Physlib.ClassicalMechanics.DampedHarmonicOscillator.Basic

For a damped harmonic oscillator with spring constant $k$, the spring constant is non-zero, i.e., $k \neq 0$.

For a damped harmonic oscillator with mass $m$, the mass is non-zero, i.e., $m \neq 0$.

For a damped harmonic oscillator with mass $m$ and spring constant $k$, the natural (undamped) angular frequency $\omega_0$ is defined as the square root of the ratio of the spring constant to the mass: $$\omega_0 = \sqrt{\frac{k}{m}}$$

For a damped harmonic oscillator, the natural angular frequency $\omega_0$ is strictly positive, i.e., $\omega_0 > 0$.

For a damped harmonic oscillator with mass $m$ and spring constant $k$, the square of the natural angular frequency $\omega_0$ is equal to the ratio of the spring constant to the mass: $$\omega_0^2 = \frac{k}{m}$$

For a damped harmonic oscillator with mass $m$, damping coefficient $\gamma$, and spring constant $k$, a position function $x : \text{Time} \to \mathbb{R}$ satisfies the equation of motion if for all times $t \in \text{Time}$, the following differential equation holds: $$m \ddot{x}(t) + \gamma \dot{x}(t) + k x(t) = 0$$ where $\dot{x}(t)$ and $\ddot{x}(t)$ represent the first and second time derivatives of $x$ at time $t$, respectively.

For a damped harmonic oscillator with mass $m$, given a position function $x: \text{Time} \to \mathbb{R}$, its kinetic energy is the function mapping each time $t$ to $$\frac{1}{2} m (\dot{x}(t))^2$$ where $\dot{x}(t)$ is the time derivative of the position $x$ at time $t$.

For a damped harmonic oscillator with spring constant $k$, given a position function $x: \text{Time} \to \mathbb{R}$, its potential energy is the function $t \mapsto \frac{1}{2} k (x(t))^2$.

For a damped harmonic oscillator with mass $m$ and spring constant $k$, given a position function $x: \text{Time} \to \mathbb{R}$, its mechanical energy is the function $t \mapsto E(t)$ defined as the sum of its kinetic and potential energies: $$E(t) = \frac{1}{2} m (\dot{x}(t))^2 + \frac{1}{2} k (x(t))^2$$ where $\dot{x}(t)$ is the time derivative of the position $x$ at time $t$.

For a damped harmonic oscillator with mass $m$, and for any position function $x: \text{Time} \to \mathbb{R}$, the kinetic energy is given by the function $t \mapsto \frac{1}{2} m (\dot{x}(t))^2$, where $\dot{x}(t)$ is the time derivative of $x$ at time $t$.

For a damped harmonic oscillator $S$ with spring constant $k$, and for any position function $x: \text{Time} \to \mathbb{R}$, the potential energy is given by the function $t \mapsto \frac{1}{2} k (x(t))^2$.

For a damped harmonic oscillator with damping coefficient $\gamma$, and for any position function $x : \text{Time} \to \mathbb{R}$, the energy dissipation rate is defined as the function $t \mapsto -\gamma (\dot{x}(t))^2$, where $\dot{x}(t)$ is the time derivative of $x$ at time $t$.

For a damped harmonic oscillator with damping coefficient $\gamma$, let $x : \text{Time} \to \mathbb{R}$ be a position function of class $C^\infty$ that satisfies the equation of motion $m \ddot{x} + \gamma \dot{x} + kx = 0$. Then the time derivative of the mechanical energy $E(t)$ at any time $t$ is given by $$\frac{dE}{dt}(t) = -\gamma (\dot{x}(t))^2,$$ where $\dot{x}(t)$ denotes the time derivative of $x$ at time $t$.

Consider a damped harmonic oscillator with mass $m$, spring constant $k$, and damping coefficient $\gamma > 0$. Let $x : \text{Time} \to \mathbb{R}$ be a position function of class $C^\infty$ that satisfies the equation of motion $m \ddot{x} + \gamma \dot{x} + kx = 0$. If the velocity at time $t$ is non-zero, $\dot{x}(t) \neq 0$, then the time derivative of the mechanical energy $E(t) = \frac{1}{2} m (\dot{x}(t))^2 + \frac{1}{2} k (x(t))^2$ is strictly negative: $$\frac{dE}{dt}(t) < 0.$$

The discriminant of a damped harmonic oscillator with mass $m$, spring constant $k$, and damping coefficient $\gamma$ is defined as $\gamma^2 - 4mk$. This value determines the damping regime (underdamped, critically damped, or overdamped) of the system.

