Physlib.ClassicalMechanics.Pendulum.SimplePendulum.Basic

For a simple pendulum with mass $m$, the mass is non-zero, i.e., $m \neq 0$.

For a simple pendulum $S$, the length $\ell$ of the rod is non-zero, i.e., $S.\ell \neq 0$.

For a simple gravity pendulum, the gravitational acceleration $g$ is non-zero ($g \neq 0$).

For a simple pendulum with a rod of length $\ell$ and subject to a uniform gravitational acceleration $g$, the angular frequency $\omega$ is defined as $$\omega = \sqrt{\frac{g}{\ell}}$$ This quantity represents the angular frequency of the pendulum's small oscillations about the bottom of its swing.

For a simple pendulum, the angular frequency $\omega$ is strictly positive ($\omega > 0$).

The angular frequency $\omega$ of the simple pendulum is not equal to zero, i.e., $\omega \neq 0$.

For a simple pendulum with rod length $\ell$ and gravitational acceleration $g$, the square of its angular frequency $\omega$ is equal to the ratio of the gravitational acceleration to the rod length: $$\omega^2 = \frac{g}{\ell}$$

For a simple pendulum with rod length $\ell$ and gravitational acceleration $g$, the inverse of the square of its angular frequency $\omega$ is given by $$(\omega^2)^{-1} = \frac{\ell}{g}$$

For a simple gravity pendulum with mass $m$ and rod length $\ell$, its moment of inertia $I$ about the pivot is defined as $I = m \ell^2$.

For a simple gravity pendulum $S$, the moment of inertia $I$ about its pivot is strictly positive, i.e., $I > 0$.

For a simple gravity pendulum, the moment of inertia $I$ about the pivot is not equal to zero ($I \neq 0$).

For a simple gravity pendulum with mass $m$, rod length $\ell$, and gravitational acceleration $g$, let $\omega$ be the angular frequency and $I$ be the moment of inertia about the pivot. The product of the square of the angular frequency and the moment of inertia is equal to $m g \ell$: $$\omega^2 I = m g \ell$$

For a simple pendulum with moment of inertia $I$, given a trajectory $\theta : \text{Time} \to \mathbb{R}^1$ representing the angular position in a 1-dimensional Euclidean space, the kinetic energy at time $t$ is defined as: $$T(t) = \frac{1}{2} I \langle \dot{\theta}(t), \dot{\theta}(t) \rangle$$ where $\dot{\theta}(t)$ denotes the time derivative of the angle $\theta$ at time $t$, and $\langle \cdot, \cdot \rangle$ is the standard inner product on the 1-dimensional Euclidean space.

For a simple gravity pendulum with mass $m$, rod length $\ell$, and gravitational acceleration $g$, the potential energy at an angular position $x$ (where $x_0$ denotes the angle $\theta$ from the downward vertical) is given by $$V(\theta) = m g \ell (1 - \cos \theta)$$ This value represents the work done against gravity to raise the bob from the equilibrium position ($\theta = 0$), where the potential energy is normalized to vanish.

For a simple pendulum, given a trajectory $\theta : \text{Time} \to \mathbb{R}^1$ representing the angular position in a 1-dimensional Euclidean space, the total energy $E$ at time $t$ is defined as the sum of the kinetic energy $T$ and the potential energy $V$: $$E(t) = T(t) + V(t)$$ where $T(t)$ is the kinetic energy of the trajectory $\theta$ at time $t$ and $V(t)$ is the potential energy at the angular position $\theta(t)$.

For a simple pendulum with moment of inertia $I$, the kinetic energy $T$ of an angular trajectory $\theta: \text{Time} \to \mathbb{R}^1$ at any time $t$ is given by $$T(\theta)(t) = \frac{1}{2} I \langle \dot{\theta}(t), \dot{\theta}(t) \rangle$$ where $\dot{\theta}(t)$ denotes the time derivative of the angle at time $t$, and $\langle \cdot, \cdot \rangle$ is the standard inner product on the 1-dimensional Euclidean space.

For a simple pendulum with mass $m$, rod length $\ell$, and gravitational acceleration $g$, the potential energy at a configuration $x \in \mathbb{R}^1$ (where $x_0$ represents the angle $\theta$ from the downward vertical) is given by: $$V(x) = m g \ell (1 - \cos \theta)$$

For a simple pendulum, given a trajectory $\theta : \text{Time} \to \mathbb{R}^1$ representing the angular position in a 1-dimensional Euclidean space, the total energy $E$ at time $t$ is the sum of the kinetic energy $T$ and the potential energy $V$: $$E(\theta)(t) = T(\theta)(t) + V(\theta(t))$$ where $T(\theta)(t)$ is the kinetic energy of the trajectory $\theta$ at time $t$ and $V(\theta(t))$ is the potential energy at the angular position $\theta(t)$.

