Damped Oscillation Flashcards

Introduction to Damped Oscillation

• Simple Harmonic Motion (SHM) is conventionally categorized as free oscillation.

• In physical reality, free oscillations rarely sustain indefinitely; they decay over time due to various resistive or dissipative forces that steadily diminish the total energy of the mechanical system.

• The inclusion of damping forces alters system dynamics in multiple ways, including the potential elimination of oscillatory behavior altogether, causing the system to return to equilibrium without oscillating.

• Damped motion is categorized based on the mathematical relationship between the natural frequency of vibration $\omega_0$ and the damping factor $\gamma$.

Simple Harmonic Motion Primer

• Definition 1: A particle executes Simple Harmonic Motion (SHM) if the net force acting upon it is directly proportional to its displacement relative to a designated origin and acts in the direction opposite to that displacement.

• The governing differential equation of motion for SHM is:

$m\ddot{x} = -kx$

• In this formulation, $m$ represents the particle mass, $x = x(t)$ denotes particle position as a function of time $t$, $\dot{x} = \frac{dx}{dt}$ represents instantaneous velocity, $\ddot{x} = \frac{d^2x}{dt^2}$ represents instantaneous acceleration, and $k > 0$ is a constant stiffness parameter.

• The differential equation can be expressed in canonical form as:

$\ddot{x} = -\omega_0^2 x$

• Here, $\omega_0 > 0$ is a constant parameter representing the natural frequency of the system, defined by $\omega_0 = \sqrt{\frac{k}{m}}$.

• Remark 1 (Linearity and Homogeneity):

  • Linearity: If $x_1(t)$ and $x_2(t)$ are two distinct solutions to $\ddot{x} = -\omega_0^2 x$, their linear combination $x(t) = c_1 x_1(t) + c_2 x_2(t)$ (where $c_1$ and $c_2$ are arbitrary constants) is also a valid solution.

  • Homogeneity: The equation contains no non-zero terms independent of $x$ or its time derivatives. Consequently, the zero solution $x(t) = 0$ represents a valid trivial solution.

• Derivation of the General Solution:

  • Substituting a trial exponential solution $x(t) = e^{\lambda t}$ yields $\dot{x}(t) = \lambda e^{\lambda t}$ and $\ddot{x}(t) = \lambda^2 e^{\lambda t}$.

  • Substitution into the canonical equation yields:

$(\lambda^2 + \omega_0^2)e^{\lambda t} = 0$

• Because $e^{\lambda t} \neq 0$ for all $t$, the characteristic equation is $(\lambda^2 + \omega_0^2) = 0$, yielding complex roots $\lambda = \pm i\omega_0$

• The general mathematical solution is:

$x(t) = c_1 e^{i\omega_0 t} + c_2 e^{-i\omega_0 t}$

• Remark 2 (Physical Reality Condition):

  • Since $x(t)$ measures real physical displacement, $x(t)$ must equal its complex conjugate $x^*(t)$, yielding:

$c_1 e^{i\omega_0 t} + c_2 e^{-i\omega_0 t} = c_1^* e^{-i\omega_0 t} + c_2^* e^{i\omega_0 t}$

$(c_1 - c_2^*)e^{i\omega_0 t} + (c_1^* - c_2)e^{-i\omega_0 t} = 0$

• Differentiating with respect to time $t$ gives:

$(c_1 - c_2^*)e^{i\omega_0 t} - (c_1^* - c_2)e^{-i\omega_0 t} = 0$

• Summing these two equations yields $2(c_1 - c_2^*)e^{i\omega_0 t} = 0$. Because $e^{i\omega_0 t} \neq 0$ for all $t$, $c_1 = c_2^*$, confirming that $c_1$ and $c_2$ are complex conjugates.

• Writing $c_1 = A e^{-i\theta}$ (where $A$ and $\theta$ are real constants) yields the physical displacement form:

$x(t) = A e^{i(\omega_0 t - \theta)} + A e^{-i(\omega_0 t - \theta)} = 2A \cos(\omega_0 t - \theta)$

• Determination of Integration Constants via Initial Conditions:

  • Case I ($x(0) = 0$, $v(0) = \frac{dx}{dt}\Big|_{t=0} = v_0$):

$0 = 2A \cos(\theta)$

$v_0 = -2A\omega_0 \sin(-\theta)$

• The first condition implies $\cos(\theta) = 0 \implies \theta = \frac{\pi}{2}$, while the second gives $A = \frac{v_0}{2\omega_0}$. The physical solution is:

$x(t) = \left(\frac{v_0}{\omega_0}\right) \sin(\omega_0 t)$

• Case II ($x(0) = x_0$, $v(0) = \frac{dx}{dt}\Big|_{t=0} = 0$):

$x_0 = 2A \cos(\theta)$

$0 = -2A\omega_0 \sin(-\theta)$

• The second condition implies $\sin(\theta) = 0 \implies \theta = 0$, while the first gives $A = \frac{x_0}{2}$. The physical solution is:

$x(t) = x_0 \cos(\omega_0 t)$

• Definition 2: A non-trivial solution $x(t)$ is classified as oscillatory if it intersects the $t$-axis more than once within a specified time interval, and non-oscillatory if it intersects the $t$-axis at most once within that interval.