A damped harmonic oscillator with mass $m$, spring constant $k$, and damping coefficient $\gamma$ is said to be **underdamped** if its discriminant $\gamma^2 - 4mk$ is strictly less than zero.

A damped harmonic oscillator with mass $m$, spring constant $k$, and damping coefficient $\gamma$ is said to be **critically damped** if its discriminant $\gamma^2 - 4mk$ is equal to zero.

A damped harmonic oscillator with mass $m$, spring constant $k$, and damping coefficient $\gamma$ is said to be **overdamped** if its discriminant $\gamma^2 - 4mk$ is strictly greater than zero.

For a damped harmonic oscillator with mass $m$, spring constant $k$, and natural angular frequency $\omega$, the spring constant is equal to the mass times the square of the angular frequency, i.e., $k = m\omega^2$.

For a damped harmonic oscillator with mass $m$, spring constant $k$, and decay rate $\beta$, if the system is critically damped (i.e., $\gamma^2 = 4mk$), then the spring constant satisfies the relation $k = m \beta^2$.

For a damped harmonic oscillator with mass $m$, spring constant $k$, and damping coefficient $\gamma$, if the system is overdamped (i.e., $\gamma^2 > 4mk$), then its natural angular frequency $\omega = \sqrt{k/m}$ is strictly less than its decay rate $\zeta = \frac{\gamma}{2m}$. That is, $\omega < \zeta$.

For a damped harmonic oscillator with mass $m$, spring constant $k$, and damping coefficient $\gamma$, if the system is underdamped (i.e., its discriminant $\Delta = \gamma^2 - 4mk < 0$), then its damped angular frequency $\omega$ is given by $$\omega = \frac{\sqrt{-\Delta}}{2m} = \frac{\sqrt{4mk - \gamma^2}}{2m}.$$

For a damped harmonic oscillator with mass $m$, spring constant $k$, and damping coefficient $\gamma$, if the system is critically damped (i.e., $\gamma^2 = 4mk$), then its angular frequency is zero.

For a damped harmonic oscillator $S$ that is overdamped, its angular frequency is given by $$\omega = \frac{\sqrt{\Delta}}{2m},$$ where $\Delta = \gamma^2 - 4mk$ is the discriminant and $m$ is the mass of the oscillator.

For an underdamped harmonic oscillator, the square of the damped angular frequency $\omega_d$ is equal to the square of the natural angular frequency $\omega$ minus the square of the decay rate $\beta$: $$\omega_d^2 = \omega^2 - \beta^2$$ where $\omega = \sqrt{k/m}$ and $\beta$ is the decay rate associated with the damping coefficient.

For a damped harmonic oscillator $S$ in the underdamped regime (where the mass $m$, spring constant $k$, and damping coefficient $\gamma$ satisfy the condition $\gamma^2 < 4mk$), the damped angular frequency $\omega_d$ is strictly positive, i.e., $\omega_d > 0$.

Consider a damped harmonic oscillator with mass $m$, spring constant $k$, and damping coefficient $\gamma$. If the oscillator is in the underdamped regime (i.e., its parameters satisfy $\gamma^2 < 4mk$), then its damped angular frequency $\Omega$ is non-zero, $\Omega \neq 0$.

For a damped harmonic oscillator $S$ in the overdamped regime (where $\gamma^2 > 4mk$), the square of the angular frequency $\Omega$ is equal to the square of the decay rate $\beta$ minus the square of the natural angular frequency $\omega$. That is: $$\Omega^2 = \beta^2 - \omega^2$$ where $\Omega$ is the angular frequency of the oscillator, $\beta$ is its decay rate, and $\omega = \sqrt{k/m}$ is its natural angular frequency.