For a simple gravity pendulum with mass $m$, rod length $\ell$, and gravitational acceleration $g$, the potential energy $V(\theta)$ at any angular position $\theta$ is non-negative: $$V(\theta) \geq 0$$ This reflects the fact that the potential energy is normalized to vanish at the bottom of the swing ($\theta = 0$), which is the lowest point of the bob's trajectory.

For a simple gravity pendulum with mass $m$, rod length $\ell$, and gravitational acceleration $g$, the potential energy $V(\theta)$ at any angular position $\theta$ satisfies the inequality $$V(\theta) \leq 2 m g \ell$$ where $V(\theta) = m g \ell (1 - \cos \theta)$. This maximum value is attained at the top of the swing.

For a simple pendulum with angle $\theta$, the potential energy $V(\theta)$ is equal to $0$ if and only if $\cos \theta = 1$.

The potential energy $V(\theta) = m g \ell (1 - \cos \theta)$ of a simple pendulum is a smooth function of the angle $\theta$. Specifically, for any $n \in \mathbb{N} \cup \{\infty\}$, the potential energy function is $n$-times continuously differentiable.

The potential energy $V(\theta) = m g \ell (1 - \cos \theta)$ of a simple pendulum, where $m$ is the mass of the bob, $\ell$ is the length of the rod, $g$ is the gravitational acceleration, and $\theta$ is the angle from the downward vertical, is a differentiable function with respect to $\theta$.

For a simple gravity pendulum with mass $m$, rod length $\ell$, and gravitational acceleration $g$, let $x$ be a point in the one-dimensional Euclidean configuration space representing the angular position, where $x_0$ denotes the angle $\theta$ from the downward vertical. The gradient of the potential energy $V$ at $x$ is given by $$\nabla V(x) = (m g \ell \sin(x_0)) \mathbf{e}_0,$$ where $\mathbf{e}_0$ is the unit basis vector of the angular coordinate.

For a simple pendulum, let $\theta(t)$ be a smooth ($C^\infty$) trajectory representing the angular position as a function of time. Then the kinetic energy $T(t) = \frac{1}{2} I \|\dot{\theta}(t)\|^2$ is differentiable with respect to time $t$.

For a simple pendulum with mass $m$, rod length $\ell$, and gravitational acceleration $g$, let $\theta(t)$ be a smooth ($C^\infty$) function mapping time to the angular position in the configuration space. Then the potential energy as a function of time, $V(\theta(t)) = m g \ell (1 - \cos \theta(t))$, is differentiable with respect to time $t$.

For a simple pendulum $S$, let $\theta : \text{Time} \to \mathbb{R}^1$ be a smooth ($C^\infty$) trajectory representing the angular position as a function of time. Then the total energy $E(t) = T(t) + V(t)$ is differentiable with respect to time $t$.

Consider a simple pendulum with moment of inertia $I$. For a smooth trajectory $\theta : \text{Time} \to \mathbb{R}^1$ representing the angular position, the time derivative of the kinetic energy $T$ is given by the inner product of the angular velocity $\dot{\theta}$ and the rate of change of the angular momentum $I \ddot{\theta}$: $$\frac{d}{dt} T(t) = \langle \dot{\theta}(t), I \ddot{\theta}(t) \rangle$$ where $\dot{\theta}(t)$ and $\ddot{\theta}(t)$ denote the first and second time derivatives of $\theta$ at time $t$, respectively.

For a simple pendulum $S$, let $\theta : \text{Time} \to \mathbb{R}^1$ be a smooth function representing the angular position over time. The time derivative of the potential energy $V$ evaluated along the trajectory $\theta(t)$ is given by the inner product of the angular velocity $\partial_t \theta(t)$ and the gradient of the potential energy $\nabla V(\theta(t))$: $$\partial_t (V(\theta(t))) = \langle \partial_t \theta(t), \nabla V(\theta(t)) \rangle$$

For a simple pendulum with moment of inertia $I$ and potential energy $V$, let $\theta : \text{Time} \to \mathbb{R}^1$ be a smooth trajectory representing the angular position. The time derivative of the total energy $E = T + V$ is given by the inner product of the angular velocity $\dot{\theta}$ and the sum of the term $I \ddot{\theta}$ and the gradient of the potential energy $\nabla V$: $$\frac{d}{dt} E(t) = \langle \dot{\theta}(t), I \ddot{\theta}(t) + \nabla V(\theta(t)) \rangle$$ where $\dot{\theta}(t)$ and $\ddot{\theta}(t)$ denote the first and second time derivatives of the trajectory $\theta$ at time $t$, respectively.

