Physlib.ClassicalMechanics.HarmonicOscillator.Basic

The spring constant $k$ of a classical harmonic oscillator is non-zero, that is, $k \neq 0$.

The mass $m$ of a classical harmonic oscillator is non-zero, that is, $m \neq 0$.

For a classical harmonic oscillator with mass $m$ and spring constant $k$, the angular frequency $\omega$ is defined as the square root of the ratio of the spring constant to the mass: $$\omega = \sqrt{\frac{k}{m}}$$

The angular frequency $\omega$ of a classical harmonic oscillator is strictly positive, that is, $\omega > 0$.

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

For a classical harmonic oscillator, the angular frequency $\omega$ is not equal to zero, i.e., $\omega \neq 0$.

For a classical harmonic oscillator with mass $m$, spring constant $k$, and angular frequency $\omega$, the inverse of the square of the angular frequency is equal to the ratio of the mass to the spring constant: $$\frac{1}{\omega^2} = \frac{m}{k}$$

For a classical harmonic oscillator with mass $m$, the kinetic energy of a trajectory $x(t)$ in a 1-dimensional Euclidean space is a function of time $t$ defined by: $$\frac{1}{2} m \langle \dot{x}(t), \dot{x}(t) \rangle$$ where $\dot{x}(t)$ denotes the time derivative of the position $x(t)$ and $\langle \cdot, \cdot \rangle$ is the real inner product. This is equivalent to the expression $\frac{1}{2} m \|\dot{x}(t)\|^2$.

The potential energy of a classical harmonic oscillator with spring constant $k$ at position $x \in \mathbb{R}^1$ is defined as $\frac{1}{2} k \langle x, x \rangle$, where $\langle \cdot, \cdot \rangle$ denotes the real inner product on the 1-dimensional Euclidean space.

For a classical harmonic oscillator $S$ and a trajectory $x(t)$ in a 1-dimensional Euclidean space, the total energy $E(t)$ at time $t$ is defined as the sum of the kinetic energy $T(t)$ and the potential energy $V(x(t))$: $$E(t) = T(t) + V(x(t))$$ where $T(t)$ is the kinetic energy of the trajectory and $V(x(t))$ is the potential energy at position $x(t)$.

For a classical harmonic oscillator with mass $m$ and a trajectory $x(t)$ in a 1-dimensional Euclidean space, the kinetic energy at time $t$ is given by $$T(t) = \frac{1}{2} m \langle \dot{x}(t), \dot{x}(t) \rangle$$ where $\dot{x}(t)$ denotes the time derivative of the position $x(t)$ and $\langle \cdot, \cdot \rangle$ represents the standard real inner product.

For a classical harmonic oscillator $S$ with spring constant $k$, the potential energy at a position $x \in \mathbb{R}^1$ is given by $V(x) = \frac{1}{2} k \langle x, x \rangle$, where $\langle \cdot, \cdot \rangle$ denotes the standard real inner product on the 1-dimensional Euclidean space.

For a classical harmonic oscillator $S$ and a trajectory $x(t)$ in a 1-dimensional Euclidean space, the total energy $E(t)$ at any time $t$ is the sum of the kinetic energy $T(t)$ of the trajectory and the potential energy $V(x(t))$ at position $x(t)$: $$E(t) = T(t) + V(x(t))$$

For a classical harmonic oscillator $S$ with mass $m$, if a trajectory $x(t)$ in 1-dimensional Euclidean space $\mathbb{R}^1$ is a smooth ($C^\infty$) function of time, then the kinetic energy $T(t) = \frac{1}{2} m \langle \dot{x}(t), \dot{x}(t) \rangle$ is differentiable with respect to $t$.

Consider a classical harmonic oscillator $S$ with spring constant $k$. Let $x(t) \in \mathbb{R}^1$ be a smooth ($C^\infty$) trajectory. Then the potential energy as a function of time, $V(x(t)) = \frac{1}{2} k \langle x(t), x(t) \rangle$, is differentiable with respect to $t$.

Consider a classical harmonic oscillator $S$. Let $x(t)$ be a trajectory in 1-dimensional Euclidean space that is a smooth ($C^\infty$) function of time. Then the total energy $E(t)$, defined as the sum of the kinetic energy and the potential energy, is differentiable with respect to time $t$.