For a damped harmonic oscillator $S$ with mass $m$, spring constant $k$, and damping coefficient $\gamma$, if the oscillator is overdamped (i.e., $\gamma^2 > 4mk$), then its associated angular frequency is strictly positive ($0 < \text{angularFrequency}$).

For a damped harmonic oscillator $S$ with mass $m$, spring constant $k$, and damping coefficient $\gamma$, if the system is overdamped (meaning $\gamma^2 > 4mk$), then its associated angular frequency $\Omega$ is non-zero ($\Omega \neq 0$).

Given a damped harmonic oscillator $S$ characterized by mass $m$, spring constant $k$, and damping coefficient $\gamma$, this function constructs the corresponding `HarmonicOscillator` (an undamped system) provided a proof $hS$ that the system is undamped (i.e., $\gamma = 0$). The resulting harmonic oscillator retains the same mass $m$ and spring constant $k$ as the original system.

Consider a damped harmonic oscillator $S$ with mass $m$, spring constant $k$, and damping coefficient $\gamma$. If $S$ is undamped (i.e., $\gamma = 0$), then for any smooth trajectory $x: \text{Time} \to \mathbb{R}^1$, $x$ satisfies the damped equation of motion $$m \ddot{x}(t) + \gamma \dot{x}(t) + k x(t) = 0$$ if and only if it satisfies the equation of motion for the corresponding undamped harmonic oscillator.

The Caldirola–Kanai Lagrangian for a damped harmonic oscillator with mass $m$, spring constant $k$, and damping coefficient $\gamma$ is a real-valued function of time $t$, position $x \in \mathbb{R}^1$, and velocity $v \in \mathbb{R}^1$. It is defined as the Lagrangian of the corresponding undamped harmonic oscillator multiplied by a time-dependent exponential factor: $$L(t, x, v) = e^{\frac{\gamma}{m} t} \left( \frac{1}{2} m \langle v, v \rangle - \frac{1}{2} k \langle x, x \rangle \right)$$ where $\langle \cdot, \cdot \rangle$ denotes the standard inner product on the 1-dimensional Euclidean space.

For a damped harmonic oscillator with mass $m$, spring constant $k$, and damping coefficient $\gamma$, the Lagrangian $L$ at time $t$, position $x \in \mathbb{R}^1$, and velocity $v \in \mathbb{R}^1$ is given by: $$L(t, x, v) = \exp\left(\frac{\gamma}{m} t\right) \left( \frac{1}{2} m \langle v, v \rangle - \frac{1}{2} k \langle x, x \rangle \right)$$ where $\langle \cdot, \cdot \rangle$ denotes the standard inner product on the 1-dimensional Euclidean space.

For a damped harmonic oscillator with mass $m$, spring constant $k$, and damping coefficient $\gamma$, the Caldirola–Kanai Lagrangian $L$ evaluated at time $t$, position $x(t) \in \mathbb{R}^1$, and velocity $\dot{x}(t) \in \mathbb{R}^1$ is equal to the product of an exponential factor $e^{\frac{\gamma}{m} t}$ and the difference between the kinetic energy $T(t)$ and potential energy $V(x(t))$: $$L(t, x(t), \dot{x}(t)) = e^{\frac{\gamma}{m} t} \left( T(t) - V(x(t)) \right)$$ where $T(t) = \frac{1}{2} m \langle \dot{x}(t), \dot{x}(t) \rangle$ and $V(x(t)) = \frac{1}{2} k \langle x(t), x(t) \rangle$.

Let $S$ be a damped harmonic oscillator with mass $m$, spring constant $k$, and damping coefficient $\gamma$. If $S$ is undamped (i.e., $\gamma = 0$), then its Caldirola–Kanai Lagrangian $L(t, x, v)$ is equal to the Lagrangian of the corresponding simple harmonic oscillator: $$L(t, x, v) = \frac{1}{2} m \langle v, v \rangle - \frac{1}{2} k \langle x, x \rangle,$$ where $t$ is time, $x \in \mathbb{R}^1$ is position, $v \in \mathbb{R}^1$ is velocity, and $\langle \cdot, \cdot \rangle$ denotes the standard inner product on 1-dimensional Euclidean space.