The Lagrangian $L(t, x, v)$ of a simple gravity pendulum is a function of time $t$, angular position $x$, and angular velocity $v$, defined as the kinetic energy minus the potential energy: $$L(t, x, v) = \frac{1}{2} I \langle v, v \rangle - V(x)$$ where $I$ is the moment of inertia of the bob about the pivot, $\langle v, v \rangle$ is the squared norm of the angular velocity vector $v$ in the one-dimensional Euclidean space, and $V(x)$ is the potential energy at angular position $x$. The Lagrangian is autonomous, meaning it does not explicitly depend on the time parameter $t$.

The Lagrangian $L(t, x, v)$ of a simple pendulum with mass $m$, rod length $\ell$, gravitational acceleration $g$, and moment of inertia $I$, is given by: $$L(t, x, v) = \frac{1}{2} I \langle v, v \rangle - m g \ell (1 - \cos x_0)$$ where $x$ and $v$ are the angular position and velocity vectors in a one-dimensional Euclidean space, and $x_0$ denotes the single component of the position vector.

For a simple gravity pendulum, let $\theta: \text{Time} \to \mathbb{R}^1$ be a trajectory representing the lift of the angular position. At any time $t$, the Lagrangian $L$ evaluated at the position $\theta(t)$ and angular velocity $\dot{\theta}(t)$ is equal to the kinetic energy $T$ of the trajectory at that time minus the potential energy $V$ at that position: $$L(t, \theta(t), \dot{\theta}(t)) = T(\theta, t) - V(\theta(t))$$ where $\dot{\theta}(t)$ denotes the time derivative of $\theta$ at time $t$.

The Lagrangian $L(t, x, v)$ of a simple gravity pendulum, considered as a function of time $t \in \text{Time}$, angular position $x \in \mathbb{R}^1$, and angular velocity $v \in \mathbb{R}^1$, is $C^\infty$ smooth. Specifically, for any $n \in \{0, 1, \dots, \infty\}$, the uncurried mapping $(t, x, v) \mapsto L(t, x, v)$ is $n$-times continuously differentiable over $\mathbb{R}$.

For a simple gravity pendulum with mass $m$, rod length $\ell$, and gravitational acceleration $g$, let $L(t, x, v)$ be the Lagrangian. For any time $t$, angular position $x \in \mathbb{R}^1$, and angular velocity $v \in \mathbb{R}^1$, the gradient of the Lagrangian with respect to the angular position $x$ is given by $$\nabla_x L(t, x, v) = -(m g \ell \sin \theta) \mathbf{e}_0$$ where $\theta = x_0$ is the angular coordinate (the first component of the vector $x$) and $\mathbf{e}_0$ is the unit basis vector of the one-dimensional Euclidean space.

For a simple gravity pendulum with moment of inertia $I$, given a time $t$, an angular position $x$, and an angular velocity $v$ (both represented as vectors in a one-dimensional Euclidean space), the gradient of the Lagrangian $L$ with respect to the velocity $v$ is equal to the product of the moment of inertia and the velocity: $$\nabla_v L(t, x, v) = I v.$$ This quantity represents the angular momentum of the pendulum about its pivot.

For a simple gravity pendulum, the torque (or generalized force) $\tau$ at an angular configuration $x$ is defined as the negative gradient of the potential energy $V$ with respect to $x$: $$\tau(x) = -\nabla V(x)$$ where $x$ is represented as an element of a one-dimensional Euclidean space (the angle $\theta$).

For a simple gravity pendulum with mass $m$, rod length $\ell$, and gravitational acceleration $g$, the torque $\tau$ at an angular configuration $x$ (where $x_0$ is the angle $\theta$ from the downward vertical) is given by $$\tau(x) = -(m g \ell \sin x_0) \mathbf{e}_0$$ where $\mathbf{e}_0$ is the unit vector of the angular coordinate.

