--- marp: true theme: graph_paper paginate: true title: Day 28 - Workshop Session description: Slides for PHY 321 Spring 2026, Day 28: Workshop Session author: Prof. Danny Caballero keywords: classical mechanics, differential equations, motion, oscillations, resonance url: https://dannycaballero.info/phy321msu/slides/day-28-workshop-session.html --- # Day 28 - Workshop Session ![bg right width:600px](../images/notes/week10/trojan_horse_maths.jpeg) --- ## Exercise 2 Where does the energy go? The damped harmonic oscillator is described by the equation of motion is: $$m\ddot{x} + b\dot{x} + kx = 0$$ where $m$ is the mass, $b$ is the damping coefficient, and $k$ is the spring constant. The damping term ($F_{damp} = - b\dot{x}$) models the dissipative forces acting on the oscillator. The total energy for the oscillator is given by the sum of the kinetic and potential energies, $$E = \frac{1}{2}m\dot{x}^2 + \frac{1}{2}kx^2.$$ * 2a What is the energy per unit time dissipated by the damping force? --- ## Exercise 2 Where does the energy go? $$m\ddot{x} + b\dot{x} + kx = 0$$ $$E = \frac{1}{2}m\dot{x}^2 + \frac{1}{2}kx^2$$ * 2b Take the time derivative of the total energy and show that it is equal (in magnitude) to the energy dissipated by the damping force. * 2c What is the sign relationship between the energy dissipated by the damping force and the time derivative of the total energy? --- ## Exercise 3, Unpacking the critically damped solution The solution for critical damping ($\beta = \omega_0$) is given by, $$x(t) = x_1(t) + x_2(t) = Ae^{-\beta t} + Bte^{-\beta t},$$ where $A$ and $B$ are constants. Notice the second solution $x_2(t) = Bte^{-\omega_0t}$ has an additional linear term $t$. We glossed over this solution in class, but it is important to understand why this term is present because it tells us about solving differential equations with pathologically difficult-to-see solutions. ---- ## Exercise 3, Unpacking the critically damped solution Start with the under damped solutions, $$y_1(t) = e^{-\beta t}\cos(\omega_1 t) \quad \text{and} \quad y_2(t) = e^{-\beta t}\sin(\omega_1 t),$$ where we have used the notation $\omega_1 = \sqrt{\omega_0^2 - \beta^2}$. * 3a (3pt). Show that you can recover the first solution $x_1(t)$ by taking the limit of $\beta \rightarrow \omega_0$ of $y_1(t)$. * 3b (3pt). Show that you cannot recover the second solution $x_2(t)$ by taking the limit of $\beta \rightarrow \omega_0$ of $y_2(t)$ directly. What do you get? * 3c (4pt). If $\beta \neq \omega_0$, you can divide $y_2(t)$ by $\omega_1$. Now show that in the limit $\beta \rightarrow \omega_0$ of $y_2(t)/\omega_1$, we recover the form of $x_2(t)$.