0 \cos(\omega t)$:

$$m x'' + c x' + k x = F_0 \cos(\omega t)$$

The general solution consists of a transient complementary solution $x_c(t)$ (which dies out exponentially as $t \to \infty$) and a persistent steady-state particular solution $x_p(t) = x_{ss}(t)$:

$$x_{ss}(t) = X(\omega) \cos(\omega t - \phi)$$

Using the method of undetermined coefficients, the steady-state amplitude is derived exactly as:

$$X(\omega) = \frac{F_0/m}{\sqrt{(\omega_0^2 - \omega^2)^2 + \left(\frac{c\omega}{m}\right)^2}} = \frac{F_0/k}{\sqrt{\left(1 - \frac{\omega^2}{\omega_0^2}\right)^2 + \left(2\zeta \frac{\omega}{\omega_0}\right)^2}}$$

and the phase lag is $\phi = \arctan\left(\frac{c\omega/m}{\omega_0^2 - \omega^2}\right) = \arctan\left(\frac{2\zeta (\omega/\omega_0)}{1 - (\omega/\omega_0)^2}\right)$.

2. Resonance Phenomena

  • Pure Secular Resonance (No Damping, $\zeta = 0$): When $\omega = \omega_0$, the denominator vanishes. The particular solution undergoes linear secular growth: $$x_p(t) = \frac{F_0}{2 m \omega_0} t \sin(\omega_0 t)$$ The amplitude grows without bound as $t \to \infty$, leading to structural destruction.
  • Resonance with Damping ($\zeta < 1/\sqrt{2}$): Maximizing $X(\omega)$ by minimizing the radicand yields the resonant frequency $\omega_r$: $$\omega_r = \omega_0 \sqrt{1 - 2\zeta^2}$$ At resonance, the peak amplitude is $X_{\text{max}} = \frac{F_0/k}{2\zeta\sqrt{1 - \zeta^2}}$. The sharpness of the resonance peak is measured by the Quality Factor $Q$: $$Q = \frac{1}{2\zeta} = \frac{\sqrt{km}}{c}$$

ยง8.3 Series RLC Electrical Networks & Variable-Mass Rocket Dynamics

1. Series RLC Electrical Network Modeling

By Kirchhoff's voltage law, the sum of voltage drops across an inductor $L$, resistor $R$, and capacitor $C$ in a closed loop equals the applied electromotive force $E(t)$:

$$V_L + V_R + V_C = L \frac{dI}{dt} + R I + \frac{1}{C} q = E(t)$$

Since electric current is the time derivative of charge, $I(t) = \frac{dq}{dt}$, substituting yields the second-order linear ODE for charge $q(t)$:

$$L \frac{d^2q}{dt^2} + R \frac{dq}{dt} + \frac{1}{C} q = E(t)$$

Differentiating with respect to $t$ yields the dual equation for the electric current $I(t)$:

$$L \frac{d^2I}{dt^2} + R \frac{dI}{dt} + \frac{1}{C} I = \frac{dE}{dt}$$

Mechanical-Electrical Analogy: Inductance $L \leftrightarrow$ Mass $m$; Resistance $R \leftrightarrow$ Damping $c$; Inverse capacitance $1/C \leftrightarrow$ Spring stiffness $k$; Charge $q \leftrightarrow$ Position $x$; Current $I \leftrightarrow$ Velocity $v$.

2. Rocket Motion Dynamics (Variable Mass Systems)

A rocket burning propellant is an open variable-mass mechanical system. Let $M(t) = M_0 - \alpha t$ denote the total mass of the rocket at time $t$, where $\alpha = -\frac{dM}{dt}$ is the constant mass burn rate, and let $u$ denote the exhaust velocity of gases relative to the rocket.