For a damped harmonic oscillator $S$ with mass $m$, spring constant $k$, and damping coefficient $\gamma$, the Lagrangian function $L(t, x, v)$ (defined as the Caldirola–Kanai Lagrangian $L(t, x, v) = e^{\frac{\gamma}{m} t} \left( \frac{1}{2} m \langle v, v \rangle - \frac{1}{2} k \langle x, x \rangle \right)$) is $C^n$ (continuously differentiable to order $n$) for any $n \in \mathbb{N} \cup \{\infty\}$ with respect to time $t$, position $x$, and velocity $v$.

For a damped harmonic oscillator with mass $m$, spring constant $k$, and damping coefficient $\gamma$, the gradient of the Caldirola–Kanai Lagrangian $L(t, x, v)$ with respect to the position $x \in \mathbb{R}^1$ at time $t$ and velocity $v$ is given by $$\nabla_x L(t, x, v) = -\left( e^{\frac{\gamma}{m} t} k \right) x.$$

For a damped harmonic oscillator with mass $m$ and damping coefficient $\gamma$, the gradient of the Caldirola–Kanai Lagrangian $L(t, x, v)$ with respect to the velocity $v$ at time $t$ and position $x$ is: $$\nabla_v L(t, x, v) = \left( m e^{\frac{\gamma}{m} t} \right) v$$ where $x$ and $v$ are vectors in a 1-dimensional Euclidean space $\mathbb{R}^1$.

Given a trajectory $x(t)$ mapping from time to a 1-dimensional Euclidean space $\mathbb{R}^1$, this function computes the variational gradient of the action functional $S[x] = \int L(t, x(t), \dot{x}(t)) \, dt$ for the damped harmonic oscillator. The action is defined using the Caldirola–Kanai Lagrangian: $$L(t, x, v) = e^{\frac{\gamma}{m} t} \left( \frac{1}{2} m \langle v, v \rangle - \frac{1}{2} k \langle x, x \rangle \right)$$ where $m$ is the mass, $k$ is the spring constant, and $\gamma$ is the damping coefficient. The result is a function of time corresponding to the functional derivative $\frac{\delta S}{\delta x}$ (the Euler–Lagrange expression).

For a damped harmonic oscillator with mass $m$, spring constant $k$, and damping coefficient $\gamma$, let $x : \text{Time} \to \mathbb{R}^1$ be an infinitely differentiable ($C^\infty$) trajectory. The variational gradient of the action, denoted as $\frac{\delta S}{\delta x}$, is equal to the Euler-Lagrange operator applied to the Caldirola–Kanai Lagrangian $L(t, x, v) = e^{\frac{\gamma}{m} t} \left( \frac{1}{2} m \langle v, v \rangle - \frac{1}{2} k \langle x, x \rangle \right)$. That is, $$\frac{\delta S}{\delta x} = \frac{\partial L}{\partial x} - \frac{d}{dt} \left( \frac{\partial L}{\partial \dot{x}} \right)$$

For a damped harmonic oscillator with mass $m$ and damping coefficient $\gamma$, let $x(t)$ be a smooth ($C^\infty$) trajectory in 1-dimensional Euclidean space. The variational gradient of the action functional $\frac{\delta S}{\delta x}$ based on the Caldirola–Kanai Lagrangian is given at each time $t$ by: $$\frac{\delta S}{\delta x}(t) = \exp\left(\frac{\gamma}{m} t\right) \left( F(t) - m \ddot{x}(t) \right)$$ where $F(t)$ is the total physical force (the sum of the restoring and damping forces) and $\ddot{x}(t)$ is the acceleration of the oscillator at time $t$.

For a damped harmonic oscillator with mass $m$, damping coefficient $\gamma$, and spring constant $k$, let $x: \text{Time} \to \mathbb{R}$ be an infinitely differentiable trajectory. Then $x$ satisfies the equation of motion $$m \ddot{x}(t) + \gamma \dot{x}(t) + k x(t) = 0$$ for all $t$ if and only if the variational gradient of the action $\frac{\delta S}{\delta x}$ (defined using the Caldirola–Kanai Lagrangian) is equal to zero.