• Definition 3: Because $x\left(t + \frac{2\pi}{\omega_0}\right) = x(t)$, free SHM is periodic with time period T = \frac{2\text{\pi}}{\omega_0}, where $\omega_0$ represents the natural frequency.

• Theorem 1 (Energy Conservation in SHM): The differential equation $\ddot{x} + \omega_0^2 x = 0$ conserves the total mechanical energy $E = \frac{1}{2}m\dot{x}^2 + \frac{1}{2}m\omega_0^2 x^2$

  • Proof: Multiplying $\ddot{x} + \omega_0^2 x = 0$ by $2\dot{x}$ gives $2\dot{x}\ddot{x} + 2\omega_0^2 x\dot{x} = 0$. Multiplying by mass $m$ yields:

$\frac{d}{dt}\left(\frac{1}{2}m\dot{x}^2 + \frac{1}{2}m\omega_0^2 x^2\right) = 0$

$\frac{1}{2}m\dot{x}^2 + \frac{1}{2}m\omega_0^2 x^2 = \text{constant}$

• For $x(t) = x_0 \cos(\omega_0 t)$, the energy constant equals $\frac{1}{2}m\omega_0^2 x_0^2$. Dividing energy by $\frac{1}{2}m\omega_0^2 x_0^2$ yields the normalized trajectory:

$\left(\frac{\dot{x}}{\omega_0 x_0}\right)^2 + \left(\frac{x}{x_0}\right)^2 = 1$

• Remark 3 (Phase Space Representation):

  • The coordinate system defined by plotting displacement $x(t)$ against velocity $\dot{x}(t)$ constitutes the Phase Space of the dynamical system.

  • The equation $\left(\frac{\dot{x}}{\omega_0 x_0}\right)^2 + \left(\frac{x}{x_0}\right)^2 = 1$ describes an ellipse in phase space. A closed trajectory in phase space signifies periodic motion.

• Origin Dependency and Influence of Constant Forces:

  • SHM representation depends on the chosen origin. For example, $x(t) = A \sin^2(\omega_0 t)$ describes simple harmonic oscillation centered at $x = \frac{A}{2}$ with frequency $2\omega_0$.

  • Under an additional constant force $f_0$ (acceleration unit), the equation of motion becomes:

$\ddot{x} + \omega_0^2 x = f_0$

• Here, $\omega_0$ remains unchanged, but the equilibrium position shifts from $x = 0$ to $x = \frac{f_0}{\omega_0^2}$ (such as a hanging spring-mass system in gravity where $f_0 = g$).

  • Definition 4 (Time Average): The time average of a time-dependent physical quantity $g(t)$ over period $T$ is defined as:

$\langle g(t) \rangle = \frac{1}{T} \int_0^T g(t)\,dt$

• Remark 4 (Standard Harmonic Time Averages):

$\langle \cos(\omega_0 t \pm \phi) \rangle = \langle \sin(\omega_0 t \pm \phi) \rangle = 0$

$\langle \cos^2(\omega_0 t \pm \phi) \rangle = \langle \sin^2(\omega_0 t \pm \phi) \rangle = \frac{1}{2}$

Constitutive Differential Equation of Damped Motion

• In the presence of damping forces, the total mechanical energy decreases over time:

$\frac{d}{dt}\left(\frac{1}{2}m\dot{x}^2 + \frac{1}{2}m\omega_0^2 x^2\right) < 0$

• Assuming the resistive damping force is directly proportional to velocity, $F_{\text{damping}} = \Gamma v = \Gamma \dot{x}$ (where $\Gamma > 0$ is a constant), the rate of energy loss equals dissipated power $P = F_{\text{damping}} v = \Gamma \dot{x}^2$:

$\frac{d}{dt}\left(\frac{1}{2}m\dot{x}^2 + \frac{1}{2}m\omega_0^2 x^2\right) = -\Gamma \dot{x}^2$

$m\dot{x}\ddot{x} + m\omega_0^2 x\dot{x} = -\Gamma \dot{x}^2 \implies \ddot{x} + \frac{\Gamma}{m}\dot{x} + \omega_0^2 x = 0$

$\ddot{x} + 2\gamma\dot{x} + \omega_0^2 x = 0$

• Remark 5 (Damping Factor):

  • The parameter $\gamma = \frac{\Gamma}{2m}$ is defined as the damping factor.

  • Dimensional analysis: $[\gamma][\dot{x}] = [\text{L}][\text{T}^{-2}] \implies [\gamma] = [\text{T}^{-1}]$, matching the dimensions of frequency.

