Coupled Oscillators Cheatsheet

Cheatsheet Content

Coupled Oscillators: Introduction Systems with two or more oscillators coupled together. Characterized by multiple frequencies of oscillation. Normal Modes: Specific ways a system can oscillate where all parts move with the same frequency. Normal Frequencies: The frequencies associated with normal modes. Coupled motion is crucial for understanding waves in continuous media. Physical Characteristics of Coupled Oscillators Two Coupled Pendulums (String Coupling) Pendulums with same length $l$ and equal periods. Coupling provided by a supporting string. Observation (i): In-phase motion Displace both pendulums by same amount in same direction. Oscillate with same frequency and amplitude. Frequency: $\omega_1 = \sqrt{g/l}$ (like uncoupled pendulums). Displacements: $x_a = A \cos \omega_1 t$, $x_b = A \cos \omega_1 t$ Observation (ii): Anti-phase motion Displace both pendulums by same amount in opposite directions. Oscillate with same frequency (slightly different from in-phase) and amplitude. Displacements: $x_a = B \cos \omega_2 t$, $x_b = -B \cos \omega_2 t$ Observation (iii): Energy Transfer (Beats) Displace one mass, leave other at equilibrium. Displaced mass amplitude decreases, initially at-rest mass amplitude increases. Energy is repeatedly transferred between the two masses. This behavior is a superposition of the two normal modes. Normal Modes of Oscillation (Spring Coupling) Two Pendulums Coupled by a Horizontal Spring Mass $m$, length $l$ for each pendulum. Spring constant $k$. Displacements $x_a$, $x_b$ from equilibrium. Spring is at unstretched length when pendulums are at equilibrium. Oscillations in the plane of the page. First Normal Mode (In-Phase) Masses displaced in the same direction ($x_a = x_b$). Spring retains unstretched length, plays no role. Frequency: $\omega_1 = \sqrt{g/l}$ Displacements: $$x_a = A \cos \omega_1 t$$ $$x_b = A \cos \omega_1 t$$ Second Normal Mode (Anti-Phase) Masses displaced in opposite directions ($x_a = -x_b$). Spring is alternately stretched and compressed, providing additional restoring force. Equation of motion for mass $a$: $$m \frac{d^2 x_a}{dt^2} = -\frac{mgx_a}{l} - 2kx_a$$ $$ \frac{d^2 x_a}{dt^2} + \omega_2^2 x_a = 0 $$ where $\omega_2^2 = \left(\frac{g}{l} + \frac{2k}{m}\right)$. Frequency $\omega_2 > \omega_1$ due to spring's additional restoring force. Displacements: $$x_a = B \cos \omega_2 t$$ $$x_b = -B \cos \omega_2 t$$ Characteristics of Normal Modes Both masses oscillate at the same frequency . Each mass performs Simple Harmonic Motion (SHM) with constant amplitude. Well-defined phase difference: either $0$ or $\pi$. Once initiated, the system stays in that mode and does not evolve into another. Normal modes are entirely independent of each other. Superposition of Normal Modes General motion of a coupled oscillator is a superposition of its normal modes. Introduce normal coordinates: $$q_1 = x_a + x_b$$ $$q_2 = x_a - x_b$$ Equations of motion in terms of normal coordinates: $$ \frac{d^2 q_1}{dt^2} + \omega_1^2 q_1 = 0 \quad (\text{for } q_1) $$ $$ \frac{d^2 q_2}{dt^2} + \omega_2^2 q_2 = 0 \quad (\text{for } q_2) $$ Solutions for normal coordinates: $$q_1 = C_1 \cos(\omega_1 t + \phi_1)$$ $$q_2 = C_2 \cos(\omega_2 t + \phi_2)$$ where $C_1, C_2$ are amplitudes and $\phi_1, \phi_2$ are phase angles. Displacements in terms of normal coordinates: $$x_a = \frac{1}{2}(q_1 + q_2) = \frac{1}{2}[C_1 \cos(\omega_1 t + \phi_1) + C_2 \cos(\omega_2 t + \phi_2)]$$ $$x_b = \frac{1}{2}(q_1 - q_2) = \frac{1}{2}[C_1 \cos(\omega_1 t + \phi_1) - C_2 \cos(\omega_2 t + \phi_2)]$$ Total Energy $E$ of the system: $$E = \frac{1}{2}m\left(\frac{dx_a}{dt}\right)^2 + \frac{1}{2}m\left(\frac{dx_b}{dt}\right)^2 + \frac{1}{2}mg(x_a^2+x_b^2) + \frac{1}{2}k(x_a-x_b)^2$$ In normal coordinates, this separates into two independent