Consider a classical harmonic oscillator $S$ with mass $m$. Let $x(t)$ be a smooth ($C^\infty$) trajectory in 1-dimensional Euclidean space. Then the time derivative of the kinetic energy $T(t)$ is given by: $$\frac{d}{dt} T(t) = \langle \dot{x}(t), m \ddot{x}(t) \rangle$$ where $\dot{x}(t)$ is the velocity (the first time derivative of $x(t)$), $\ddot{x}(t)$ is the acceleration (the second time derivative of $x(t)$), and $\langle \cdot, \cdot \rangle$ denotes the real inner product.

Let $S$ be a classical harmonic oscillator with spring constant $k$. For a smooth ($C^\infty$) trajectory $x(t)$ in 1-dimensional Euclidean space, the time derivative of the potential energy $V(x(t))$ is given by: $$\frac{d}{dt} V(x(t)) = \langle \dot{x}(t), k x(t) \rangle$$ where $\dot{x}(t)$ denotes the time derivative of the position (velocity) at time $t$, and $\langle \cdot, \cdot \rangle$ represents the standard inner product on the 1-dimensional Euclidean space.

Consider a classical harmonic oscillator $S$ with mass $m$ and spring constant $k$. For any smooth ($C^\infty$) trajectory $x(t)$ in 1-dimensional Euclidean space, the time derivative of the total energy $E(t)$ is given by: $$\frac{d}{dt} E(t) = \langle \dot{x}(t), m \ddot{x}(t) + k x(t) \rangle$$ where $\dot{x}(t)$ and $\ddot{x}(t)$ denote the first and second time derivatives of the position $x(t)$ (velocity and acceleration respectively), and $\langle \cdot, \cdot \rangle$ represents the standard real inner product.

For a classical harmonic oscillator with mass $m$ and spring constant $k$, the Lagrangian $L(t, x, v)$ 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 difference between the kinetic energy and the potential energy: $$L(t, x, v) = \frac{1}{2} m \langle v, v \rangle - \frac{1}{2} k \langle x, x \rangle$$ where $\langle \cdot, \cdot \rangle$ denotes the standard inner product on the 1-dimensional Euclidean space.

For a classical harmonic oscillator $S$ with mass $m$ and spring constant $k$, the Lagrangian $L(t, x, v)$ is the function of time $t$, position $x$, and velocity $v$ defined by: $$L(t, x, v) = \frac{1}{2} m \langle v, v \rangle - \frac{1}{2} k \langle x, x \rangle$$ where $x, v \in \mathbb{R}^1$ and $\langle \cdot, \cdot \rangle$ denotes the standard real inner product.

For a classical harmonic oscillator with mass $m$ and spring constant $k$, let $x(t)$ be a trajectory in 1-dimensional Euclidean space and $\dot{x}(t)$ be its time derivative. At any time $t$, the Lagrangian $L$ evaluated at position $x(t)$ and velocity $\dot{x}(t)$ is equal to the kinetic energy $T$ of the trajectory minus the potential energy $V$ of the position: $$L(t, x(t), \dot{x}(t)) = T(t) - V(x(t))$$

For any $n \in \mathbb{N} \cup \{\infty\}$, the Lagrangian $L(t, x, v)$ of the classical harmonic oscillator is $n$-times continuously differentiable ($C^n$) with respect to its arguments: time $t \in \text{Time}$, position $x \in \mathbb{R}^1$, and velocity $v \in \mathbb{R}^1$. This implies that the Lagrangian is a smooth ($C^\infty$) function of $(t, x, v)$.

In the 1-dimensional real Euclidean space $\mathbb{R}^1$ equipped with the standard inner product $\langle \cdot, \cdot \rangle$, let $\Phi: \mathbb{R}^1 \to (\mathbb{R}^1)^*$ be the Riesz representation isomorphism (the `toDual` map) which identifies a vector $v$ with the continuous linear functional $w \mapsto \langle v, w \rangle$. For any vector $x \in \mathbb{R}^1$, let $f_x$ be the continuous linear functional defined by $f_x(y) = \langle x, y \rangle$. Then the inverse of the Riesz isomorphism applied to $f_x$ returns the original vector $x$: $$\Phi^{-1}(f_x) = x$$

For any vector $x$ in the 1-dimensional real Euclidean space $\mathbb{R}^1$ (represented as `EuclideanSpace ℝ (Fin 1)`), the gradient of the function $y \mapsto \langle y, y \rangle$ (the squared $L_2$ norm) evaluated at $x$ is equal to $2x$.