• Theorem 2 (Classification of Damped Solutions): The solutions of $\ddot{x} + 2\gamma\dot{x} + \omega_0^2 x = 0$ under initial conditions $x(0) = x_0$ and $v(0) = v_0$ are classified as:

  • Non-oscillatory when $\gamma \ge \omega_0$

  • Oscillatory when $\gamma < \omega_0$

• Proof: Substituting the trial function $x(t) = e^{-\gamma t} f(t)$ into the differential equation:

$\dot{x} = e^{-\gamma t}\frac{df}{dt} - \gamma e^{-\gamma t} f(t)$

$\ddot{x} = e^{-\gamma t}\left(\frac{d^2f}{dt^2} - 2\gamma \frac{df}{dt} + \gamma^2 f\right)$

• Substitution into $\ddot{x} + 2\gamma\dot{x} + \omega_0^2 x = 0$ yields:

$e^{-\gamma t}\left[\ddot{f} + (\omega_0^2 - \gamma^2)f\right] = 0$

• Because $e^{-\gamma t} \neq 0$ for all $t \in [0, \infty]$, the reduced equation for $f(t)$ is:

$\ddot{f} + (\omega_0^2 - \gamma^2)f = 0$

Analysis of Overdamping (Large Damping: $\gamma > \omega_0$)

• When $\gamma > \omega_0$, define $\alpha = \sqrt{\gamma^2 - \omega_0^2} > 0$.

• The auxiliary equation becomes $\ddot{f} - \alpha^2 f = 0$, yielding $f(t) = A e^{\alpha t} + B e^{-\alpha t}$.

• The general displacement function $x(t) = e^{-\gamma t} f(t)$ is:

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

• Applying initial conditions $x(0) = x_0$ and $\dot{x}(0) = v_0$:

$x(0) = A + B = x_0$

$\dot{x}(0) = A(\alpha - \gamma) - B(\alpha + \gamma) = v_0$

• Solving for integration constants $A$ and $B$ yields:

$A = \frac{x_0(\alpha + \gamma) + v_0}{2\alpha}$

$B = \frac{x_0(\alpha - \gamma) - v_0}{2\alpha}$

• The complete displacement equation is:

$x(t) = \left[\frac{x_0(\alpha + \gamma) + v_0}{2\alpha}\right] e^{(\alpha - \gamma)t} + \left[\frac{x_0(\alpha - \gamma) - v_0}{2\alpha}\right] e^{-(\alpha + \gamma)t}$

• Time Axis Intersections: Setting $x(t_0) = 0$ determines when the path crosses the equilibrium axis:

$t_0 = \frac{1}{2\alpha} \ln\left(\frac{v_0 + x_0(\gamma - \alpha)}{v_0 + x_0(\gamma + \alpha)}\right)$

• Sub-case Analysis:

  • Case (a) ($x_0 = 0$, $v_0 \neq 0$):

$x(t) = \left(\frac{v_0}{\alpha}\right) e^{-\gamma t} \sinh(\alpha t)$

- Solving $t_0$ gives $t_0 = 0$ as the sole axis crossing.

• Case (b) ($x_0 \neq 0$, $v_0 = 0$):

$x(t) = x_0 e^{-\gamma t} \cosh(\alpha t) + \left(\frac{x_0 \gamma}{\alpha}\right) e^{-\gamma t} \sinh(\alpha t)$

- Evaluating $t_0$ yields $t_0 = \frac{1}{2\alpha} \ln\left(\frac{\gamma - \alpha}{\gamma + \alpha}\right) < 0$. Because $t_0$ is negative, the displacement curve never intersects the $t$-axis for $t > 0$

• Case (c) (Arbitrary Initial Conditions $x(0) = x_0 \neq 0$, $v(0) = v_0$):

$t_0 = \frac{1}{2\alpha} \ln\left[\frac{\frac{v_0}{x_0} + (\gamma - \alpha)}{\frac{v_0}{x_0} + (\gamma + \alpha)}\right]$

- If $\frac{v_0}{x_0} > 0$ ($x_0$ and $v_0$ share the same sign, meaning the particle is pushed away from origin), the logarithmic argument is strictly less than $1$, yielding $t_0 < 0$ (no physical crossing for $t > 0$).
- If $\frac{v_0}{x_0} < 0$ (particle is projected back toward origin):
  - If $\left|\frac{v_0}{x_0}\right| > \gamma + \alpha > \gamma - \alpha$, $t_0 > 0$ exists. The particle crosses the origin once before asymptotic return as $t \to \infty$
  - If $\gamma + \alpha > \left|\frac{v_0}{x_0}\right| > \gamma - \alpha$ or $\gamma + \alpha > \gamma - \alpha > \left|\frac{v_0}{x_0}\right|$, no positive real solution for $t_0$ exists.
- Expressed in terms of initial velocity $v(0) = v_0 \neq 0$:

$t_0 = \frac{1}{2\alpha} \ln\left[\frac{1 + \frac{x_0}{v_0}(\gamma - \alpha)}{1 + \frac{x_0}{v_0}(\gamma + \alpha)}\right]$

• Turning Point Dynamics:

  • For Case (a), setting $\dot{x}(t_0) = 0$ gives:

$\dot{x}(t) = \left(\frac{v_0}{\alpha}\right)\left[\alpha e^{-\gamma t} \cosh(\alpha t) - \gamma e^{-\gamma t} \sinh(\alpha t)\right] = 0$

$\tanh(\alpha t_0) = \frac{\alpha}{\gamma} \implies t_0 = \frac{1}{\alpha} \tanh^{-1}\left(\frac{\alpha}{\gamma}\right)$

• Case (b) exhibits no turning point. Highly damped non-oscillatory behavior of this type is described as dead beat.

Analysis of Critical Damping ($\gamma = \omega_0$)

• When $\gamma = \omega_0$, $\ddot{f} = 0 \implies f(t) = A t + B$.

