Physlib.ClassicalMechanics.DampedHarmonicOscillator.Solution

Given a set of initial conditions $IC$ with initial position $x_0$ and initial velocity $v_0$, and a system with angular frequency $\omega$ and decay rate $\gamma$, this function maps a time $t$ to the oscillatory component of the underdamped harmonic oscillator trajectory: $$t \mapsto x_0 \cos(\omega t) + \frac{v_0 + \gamma x_0}{\omega} \sin(\omega t)$$ This represents the part of the solution that is subsequently multiplied by the exponential factor $e^{-\gamma t}$ to form the complete trajectory. The values are elements of the 1-dimensional Euclidean space $\mathbb{R}^1$.

Given a set of initial conditions $IC$ with initial position $x_0$ and initial velocity $v_0$, and a system decay rate $\gamma$, this function maps a time $t$ to the polynomial component of the critically damped oscillator trajectory: $$t \mapsto x_0 + t(v_0 + \gamma x_0)$$ This represents the part of the solution that is subsequently multiplied by the exponential factor $e^{-\gamma t}$ to form the complete trajectory.

Given initial conditions $IC$ with initial position $x_0$ and initial velocity $v_0$, a system decay rate $\gamma$, and an angular frequency $\omega$, this function maps a time $t$ to the hyperbolic component of the overdamped oscillator trajectory: $$t \mapsto \cosh(\omega t) x_0 + \frac{\sinh(\omega t)}{\omega} (v_0 + \gamma x_0)$$ This represents the part of the solution that is subsequently multiplied by the exponential factor $e^{-\gamma t}$ to form the complete trajectory in the overdamped regime.

Given initial conditions $IC$ (representing initial position $x_0$ and initial velocity $v_0$ at $t=0$), the function `trajectory` maps a time $t$ to the position of the damped harmonic oscillator in a 1-dimensional Euclidean space $\mathbb{R}^1$. The solution is defined piecewise according to the damping regime of the system: - If the system is **underdamped**, the trajectory is $e^{-\gamma t} \cdot x_{\text{under}}(t)$, where $x_{\text{under}}(t)$ is the trigonometric base solution. - If the system is **critically damped**, the trajectory is $e^{-\gamma t} \cdot x_{\text{crit}}(t)$, where $x_{\text{crit}}(t)$ is the polynomial base solution. - Otherwise (the **overdamped** case), the trajectory is $e^{-\gamma t} \cdot x_{\text{over}}(t)$, where $x_{\text{over}}(t)$ is the hyperbolic base solution. Here, $\gamma$ denotes the decay rate of the oscillator.

Consider a damped harmonic oscillator system $S$ with initial conditions $IC$ (consisting of initial position $x_0$ and initial velocity $v_0$). If the system is underdamped (i.e., its discriminant $\gamma^2 - 4mk < 0$), then the trajectory $x(t)$ is given by the product of an exponential decay factor and a trigonometric base solution: $$x(t) = e^{-\gamma t} \cdot x_{\text{under}}(t)$$ where $\gamma$ is the decay rate of the system and $x_{\text{under}}(t)$ is the trigonometric base solution $x_0 \cos(\omega t) + \frac{v_0 + \gamma x_0}{\omega} \sin(\omega t)$.

In a critically damped harmonic oscillator system $S$, for any given initial conditions $IC$, the trajectory $x(t)$ is given by the product of an exponential decay factor and a polynomial base solution: $$x(t) = e^{-\gamma t} \cdot x_{\text{crit}}(t)$$ where $\gamma$ is the decay rate of the system and $x_{\text{crit}}(t)$ is the polynomial base solution $x_0 + t(v_0 + \gamma x_0)$ determined by the initial position $x_0$ and initial velocity $v_0$.