By conservation of momentum in a vertical gravitational field $g$ with linear atmospheric drag $-k v$:

$$M(t) \frac{dv}{dt} = -u \frac{dM}{dt} - M(t) g - k v = \alpha u - (M_0 - \alpha t) g - k v$$

Dividing by $M(t)$ yields the first-order linear differential equation for velocity $v(t)$:

$$\frac{dv}{dt} + \frac{k}{M_0 - \alpha t} v = \frac{\alpha u}{M_0 - \alpha t} - g$$

In the vacuum limit ($k = 0$), integrating directly yields the famous Tsiolkovsky rocket equation:

$$v(t) = v_0 - g t + u \ln\left(\frac{M_0}{M(t)}\right)$$
TIERED UNIVERSITY HONORS PROBLEMS

Step-by-Step Solved Examination Problems

Comprehensive analytical derivations, multi-tier solutions (Foundational, Intermediate Exam, and Honors/Proof Challenge) with complete line-by-line verification.

Tier 1 โ€ข Foundational Example 8.1: Logarithmic Decrement & Damping Extraction in a Mechanical Oscillator

An underdamped spring-mass system with mass $m = 2\text{ kg}$ and spring constant $k = 50\text{ N/m}$ undergoes free oscillations. Successive peak amplitudes on the same side are measured to be $x_1 = 8.0\text{ cm}$ and $x_2 = 5.0\text{ cm}$. Calculate the logarithmic decrement $\delta$, the damping ratio $\zeta$, the damping coefficient $c$, and the quasi-period $T_d$.

Step 1: Compute logarithmic decrement

$$\delta = \ln\left(\frac{x_1}{x_2}\right) = \ln\left(\frac{8.0}{5.0}\right) = \ln(1.6) \approx 0.4700$$



Step 2: Calculate damping ratio $\zeta$

$$\delta = \frac{2\pi \zeta}{\sqrt{1 - \zeta^2}} \implies \zeta = \frac{\delta}{\sqrt{4\pi^2 + \delta^2}} = \frac{0.4700}{\sqrt{4\pi^2 + (0.4700)^2}} = \frac{0.4700}{\sqrt{39.478 + 0.2209}} = \frac{0.4700}{6.3007} \approx 0.0746$$



Step 3: Calculate undamped natural frequency and damping coefficient $c$

$$\omega_0 = \sqrt{\frac{k}{m}} = \sqrt{\frac{50}{2}} = \sqrt{25} = 5.0\text{ rad/s}$$

Critical damping $c_c = 2\sqrt{km} = 2\sqrt{50 \cdot 2} = 2\sqrt{100} = 20\text{ N}\cdot\text{s/m}$.

$$c = \zeta c_c = (0.0746)(20) \approx 1.492\text{ N}\cdot\text{s/m}$$



Step 4: Quasi-frequency and quasi-period

$$\omega_d = \omega_0 \sqrt{1 - \zeta^2} = 5.0 \sqrt{1 - (0.0746)^2} = 5.0(0.9972) \approx 4.986\text{ rad/s}$$


$$T_d = \frac{2\pi}{\omega_d} = \frac{2\pi}{4.986} \approx 1.260\text{ seconds}$$
Final Answer & Physical Insight

$\delta = 0.470$, $\zeta = 0.0746$, $c = 1.49\text{ N}\cdot\text{s/m}$, and $T_d = 1.26\text{ s}$.

Tier 2 โ€ข Analytical University Exam Example 8.2: Steady-State Resonance Amplification & Peak Analysis

A $10\text{ kg}$ machine is supported on isolators with total stiffness $k = 4000\text{ N/m}$ and damping $c = 40\text{ N}\cdot\text{s/m}$. A harmonic force $F(t) = 100 \cos(\omega t)\text{ N}$ acts on the machine. Find the resonant frequency $\omega_r$, the maximum steady-state displacement amplitude $X_{\text{max}}$, and compare with the static deflection.

Step 1: Compute system baseline parameters

$$\omega_0 = \sqrt{\frac{k}{m}} = \sqrt{\frac{4000}{10}} = \sqrt{400} = 20\text{ rad/s}$$

Static deflection under force $F_0 = 100\text{ N}$:

$$\delta_{st} = \frac{F_0}{k} = \frac{100}{4000} = 0.025\text{ m} = 2.5\text{ cm}$$

Damping ratio $\zeta$:

$$\zeta = \frac{c}{2\sqrt{km}} = \frac{40}{2\sqrt{4000 \cdot 10}} = \frac{40}{2(200)} = \frac{40}{400} = 0.10$$