• The general displacement function is:

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

• Applying initial conditions $x(0) = x_0$ and $\dot{x}(0) = v_0$ gives $B = x_0$ and $A = v_0 + \gamma x_0$, yielding:

$x(t) = e^{-\gamma t}\left[v_0 t + x_0(1 + \gamma t)\right]$

• Time Axis Intersections: Setting $x(t_0) = 0$ yields:

$t_0 = -\frac{x_0}{v_0 + \gamma x_0}$

• Specific Cases:

  • For $x_0 = 0$, $v_0 \neq 0$: $x(t) = v_0 t e^{-\gamma t}$, crossing the time axis only at $t = 0$

  • For $x_0 \neq 0$, $v_0 = 0$: $x(t) = x_0(1 + \gamma t)e^{-\gamma t}$, yielding $t_0 = -\frac{1}{\gamma} < 0$ (no positive intersection).

• Parametric Crossing Conditions:

  • For $x_0 = -\zeta_0$ ($\zeta_0 > 0$) and $v_0 > 0$:

$t_0 = \frac{\zeta_0}{v_0 - \gamma \zeta_0}$

- Positivity of $t_0$ requires $\gamma < \frac{v_0}{\zeta_0}$.

• For $x_0 = \xi_0 > 0$ and $v_0 = -\eta_0 < 0$ ($\eta_0 > 0$):

$t_0 = \frac{\xi_0}{\eta_0 - \gamma \xi_0}$

- Positivity of $t_0$ requires $\gamma < \frac{\eta_0}{x_0}$.

• Turning Point Analysis: For $x_0 = 0$ and $v_0 \neq 0$:

$\dot{x}(t) = v_0\left[e^{-\gamma t} - \gamma t e^{-\gamma t}\right] = 0 \implies t_0 = \frac{1}{\gamma} = \frac{1}{\omega_0}$

• Critical damping provides the fastest rate of decay to equilibrium without overshoot or oscillation.

Analysis of Weak Damping ($\gamma < \omega_0$)

• When $\gamma < \omega_0$, define $\tilde{\omega} = \sqrt{\omega_0^2 - \gamma^2} > 0$

• The differential equation for $f(t)$ becomes $\ddot{f} = -\tilde{\omega}^2 f$, with general solution $f(t) = A e^{i\tilde{\omega}t} + B e^{-i\tilde{\omega}t}$.

• Applying initial conditions $x(0) = x_0$ and $\dot{x}(0) = v_0$:

$x(0) = A + B = x_0$

$\dot{x}(0) = -A(\gamma - i\tilde{\omega}) - B(\gamma + i\tilde{\omega}) = v_0$

• Solving for $A$ and $B$ yields:

$A = \frac{x_0(i\tilde{\omega} + \gamma) + v_0}{2i\tilde{\omega}}$

$B = \frac{x_0(i\tilde{\omega} - \gamma) - v_0}{2i\tilde{\omega}}$

• Substituting $A$ and $B$ into $x(t) = e^{-\gamma t}(A e^{i\tilde{\omega}t} + B e^{-i\tilde{\omega}t})$ produces:

$x(t) = e^{-\gamma t}\left[x_0 \cos(\tilde{\omega} t) + \left(\frac{x_0\gamma + v_0}{\tilde{\omega}}\right) \sin(\tilde{\omega} t)\right]$

$x(t) = C e^{-\gamma t} \cos(\tilde{\omega} t - \theta_0)$

• In this expression, amplitude envelope $C$ and phase angle $\theta_0$ are defined by:

$C = \frac{1}{\tilde{\omega}}\sqrt{x_0^2 \tilde{\omega}^2 + (x_0\gamma + v_0)^2}$

$\theta_0 = \tan^{-1}\left(\frac{x_0\gamma + v_0}{x_0\tilde{\omega}}\right)$

• Specific Initial Condition Examples:

  • Case 1 ($x(0) = 0$, $v(0) = v_0 \neq 0$):

$x(t) = \left(\frac{v_0}{\tilde{\omega}}\right) e^{-\gamma t} \sin(\tilde{\omega} t)$

• Case 2 ($x(0) = x_0 \neq 0$, $v(0) = 0$):

$x(t) = \left(\frac{x_0 \omega_0}{\tilde{\omega}}\right) e^{-\gamma t} \cos\left(\tilde{\omega} t - \tan^{-1}\left(\frac{\gamma}{\tilde{\omega}}\right)\right)$

• Quasi-Periodicity and Extrema Characteristics:

  • Zero-crossings occur at time intervals separated by $\Delta t = \frac{\pi}{\tilde{\omega}}$. The quasi-period is $\tilde{T} = \frac{2\pi}{\tilde{\omega}}$.

  • The time interval between a maximum and the subsequent minimum is $\frac{\pi}{\tilde{\omega}}$, though extrema do not fall precisely halfway between zero-crossings.

  • Extrema occur when $\dot{x}(t) = 0$:

$\cos(\tilde{\omega} t) - \left(\frac{\gamma}{\tilde{\omega}}\right) \sin(\tilde{\omega} t) = 0 \implies \tan(\tilde{\omega} t) = \frac{\tilde{\omega}}{\gamma}$

• In the limit as $\frac{\gamma}{\tilde{\omega}} \to 0$, $\tilde{\omega} t \to \frac{\pi}{2} \implies t \to \frac{\pi}{2\tilde{\omega}}$.