Let $c$ be a real constant and $x$ be a vector in the 1-dimensional real Euclidean space $\mathbb{R}^1$. The gradient of the function $f(y) = c \langle y, y \rangle$ evaluated at $x$ is equal to $(2c)x$, where $\langle \cdot, \cdot \rangle$ denotes the standard inner product on $\mathbb{R}^1$.

For a classical harmonic oscillator with spring constant $k$, let $L(t, x, v)$ be the Lagrangian defined as the difference between kinetic and potential energy. For any time $t$, position $x \in \mathbb{R}^1$, and velocity $v \in \mathbb{R}^1$, the gradient of the Lagrangian with respect to the position $x$ is equal to $-kx$: $$\nabla_x L(t, x, v) = -k x$$ where $\mathbb{R}^1$ denotes the 1-dimensional real Euclidean space.

For a classical harmonic oscillator with mass $m$ and spring constant $k$, the gradient of the Lagrangian $L(t, x, v) = \frac{1}{2} m \langle v, v \rangle - \frac{1}{2} k \langle x, x \rangle$ with respect to the velocity $v$ is equal to $m v$. Here, $t$ denotes time, $x$ denotes the position in 1-dimensional Euclidean space $\mathbb{R}^1$, and $\langle \cdot, \cdot \rangle$ is the standard inner product.

Given a trajectory $x: \text{Time} \to \mathbb{R}^1$, this definition computes the variational gradient of the action functional $S[x] = \int L(t, x(t), \dot{x}(t)) \, dt$, where $L$ is the Lagrangian of the harmonic oscillator. This result is a function of time representing the Euler-Lagrange operator evaluated along the trajectory: $$t \mapsto \frac{\partial L}{\partial x} - \frac{d}{dt} \left( \frac{\partial L}{\partial \dot{x}} \right)$$ where the derivatives of the Lagrangian are evaluated at position $x(t)$ and velocity $\dot{x}(t)$.

For a classical harmonic oscillator $S$ and a smooth ($C^\infty$) trajectory $x: \text{Time} \to \mathbb{R}^1$, the variational gradient of the action functional $\frac{\delta S}{\delta x}$ is equal to the Euler-Lagrange operator applied to the Lagrangian $L$ of the harmonic oscillator. That is, $$\frac{\delta S}{\delta x}(t) = \frac{\partial L}{\partial x}(t, x(t), \dot{x}(t)) - \frac{d}{dt} \left( \frac{\partial L}{\partial \dot{x}}(t, x(t), \dot{x}(t)) \right)$$ where $L$ is the Lagrangian $L(t, x, v) = \frac{1}{2} m \|v\|^2 - \frac{1}{2} k \|x\|^2$, and $\dot{x}(t)$ denotes the time derivative of the trajectory.

For a trajectory $x: \text{Time} \to \mathbb{R}^1$ of a classical harmonic oscillator, the property `EquationOfMotion` is satisfied if the variational gradient of the action (the Euler-Lagrange operator) is zero for all $t$. This corresponds to the Euler-Lagrange equation: $$\frac{\delta S}{\delta x} = \frac{\partial L}{\partial x} - \frac{d}{dt} \left( \frac{\partial L}{\partial \dot{x}} \right) = 0$$ where $L$ is the Lagrangian of the harmonic oscillator, and the derivatives are evaluated along the trajectory $x(t)$.

For a classical harmonic oscillator $S$ and a trajectory $x: \text{Time} \to \mathbb{R}^1$, the property that $x$ satisfies the equation of motion is equivalent to the condition that the variational gradient of the action $\frac{\delta S}{\delta x}$ is identically zero for all $t$. Here, the variational gradient (the Euler-Lagrange operator) is defined as: $$\frac{\delta S}{\delta x} = \frac{\partial L}{\partial x} - \frac{d}{dt} \left( \frac{\partial L}{\partial \dot{x}} \right)$$ where $L$ is the Lagrangian of the harmonic oscillator.