For a damped harmonic oscillator in the overdamped regime (where the discriminant $\gamma^2 - 4mk > 0$), the trajectory $x(t)$ for a given set of initial conditions $IC$ is the product of an exponential decay factor and the hyperbolic base solution. Specifically, for any time $t$, the trajectory is given by: $$x(t) = e^{-\gamma t} \cdot x_{\text{over}}(t)$$ where $\gamma$ is the decay rate of the oscillator and $x_{\text{over}}(t)$ is the hyperbolic base solution defined as $\cosh(\omega t) x_0 + \frac{\sinh(\omega t)}{\omega} (v_0 + \gamma x_0)$ for initial position $x_0$ and initial velocity $v_0$.

For any initial conditions $IC$ of a damped harmonic oscillator system $S$, the resulting trajectory $x(t)$ is infinitely differentiable with respect to time, i.e., $x \in C^\infty(\mathbb{R}, \mathbb{R}^1)$.

For a damped harmonic oscillator with given initial conditions $IC$, let $x(t)$ denote the trajectory of the system as a function of time. The position of the oscillator at time $t=0$ is equal to the initial position $x_0$ specified in the initial conditions: $$x(0) = x_0$$

Consider a damped harmonic oscillator system with mass $m$, damping coefficient $\gamma$, and spring constant $k$. For any given initial conditions $IC$ (specifying position and velocity at $t=0$), if the system is critically damped (i.e., its discriminant $\gamma^2 - 4mk = 0$), then the trajectory $x(t)$ associated with $IC$ satisfies the damped equation of motion: $$m \ddot{x}(t) + \gamma \dot{x}(t) + k x(t) = 0$$ for all times $t$.

Consider a damped harmonic oscillator system $S$ with mass $m$, damping coefficient $\gamma$, and spring constant $k$. If the system is underdamped (that is, the discriminant satisfies $\gamma^2 < 4mk$), then for any initial conditions $IC$ (initial position $x_0$ and initial velocity $v_0$), the resulting trajectory $x(t)$ satisfies the equation of motion: $$m \ddot{x}(t) + \gamma \dot{x}(t) + k x(t) = 0$$ where $\dot{x}(t)$ and $\ddot{x}(t)$ are the first and second time derivatives of the position $x(t)$ with respect to time $t$.

For a damped harmonic oscillator with mass $m$, damping coefficient $\gamma$, and spring constant $k$, if the system is overdamped (i.e., $\gamma^2 > 4mk$), then the trajectory $x(t)$ corresponding to the initial conditions $IC$ satisfies the equation of motion: $$m \ddot{x}(t) + \gamma \dot{x}(t) + k x(t) = 0$$ where $\dot{x}(t)$ and $\ddot{x}(t)$ are the first and second time derivatives of the position, respectively.

For a damped harmonic oscillator with mass $m$, damping coefficient $\gamma$, and spring constant $k$, the trajectory $x(t)$ defined for any given initial conditions $IC$ satisfies the equation of motion: $$m \ddot{x}(t) + \gamma \dot{x}(t) + k x(t) = 0$$ where $\dot{x}(t)$ and $\ddot{x}(t)$ denote the first and second time derivatives of the position $x(t)$, respectively. This holds regardless of whether the system is underdamped, critically damped, or overdamped.

Let $x, y : \text{Time} \to \mathbb{R}^1$ be two infinitely differentiable ($C^\infty$) functions. If both $x$ and $y$ satisfy the equation of motion for a damped harmonic oscillator $m \ddot{x} + \gamma \dot{x} + kx = 0$, and share the same initial position $x(0) = y(0)$ and initial velocity $\dot{x}(0) = \dot{y}(0)$, then the trajectories are identical, $x = y$.

Consider a damped harmonic oscillator with initial conditions $IC$ specifying an initial position $x_0$ and an initial velocity $v_0$. Let $x: \text{Time} \to \mathbb{R}$ be a smooth function (of class $C^\infty$) that satisfies the equation of motion $m \ddot{x}(t) + \gamma \dot{x}(t) + k x(t) = 0$. If $x$ satisfies the initial conditions $x(0) = x_0$ and the time derivative at $t=0$ is $\dot{x}(0) = v_0$, then $x$ is equal to the uniquely defined trajectory for those initial conditions.