Energy, Power Dissipation, and Characteristics of Weakly Damped Oscillations

• Energy Derivation under Weak Damping ($\gamma \ll \omega_0$):

  • Writing displacement as $x(t) = A(t) \sin(\tilde{\omega} t)$, where $A(t) = \left(\frac{v_0}{\tilde{\omega}}\right) e^{-\gamma t}$ is the time-decaying amplitude.

  • The total instantaneous energy $E(t) = \frac{1}{2}m\dot{x}^2 + \frac{1}{2}m\omega_0^2 x^2$ expands to:

$E(t) = \frac{1}{2}m\left(\frac{v_0^2}{\tilde{\omega}^2}\right) e^{-2\gamma t}\left[\tilde{\omega}^2 \cos^2(\tilde{\omega} t) + \gamma^2 \sin^2(\tilde{\omega} t) - 2\tilde{\omega}\gamma \cos(\tilde{\omega} t) \sin(\tilde{\omega} t)\right] + \frac{1}{2}m\omega_0^2\left(\frac{v_0^2}{\tilde{\omega}^2}\right) e^{-2\gamma t} \sin^2(\tilde{\omega} t)$

• Over one quasi-period $\tilde{T}$, assuming $e^{-2\gamma t}$ changes negligibly over single cycles ($\gamma \ll \omega_0$), time averaging yields:

$\langle \cos^2(\tilde{\omega} t) \rangle \approx \frac{1}{2}$

$\langle \sin^2(\tilde{\omega} t) \rangle \approx \frac{1}{2}$

$\langle \sin(2\tilde{\omega} t) \rangle \approx 0$

• The cycle-averaged total energy is:

$\langle E(t) \rangle = \frac{1}{\tilde{T}} \int_0^{\tilde{T}} E(t)\,dt = \frac{m v_0^2}{4\tilde{\omega}^2} e^{-2\gamma t}\left[\tilde{\omega}^2 + \omega_0^2 + \gamma^2\right] = \frac{1}{2}m A_0^2 \omega_0^2 e^{-2\gamma t}$

• In this expression, $A_0^2 = \frac{v_0^2}{\tilde{\omega}^2}$ and $\tilde{\omega}^2 = \omega_0^2 - \gamma^2$

• Power Dissipation:

  • Instantaneous power dissipated by resistive force: $P_{\text{damping}} = F_{\text{damping}} \cdot v = 2m\gamma \dot{x}^2$

  • Cycle-averaged dissipated power over $\tilde{T}$:

$\langle P(t) \rangle = \frac{1}{2}m A_0^2 \omega_0^2 (2\gamma) e^{-2\gamma t} = 2\gamma \langle E(t) \rangle$

• Logarithmic Decrement:

  • Consider displacement peaks at quarter-period offsets $t = \frac{\tilde{T}}{4}, \frac{3\tilde{T}}{4}, \frac{5\tilde{T}}{4}, \dots$

• Absolute amplitude values at successive half-period intervals:

$A_1 = \left| A\left(t = \frac{\tilde{T}}{4}\right) \right| = A_0 e^{-\frac{\gamma \tilde{T}}{4}}$

$A_2 = \left| A\left(t = \frac{3\tilde{T}}{4}\right) \right| = A_0 e^{-\frac{3\gamma \tilde{T}}{4}}$

$A_3 = \left| A\left(t = \frac{5\tilde{T}}{4}\right) \right| = A_0 e^{-\frac{5\gamma \tilde{T}}{4}}$

• Ratio of successive amplitudes:

$\frac{A_1}{A_2} = \frac{A_2}{A_3} = \frac{A_3}{A_4} = \dots = \frac{A_{n-1}}{A_n} = e^{\frac{\gamma \tilde{T}}{2}} = \Delta$

$\ln(\Delta) = \frac{\gamma \tilde{T}}{2}$

• For $n$ consecutive peaks:

$\frac{A_1}{A_n} = \frac{A_1}{A_2} \cdot \frac{A_2}{A_3} \dots \frac{A_{n-1}}{A_n} = \Delta^{n-1} \implies \ln\left(\frac{A_1}{A_n}\right) = (n-1)\ln(\Delta)$

$\ln(\Delta) = \frac{1}{n-1}\ln\left(\frac{A_1}{A_n}\right)$

• Definition 5: Logarithmic decrement $\lambda$ is the natural logarithm of the ratio of two successive amplitudes separated by a half period $\frac{\tilde{T}}{2}$:

$\lambda = \ln(\Delta) = \frac{\gamma \tilde{T}}{2}$

• Relaxation Time:

  • Definition 6: Relaxation time $\tau = \frac{1}{\gamma}$ is the time required for the amplitude to decay to $\frac{1}{e}$ of its initial magnitude.