For a classical harmonic oscillator $S$, the force at a position $x \in \mathbb{R}^1$ is defined as the negative gradient of the potential energy $U(x)$, denoted as $-\nabla U(x)$.

For a classical harmonic oscillator $S$ with spring constant $k$, the force at a position $x$ in the 1-dimensional Euclidean space $\mathbb{R}^1$ is given by $F(x) = -k x$.

For a classical harmonic oscillator $S$ with mass $m$, and a smooth ($C^\infty$) trajectory $x: \text{Time} \to \mathbb{R}^1$, the variational gradient of the action $\frac{\delta S}{\delta x}$ (also known as the Euler-Lagrange operator) at any time $t$ is equal to the force $F(x(t))$ minus the product of the mass $m$ and the second time derivative of the trajectory $\ddot{x}(t)$: $$\frac{\delta S}{\delta x}(t) = F(x(t)) - m \ddot{x}(t)$$ where $F(x) = -kx$ is the restoring force of the oscillator and $\ddot{x}(t)$ denotes the acceleration $\frac{d^2 x}{dt^2}$.

For a classical harmonic oscillator $S$ with mass $m$, let $x: \text{Time} \to \mathbb{R}^1$ be a smooth ($C^\infty$) trajectory. The trajectory satisfies the equation of motion (defined by the vanishing of the variational gradient of the action, or the Euler-Lagrange equation) if and only if it satisfies Newton's second law for all times $t$: $$m \ddot{x}(t) = F(x(t))$$ where $\ddot{x}(t)$ is the second time derivative of the position and $F(x(t))$ is the force acting on the oscillator at position $x(t)$.

For a classical harmonic oscillator $S$, let $x: \text{Time} \to \mathbb{R}^1$ be a smooth ($C^\infty$) trajectory. If $x(t)$ satisfies the equation of motion for the system, then the time derivative of the total energy $E(t)$ is zero for all $t$: $$\frac{d}{dt} E(t) = 0$$

For a classical harmonic oscillator $S$, let $x: \text{Time} \to \mathbb{R}^1$ be a smooth ($C^\infty$) trajectory. If $x(t)$ satisfies the equation of motion for the system, then for any time $t$, the total energy $E(t)$ is equal to the initial energy at time $0$: $$E(t) = E(0)$$ where $E(t)$ is the sum of the kinetic and potential energies of the oscillator at time $t$.

Given a classical harmonic oscillator with mass $m$, at a specific time $t$ and position $x \in \mathbb{R}^1$, this definition establishes a linear isomorphism between the velocity space and the canonical momentum space (both modeled as $\mathbb{R}^1$). The mapping is defined by the gradient of the Lagrangian with respect to velocity, $v \mapsto \nabla_v L(t, x, v)$, which for this system is $p = m v$. The inverse mapping is $p \mapsto \frac{1}{m} p$.

For a classical harmonic oscillator with mass $m$ and spring constant $k$, the canonical momentum $p$ at time $t$ and position $x \in \mathbb{R}^1$ corresponding to a velocity $v \in \mathbb{R}^1$ is given by the product of the mass and the velocity: $$p = mv$$ where $p, v, x \in \mathbb{R}^1$.

For a classical harmonic oscillator with mass $m$ and spring constant $k$, the Hamiltonian $H(t, p, x)$ is a real-valued function of time $t$, momentum $p \in \mathbb{R}^1$, and position $x \in \mathbb{R}^1$. It is defined via the Legendre transformation of the Lagrangian $L$ as: $$H(t, p, x) = \langle p, v \rangle - L(t, x, v)$$ where $\langle \cdot, \cdot \rangle$ denotes the standard inner product on the 1-dimensional Euclidean space, and the velocity $v$ is expressed as a function of momentum $p$ using the inverse of the canonical momentum mapping $v = \text{toCanonicalMomentum}^{-1}(p)$.

For a classical harmonic oscillator $S$ with mass $m$ and spring constant $k$, the Hamiltonian $H$ as a function of time $t$, momentum $p$, and position $x$ is given by: $$H(t, p, x) = \frac{1}{2m} \langle p, p \rangle + \frac{1}{2} k \langle x, x \rangle$$ where $p, x \in \mathbb{R}^1$ and $\langle \cdot, \cdot \rangle$ denotes the standard real inner product.