harmonic oscillator energies: $$E = \frac{1}{4}m\left[\left(\frac{dq_1}{dt}\right)^2 + \omega_1^2 q_1^2\right] + \frac{1}{4}m\left[\left(\frac{dq_2}{dt}\right)^2 + \omega_2^2 q_2^2\right]$$ This shows no "cross terms" between $q_1$ and $q_2$, confirming independence. Beats Phenomenon Example ($x_a=A, x_b=0$ at $t=0$): $$x_a = A \cos\left(\frac{\omega_2 - \omega_1}{2}t\right) \cos\left(\frac{\omega_2 + \omega_1}{2}t\right)$$ $$x_b = A \sin\left(\frac{\omega_2 - \omega_1}{2}t\right) \sin\left(\frac{\omega_2 + \omega_1}{2}t\right)$$ This shows a high-frequency oscillation modulated by a low-frequency term, analogous to beats in sound waves. Energy is transferred between oscillators. Oscillating Masses Coupled by Springs Two Masses with Three Springs Two identical masses $m$ connected by three identical springs of constant $k$. Outer springs connected to rigid walls. Equations of motion for masses $a$ and $b$: $$m \frac{d^2 x_a}{dt^2} = -k x_a + k(x_b - x_a) = k x_b - 2k x_a$$ $$m \frac{d^2 x_b}{dt^2} = -k(x_b - x_a) - k x_b = k x_a - 2k x_b$$ Assume normal mode solutions $x_a = A \cos \omega t$, $x_b = B \cos \omega t$. Substituting and solving yields a quadratic equation for $\omega^2$: $$(2k - m\omega^2)^2 = k^2$$ Normal Frequencies: $\omega_1^2 = k/m$ (First Normal Mode: $A=B$, in-phase) $\omega_2^2 = 3k/m$ (Second Normal Mode: $A=-B$, anti-phase) General motion: $$x_a = C_1 \cos \omega_1 t + C_2 \cos \omega_2 t$$ $$x_b = C_1 \cos \omega_1 t - C_2 \cos \omega_2 t$$ Matrix Approach (Eigenvalue Problem): $$\begin{pmatrix} 2k/m - \omega^2 & -k/m \\ -k/m & 2k/m - \omega^2 \end{pmatrix} \begin{pmatrix} A \\ B \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}$$ Non-zero solutions exist if the determinant is zero, leading to the same $\omega^2$ values. Forced Oscillations of Coupled Oscillators When a periodic driving force is applied to a coupled oscillator. The system exhibits large amplitude oscillations when the driving frequency $\omega$ is close to either of the normal frequencies ($\omega_1$ or $\omega_2$). Consider two masses with three springs, one outer spring driven harmonically: $\xi = a \cos \omega t$. $$m \frac{d^2 x_a}{dt^2} = -k(x_a - \xi) + k(x_b - x_a)$$ $$m \frac{d^2 x_b}{dt^2} = -k(x_b - x_a) - k x_b$$ Transforming to normal coordinates $q_1 = x_a + x_b$ and $q_2 = x_a - x_b$: $$\frac{d^2 q_1}{dt^2} + \frac{k}{m} q_1 = \frac{F_0}{m} \cos \omega t$$ $$\frac{d^2 q_2}{dt^2} + \frac{3k}{m} q_2 = \frac{F_0}{m} \cos \omega t$$ where $F_0 = ka$. These are independent forced SHM equations. Steady-state solutions $q_1 = C_1 \cos \omega t$, $q_2 = C_2 \cos \omega t$: $$C_1 = \frac{F_0/m}{(\omega_1^2 - \omega^2)}$$ $$C_2 = \frac{F_0/m}{(\omega_2^2 - \omega^2)}$$ Amplitudes $C_1, C_2$ become infinitely large (in the absence of damping) when $\omega = \omega_1$ or $\omega = \omega_2$. When $\omega \approx \omega_1$: $x_a \approx x_b$ (in-phase oscillation). When $\omega \approx \omega_2$: $x_a \approx -x_b$ (anti-phase oscillation). Transverse Oscillations Single Mass with Two Springs (Transverse Displacement) Mass $m$ connected by two springs (constant $k$, length $l$, tension $T$). Displaced transversely by $y$. For small displacements, tension $T$ is approximately constant. Equation of motion: $$m \frac{d^2 y}{dt^2} = -2T \sin \theta \approx -2T \frac{y}{l}$$ Frequency: $\omega = \sqrt{2T/(ml)}$ Two Masses with Three Springs (Transverse Displacements) Two equal masses $m$ connected by three identical springs of length $l$ and under tension $T$. Displacements $y_a, y_b$. Equations of motion: $$m \frac{d^2 y_a}{dt^2} = \frac{T}{l}(y_b - 2y_a)$$ $$m \frac{d^2 y_b}{dt^2} = \frac{T}{l}(y_a - 2y_b)$$ Normal Frequencies: $\omega_1^2 = T/(ml)$ (First Normal Mode: $A=B$, in-phase) $\omega_2^2 = 3T/(ml)$ (Second Normal Mode: $A=-B$, anti-phase) These normal modes resemble standing waves on a taut string.