  • Remark 9: Relationship between logarithmic decrement $\lambda$ and relaxation time $\tau$:

$\lambda \tau = \frac{\tilde{T}}{2}$

• Quality Factor ($Q$-Value):

  • From $\langle E(t) \rangle = (2\gamma)^{-1} \langle P(t) \rangle$, multiplying both sides by $\tilde{T}$ yields:

$\tilde{T} \langle E(t) \rangle = \tilde{T}(2\gamma)^{-1} \langle P(t) \rangle$

$\left(\frac{2\pi}{\tilde{\omega}}\right) \langle E(t) \rangle = \left(\frac{2\pi}{\tilde{\omega}}\right) (2\gamma)^{-1} \langle P(t) \rangle \implies \frac{\tilde{\omega}}{2\gamma} = \frac{2\pi \langle E(t) \rangle}{\tilde{T} \langle P(t) \rangle}$

• Definition 7: Quality factor $Q = \frac{\tilde{\omega}}{2\gamma}$ is defined physically as:

$Q = 2\pi \left(\frac{\text{Average energy stored in one period}}{\text{Average energy lost in one period}}\right)$

Solved Worked Examples and Practice Problems

• Problem 1: The governing differential equation of a damped system is given by $5\ddot{x} + 0.01\dot{x} + \alpha^2 x = 0$

  • (i) Find damping factor $\gamma$ and natural frequency $\omega_0$ for $\alpha = 5\sqrt{5}$:

  • Divide equation by $5$:

$\ddot{x} + 0.002\dot{x} + 25 x = 0$

- Comparing with $\ddot{x} + 2\gamma\dot{x} + \omega_0^2 x = 0$ gives $2\gamma = 0.002 \implies \gamma = 0.001\,\text{s}^{-1}$ and $\omega_0^2 = 25 \implies \omega_0 = 5\,\text{s}^{-1}$.

• (ii) Find conditions on $\alpha$ for large, critical, and weak damping:

  • Large damping ($\gamma > \omega_0$): $\alpha > 0.001$

  • Critical damping ($\gamma = \omega_0$): $\alpha = 0.001$

  • Weak damping ($\gamma < \omega_0$): $\alpha < 0.001$

• (iii) Find $x(t)$ for $\alpha = 5\sqrt{5}$ if $x(0) = 3$ and $v(0) = 0$:

  • Weakly damped general solution: $x(t) = e^{-\gamma t}\left[A \cos(\tilde{\omega} t) + B \sin(\tilde{\omega} t)\right]$

  • Here $\tilde{\omega} = \sqrt{\omega_0^2 - \gamma^2} = \sqrt{25 - (0.001)^2} \approx 5\,\text{s}^{-1}$.

  • Applying initial conditions: $x(0) = 3 = A$, and $v(0) = -\gamma A + B\tilde{\omega} = 0 \implies B = \frac{\gamma A}{\tilde{\omega}} = \frac{0.001 \times 3}{5} = 6 \times 10^{-4}$.

  • Displacement function:

$x(t) \approx e^{-0.001 t}\left[3 \cos(5t) + 6 \times 10^{-4} \sin(5t)\right]$

• (iv) Repeat (iii) for initial conditions $x(0) = 5$ and $v(0) = 3$:

  • $A = 5$, and 3 = -\gamma A + \tilde{\omega} B \implies 3 = -(0.001)(5) + 5 B \implies B = \frac{3.005}{5} = 0.601$.\n - Solution: x(t) = e^{-0.001 t}\left[5 \cos(5t) + 0.601 \sin(5t)\right].\n\n- **Problem 2:** A massless string suspended from a rigid rod carries a mass of 200\,\text{g}$at its lower end. The system oscillates with a period of$0.2\,\text{s}$, and amplitude reduces to half its initial value in$30\,\text{s}. Calculate (a) relaxation time, (b) quality factor, and (c) logarithmic decrement.\n - **Solution:**\n - **(a) Relaxation Time:** Using A(t) = A(0)e^{-\gamma t}$,$\gamma t = \ln\left(\frac{A(0)}{A(t)} ight)$. For$t = 30\,\text{s}$and$\frac{A(0)}{A(30)} = 2:\n\n\gamma (30) = \ln(2) \implies \gamma = \frac{\ln(2)}{30} \approx 0.023\,\text{s}^{-1}\n\n\tau = \frac{1}{\gamma} = \frac{30}{\ln(2)} \approx 43.28\,\text{s}\n\n - **(b) Quality Factor:** Natural frequency \omega_0 = \frac{2\pi}{T} = \frac{2\pi}{0.2} = 31.4\,\text{s}^{-1}.\n\nQ = \frac{\tilde{\omega}}{2\gamma} = \frac{\sqrt{\omega_0^2 - \gamma^2}}{2\gamma} \approx \frac{31.4}{2 \times 0.023} \approx 683\n\n - **(c) Logarithmic Decrement:**\n\n\lambda = \frac{\gamma \tilde{T}}{2} = \frac{\gamma \pi}{\tilde{\omega}} \approx \frac{0.023 \times \pi}{31.4} \approx 0.0025\n\n- **Problem 3:** A critically damped mechanical system is displaced by distance x_0$from equilibrium and released from rest. Show that$x(t) = x_0(1 + \omega_0 t)e^{-\omega_0 t}.\n - **Solution:**\n - For critical damping (\gamma = \omega_0$),$x(t) = e^{-\gamma t}(A t + B).\n - Applying x(0) = x_0 \implies B = x_0\n - Applying v(0) = 0 = \left[-\gamma e^{-\gamma t}(A t + B) + e^{-\gamma t} A\right]{t=0} = -\gamma B + A \implies A = \gamma B = \gamma x_0 = \omega_0 x_0\n - Substituting constants yields:\n\nx(t) = x_0(1 + \omega_0 t)e^{-\omega_0 t}\n\n- **Problem 4:** A weakly damped oscillator has mass m$, oscillation frequency$\alpha$, and logarithmic decrement$\lambda. Calculate the stiffness force and resistive force.\n - **Solution:**\n - Given damped frequency \tilde{\omega} = \alpha$and logarithmic decrement$\lambda = \frac{\gamma \pi}{\alpha} \implies \gamma = \frac{\lambda \alpha}{\pi}.\n - Natural frequency relation: \omega_0^2 = \tilde{\omega}^2 + \gamma^2 = \alpha^2 + \frac{\lambda^2 \alpha^2}{\pi^2} = \alpha^2\left(1 + \frac{\lambda^2}{\pi^2}\right).\n - Stiffness force parameter:\n\nk x = m \omega_0^2 x = m \alpha^2 \left(1 + \frac{\lambda^2}{\pi^2}\right) x\n\n - Resistive damping force parameter:\n\n2 m \gamma v = \frac{2 m \lambda \alpha v}{\pi}\n\n- **Problem 5:** The ratio between frequency \tilde{\omega}$of a weakly damped oscillator and natural frequency$\omega_0$is$r : 1$(i.e.$\frac{\tilde{\omega}}{\omega_0} = r).\n - **(i) Logarithmic decrement \lambda$and Quality factor$Q:**\n - Damping factor \gamma = \sqrt{\omega_0^2 - \tilde{\omega}^2} = \omega_0 \sqrt{1 - r}\n - Logarithmic decrement:\n\n\lambda = \frac{\gamma \pi}{\tilde{\omega}} = \frac{\pi \sqrt{1 - r}}{r}\n\n - Quality factor:\n\nQ = \frac{\tilde{\omega}}{2\gamma} = \frac{r}{2\sqrt{1 - r}}\n\n - **(ii) System energy and time taken to reduce to \frac{1}{e}$of initial energy for maximum initial displacement$x{\text{max}}:**\n - Energy expression:\n\nE(t) = \frac{1}{2} m \omega_0^2 x_{\text{max}}^2 e^{-2\gamma t} = \frac{1}{2} m \omega_0^2 x_{\text{max}}^2 e^{-2\omega_0 \sqrt{1 - r} t}\n\n - Time t_1$to reach$\frac{1}{e} initial energy:\n\n\frac{E(t_1)}{E(0)} = \frac{1}{e} = e^{-2\gamma t_1} \implies t_1 = \frac{1}{2\gamma} = \frac{1}{2\omega_0 \sqrt{1 - r}}\n\n - **(iii) Energy loss during the first cycle:**\n - Time for one cycle t = \frac{2\pi}{\tilde{\omega}}. Energy remaining after first cycle:\n\nE\left(\frac{2\pi}{\tilde{\omega}}\right) = E(0) e^{-\frac{4\pi \gamma}{\tilde{\omega}}}\n\n - Energy lost during first cycle:\n\n\Delta E = E(0) - E\left(\frac{2\pi}{\tilde{\omega}}\right) = E(0)\left(1 - e^{-\frac{4\pi}{r}\sqrt{1-r}}\right)$</p></li></ul></li></ul><h5>Introduction to Damped Oscillation</h5><ul><li><p>Simple Harmonic Motion (SHM) is also known as free oscillation, where a system oscillates back and forth without stopping.</p></li><li><p>In real life, free oscillations do not last forever because friction, air resistance, and other opposing forces slow them down and reduce their total energy over time.</p></li><li><p>These opposing forces are called damping forces. Damping can slow down the motion or even stop the system from swinging altogether, causing it to return to its rest position smoothly.</p></li><li><p>Damped motion is divided into different types based on the relationship between the natural frequency$\omega_0$and the damping factor$\gamma$.</p></li></ul><h5>Simple Harmonic Motion Primer</h5><ul><li><p><strong>Definition 1:</strong> A particle undergoes Simple Harmonic Motion (SHM) when the total force pulling it back to its starting point is directly proportional to how far it has moved, and acts in the opposite direction of that movement.