For a classical harmonic oscillator $S$, the Hamiltonian function $H(t, p, x)$, considered as a function of time $t$, momentum $p$, and position $x$, is $n$-times continuously differentiable over $\mathbb{R}$ for any $n \in \mathbb{N} \cup \{\infty\}$.

For a classical harmonic oscillator with spring constant $k$, the gradient of the Hamiltonian $H(t, p, x)$ with respect to the position $x$ is equal to $k x$, where $t$ is the time and $p$ is the momentum. Here, $x$ and $p$ are vectors in the 1-dimensional real Euclidean space $\mathbb{R}^1$.

For a classical harmonic oscillator with mass $m$, the gradient of the Hamiltonian $H(t, p, x)$ with respect to the momentum $p$ at a given time $t$ and position $x \in \mathbb{R}^1$ is equal to the momentum divided by the mass: $$\nabla_p H(t, p, x) = \frac{1}{m} p$$ where $p \in \mathbb{R}^1$ is the canonical momentum.

For a classical harmonic oscillator with mass $m$ and spring constant $k$, and for any trajectory $x(t)$ in a 1-dimensional Euclidean space $\mathbb{R}^1$, the Hamiltonian $H$ evaluated at time $t$, position $x(t)$, and the canonical momentum $p(t) = m \dot{x}(t)$ is equal to the total energy $E(t)$ of the system at that time. That is, $$H(t, p(t), x(t)) = E(t)$$ where $p(t)$ is the canonical momentum corresponding to the velocity $\dot{x}(t)$ and $E(t)$ is the sum of the kinetic and potential energies.

For a classical harmonic oscillator with Hamiltonian $H$, the Hamilton equation operator maps a momentum trajectory $p: \text{Time} \to \mathbb{R}^1$ and a position trajectory $q: \text{Time} \to \mathbb{R}^1$ to a function of time $t$ defined by the pair: $$\left( \dot{q}(t) - \nabla_p H(t, p(t), q(t)), \ -\dot{p}(t) - \nabla_q H(t, p(t), q(t)) \right)$$ In this expression, $\dot{q}$ and $\dot{p}$ denote the time derivatives of the position and momentum trajectories, while $\nabla_p H$ and $\nabla_q H$ denote the gradients of the oscillator's Hamiltonian with respect to momentum and position, respectively. The vanishing of this operator for a given pair $(p, q)$ signifies that the trajectories satisfy the classical Hamilton's equations.

For a classical harmonic oscillator $S$, let $x: \text{Time} \to \mathbb{R}^1$ be a smooth ($C^\infty$) trajectory. Let $p(t)$ be the canonical momentum trajectory corresponding to the velocity $\dot{x}(t)$ at position $x(t)$, defined by $p(t) = m \dot{x}(t)$. The trajectory $x$ satisfies the equation of motion (the Euler-Lagrange equation) if and only if the pair $(p, x)$ satisfies Hamilton's canonical equations, which is equivalent to the vanishing of the Hamilton equation operator: $$\text{hamiltonEqOp}(S, p, x) = 0$$

For a classical harmonic oscillator $S$ with mass $m$, let $x: \text{Time} \to \mathbb{R}^1$ be a smooth ($C^\infty$) trajectory. The following statements are equivalent: 1. The trajectory $x$ satisfies the equation of motion (the Euler-Lagrange equation). 2. The trajectory $x$ satisfies Newton's second law for all $t$: $m \ddot{x}(t) = F(x(t))$, where $\ddot{x}$ is the second time derivative of the position and $F$ is the force of the oscillator. 3. The phase-space trajectory $(p, x)$ satisfies Hamilton's equations (i.e., the Hamilton equation operator vanishes), where $p(t) = m \dot{x}(t)$ is the canonical momentum. 4. The variational gradient of the action functional $S[q] = \int L(t, q(t), \dot{q}(t)) \, dt$ is zero at $x$, where $L$ is the Lagrangian of the harmonic oscillator. 5. The variational gradient of the phase-space action functional $S[p, q] = \int \left( \langle p(t), \dot{q}(t) \rangle - H(t, p(t), q(t)) \right) dt$ is zero at the trajectory $(p, x)$, where $H$ is the Hamiltonian and $\langle \cdot, \cdot \rangle$ is the standard inner product.