Step 2: Resonant frequency $\omega_r$

$$\omega_r = \omega_0 \sqrt{1 - 2\zeta^2} = 20 \sqrt{1 - 2(0.10)^2} = 20 \sqrt{1 - 0.02} = 20 \sqrt{0.98} \approx 19.80\text{ rad/s}$$



Step 3: Maximum amplitude $X_{\text{max}}$

$$X_{\text{max}} = \frac{\delta_{st}}{2\zeta\sqrt{1 - \zeta^2}} = \frac{0.025}{2(0.10)\sqrt{1 - 0.01}} = \frac{0.025}{0.20 \cdot 0.99498} = \frac{0.025}{0.1990} \approx 0.1256\text{ m} = 12.56\text{ cm}$$



Step 4: Dynamic amplification factor

$$Q \approx \frac{X_{\text{max}}}{\delta_{st}} = \frac{12.56\text{ cm}}{2.5\text{ cm}} \approx 5.025$$
Final Answer & Physical Insight

$\omega_r \approx 19.80\text{ rad/s}$, $X_{\text{max}} \approx 12.56\text{ cm}$, with dynamic amplification factor $M \approx 5.03$ times static deflection.

Tier 3 โ€ข Honors Challenge Example 8.3: Variable-Mass Rocket Ascent Under Gravity & Burnout Velocity

A sounding rocket of initial mass $M_0 = 1000\text{ kg}$ burns propellant at a constant rate $\alpha = 20\text{ kg/s}$ with constant exhaust velocity $u = 2400\text{ m/s}$. The propellant accounts for $800\text{ kg}$ of total initial mass. Assuming vertical ascent from rest ($v(0) = 0$) under uniform gravity $g = 9.81\text{ m/s}^2$ and negligible atmospheric resistance, calculate the burnout time $t_b$, burnout altitude $y(t_b)$, and maximum velocity attained.

Step 1: Determine burnout time and mass ratio
Fuel mass is $M_f = 800\text{ kg}$, dry mass is $M_b = 200\text{ kg}$.

$$t_b = \frac{M_f}{\alpha} = \frac{800}{20} = 40\text{ seconds}$$

At burnout, $M(t_b) = 1000 - 20(40) = 200\text{ kg}$.
Mass ratio $\frac{M_0}{M_b} = \frac{1000}{200} = 5$.

Step 2: Compute burnout velocity via Tsiolkovsky equation

$$v(t) = -gt + u \ln\left(\frac{M_0}{M_0 - \alpha t}\right)$$

At burnout $t = 40\text{ s}$:

$$v(t_b) = -9.81(40) + 2400 \ln(5) = -392.4 + 2400(1.60944) = -392.4 + 3862.65 = 3470.25\text{ m/s}$$



Step 3: Integrate for burnout altitude $y(t_b)$

$$y(t_b) = \int_0^{t_b} v(t) dt = \int_0^{t_b} \left[ -gt + u \ln M_0 - u \ln(M_0 - \alpha t) \right] dt$$

Evaluating the integral:

$$\int_0^{t_b} -gt dt = -\frac{1}{2} g t_b^2 = -\frac{1}{2}(9.81)(1600) = -7848\text{ m}$$

Using $\int \ln(A - \alpha t) dt = -\frac{1}{\alpha}[(A - \alpha t)\ln(A - \alpha t) - (A - \alpha t)]$:

$$\int_0^{t_b} u \ln\left(\frac{M_0}{M_0 - \alpha t}\right) dt = u t_b - \frac{u M_b}{\alpha} \ln\left(\frac{M_0}{M_b}\right) = (2400)(40) - \frac{(2400)(200)}{20} \ln(5) = 96000 - 24000(1.60944) = 96000 - 38626.5 = 57373.5\text{ m}$$

Thus total altitude at burnout is:

$$y(t_b) = 57373.5 - 7848 = 49525.5\text{ meters} \approx 49.53\text{ km}$$
Final Answer & Physical Insight

Burnout time: $t_b = 40\text{ s}$. Maximum burnout velocity: $v(t_b) \approx 3470\text{ m/s}$ ($3.47\text{ km/s}$). Burnout altitude: $y(t_b) \approx 49.53\text{ km}$.