</p></li><li><p>The basic equation of motion for SHM is:<br>$m\ddot{x} = -k x$</p></li><li><p>In this equation,$m$is the mass of the particle,$x(t)$is its position at time$t$,$\dot{x} = \frac{dx}{dt}$is its velocity,$\ddot{x} = \frac{d^2 x}{dt^2}$is its acceleration, and$k > 0$is the stiffness constant of the spring or restoring force.</p></li><li><p>We can rewrite this equation in standard canonical form as:<br>$\ddot{x} = -\omega_0^2 x$</p></li><li><p>Here,$\omega0 > 0$is the natural frequency of the system, defined as$\omega0 = \sqrt{\frac{k}{m}}$.</p></li><li><p><strong>Remark 1 (Linearity and Homogeneity):</strong></p><ul><li><p>Linearity: If$x1(t)$and$x2(t)$are two different valid solutions to the equation, any linear combination$x(t) = c1 x1(t) + c2 x2(t)$(where$c1$and$c2$are constants) is also a valid solution.</p></li><li><p>Homogeneity: All terms depend on$x$or its time derivatives, meaning$x(t) = 0$(staying completely still) is also a valid solution.</p></li></ul></li><li><p><strong>Derivation of the General Solution:</strong></p><ul><li><p>We try a trial solution of the form$x(t) = e^{\lambda t}$. This gives velocity$\dot{x}(t) = \lambda e^{\lambda t}$and acceleration$\ddot{x}(t) = \lambda^2 e^{\lambda t}$.</p></li><li><p>Substituting these into the standard equation gives:<br>$(\lambda^2 + \omega_0^2) e^{\lambda t} = 0$</p></li><li><p>Since$e^{\lambda t}$is never zero, we get the characteristic equation$\lambda^2 + \omega0^2 = 0$, which yields complex roots$\lambda = \pm i\omega0$</p></li><li><p>The general solution is:<br>$x(t) = c1 e^{i\omega0 t} + c2 e^{-i\omega0 t}$</p></li></ul></li><li><p><strong>Remark 2 (Physical Reality Condition):</strong></p><ul><li><p>Since physical displacement$x(t)$must be a real number,$x(t)$must equal its complex conjugate$x^*(t)$.</p></li><li><p>This requirement forces$c1$and$c2$to be complex conjugates of each other ($c1 = c2^*$).</p></li><li><p>Setting$c1 = A e^{-i\theta}$(where$A$and$\theta$are real constants) gives the physical displacement formula:$x(t) = 2A \cos(\omega0 t - \theta)$</p></li></ul></li><li><p><strong>Finding Constants using Initial Conditions:</strong></p><ul><li><p><strong>Case I (Starts at origin with initial velocity$v(0) = v0$and position$x(0) = 0$):</em></strong><em><br>$x(t) = \left(\frac{v0}{\omega0}\right) \sin(\omega0 t)$</p></li><li><p><strong>Case II (Released from rest at position$x(0) = x0$with velocity$v(0) = 0$):</em></strong><em><br>$x(t) = x0 \cos(\omega_0 t)$</p></li></ul></li><li><p><strong>Energy Conservation in SHM:</strong></p><ul><li><p>The total energy$E$of the system remains constant over time:<br>$\frac{d}{dt}\left(\frac{1}{2}m\dot{x}^2 + \frac{1}{2}m\omega0^2 x^2\right) = 0$$\frac{1}{2}m\dot{x}^2 + \frac{1}{2}m\omega0^2 x^2 = \text{constant}$</p></li><li><p>Dividing by total energy gives the normalized path equation:<br>$\left(\frac{\dot{x}}{\omega0 x0}\right)^2 + \left(\frac{x}{x_0}\right)^2 = 1$</p></li></ul></li><li><p><strong>Remark 3 (Phase Space Representation):</strong></p><ul><li><p>A graph plotting position$x(t)$against velocity$\dot{x}(t)$is called Phase Space.</p></li><li><p>For SHM, the trajectory forms an ellipse in phase space. A closed loop in phase space means the motion repeats periodically.</p></li></ul></li><li><p><strong>Definition 4 (Time Average):</strong></p><ul><li><p>The time average of a quantity$g(t)$over a time period$T$is:<br>$\langle g(t) \rangle = \frac{1}{T} \int_0^T g(t)\,dt$</p></li><li><p>Useful standard averages:<br>$\langle \cos(\omega0 t \pm \phi) \rangle = \langle \sin(\omega0 t \pm \phi) \rangle = 0$<br>$\langle \cos^2(\omega0 t \pm \phi) \rangle = \langle \sin^2(\omega0 t \pm \phi) \rangle = \frac{1}{2}$</p></li></ul></li></ul><h5>Constitutive Differential Equation of Damped Motion</h5><ul><li><p>When damping is present, total mechanical energy decreases over time due to drag forces, which are proportional to velocity ($F_{\text{damping}} = \Gamma \dot{x}$).</p></li><li><p>The rate of energy loss equals the dissipated power$P = \Gamma \dot{x}^2$:<br>$\frac{d}{dt}\left(\frac{1}{2}m\dot{x}^2 + \frac{1}{2}m\omega_0^2 x^2\right) = -\Gamma \dot{x}^2$</p></li><li><p>This simplifies to the fundamental differential equation for damped motion:<br>$\ddot{x} + 2\gamma \dot{x} + \omega_0^2 x = 0$</p></li><li><p><strong>Remark 5 (Damping Factor):</strong></p><ul><li><p>The parameter$\gamma = \frac{\Gamma}{2m}$is called the damping factor, and it has units of frequency ($\text{s}^{-1}$).</p></li></ul></li><li><p><strong>Theorem 2 (Classification of Damped Motion):</strong></p><ul><li><p><strong>Overdamping ($\gamma > \omega_0$):</strong> Strong damping causes the system to return slowly to rest without oscillating.</p></li><li><p><strong>Critical Damping ($\gamma = \omega_0$):</strong> The system returns to rest as fast as possible without oscillating.</p></li><li><p><strong>Weak Damping / Underdamping ($\gamma < \omega_0$):</strong> The system continues to oscillate back and forth, but its amplitude decays exponentially over time according to$e^{-\gamma t}$$.