For a simple gravity pendulum with mass $m$, rod length $\ell$, and gravitational acceleration $g$, let $x$ be the configuration in the one-dimensional Euclidean space representing the angle $\theta = x_0$ from the downward vertical. The single component of the torque $\tau(x)$ is given by $$\tau(x)_0 = -m g \ell \sin \theta$$

Let $\theta : \text{Time} \to \mathbb{R}^1$ be a function representing the angle of a simple pendulum over time. The equation of motion is the condition that at every instant $t$, the product of the moment of inertia $I$ and the second time derivative of the angle $\ddot{\theta}(t)$ is equal to the torque $\tau(\theta(t))$ exerted by gravity: $$I \ddot{\theta}(t) = \tau(\theta(t))$$ where $\tau(\theta) = -m g \ell \sin \theta$ is the gravitational torque about the pivot.

For a simple gravity pendulum with moment of inertia $I$ and potential energy $V$, a function $\theta : \text{Time} \to \mathbb{R}^1$ representing the angle satisfies the equation of motion if and only if, at every instant $t$, the sum of the product of the moment of inertia and the second time derivative of the angle, $I \ddot{\theta}(t)$, and the gradient of the potential energy $\nabla V(\theta(t))$ vanishes: $$I \ddot{\theta}(t) + \nabla V(\theta(t)) = 0$$ This represents the rotational form of Newton's second law for the pendulum.

A function $\theta : \text{Time} \to \mathbb{R}^1$ representing the angle of a simple pendulum $S$ is a solution if it satisfies two conditions: 1. It is infinitely differentiable, i.e., $\theta \in C^\infty(\text{Time}, \mathbb{R}^1)$. 2. It satisfies the equation of motion $I \ddot{\theta}(t) = \tau(\theta(t))$ for all $t \in \text{Time}$, where $I$ is the moment of inertia and $\tau(\theta) = -m g \ell \sin \theta$ is the gravitational torque about the pivot.

Let $S$ be a simple pendulum. If a function $\theta : \text{Time} \to \mathbb{R}^1$ is a solution to the simple pendulum's equation of motion (as defined by `IsSolution`), then $\theta$ is infinitely differentiable, which is denoted as $\theta \in C^\infty(\text{Time}, \mathbb{R}^1)$.

Let $S$ be a simple pendulum with mass $m$, rod length $\ell$, and gravitational acceleration $g$. If a function $\theta : \text{Time} \to \mathbb{R}^1$ is a solution for $S$, then $\theta$ satisfies the equation of motion $$I \ddot{\theta}(t) = \tau(\theta(t))$$ at every instant $t$, where $I = m \ell^2$ is the moment of inertia of the bob about the pivot and $\tau(\theta) = -m g \ell \sin \theta$ is the gravitational torque.

Let $S$ be a simple pendulum with angular frequency $\omega = \sqrt{g/\ell}$. For a function $\theta : \text{Time} \to \mathbb{R}^1$ representing the angular configuration of the pendulum, the equation of motion (defined by the balance of angular momentum and gravitational torque, $I \ddot{\theta} = \tau(\theta)$) holds if and only if at every time $t$, the scalar equation $$\ddot{\theta}(t) + \omega^2 \sin(\theta(t)) = 0$$ is satisfied, where $\ddot{\theta}(t)$ is the second time derivative of the angle and $\theta(t)$ is the scalar component of the angular lift.

Consider two simple pendulums, $S$ and $S'$, and let $\omega_S$ and $\omega_{S'}$ be their respective angular frequencies (defined by $\omega = \sqrt{g/\ell}$). If $\omega_{S'} = \omega_S$, then for any function $\theta: \text{Time} \to \mathbb{R}^1$ representing the angle of the pendulum over time, $\theta$ satisfies the equation of motion for $S'$ if and only if it satisfies the equation of motion for $S$. This implies that the angular motion of a simple pendulum is determined solely by its angular frequency $\omega$, and is independent of its mass $m$.

For a simple pendulum $S$, let $\theta: \text{Time} \to \mathbb{R}^1$ be a function representing the angular position of the bob over time. The variational gradient is the function mapping $t \in \text{Time}$ to the functional derivative $\frac{\delta \mathcal{S}}{\delta \theta}(t)$ of the action integral $\mathcal{S}[\theta] = \int L(t, \theta(t), \dot{\theta}(t)) \, dt$, where $L$ is the Lagrangian of the pendulum. If the path $\theta$ is not sufficiently smooth for the gradient to exist, the function returns the zero function.

For a simple pendulum $S$ and a smooth trajectory of the angle $\theta: \text{Time} \to \mathbb{R}^1$, the variational gradient of the action $\frac{\delta \mathcal{S}}{\delta \theta}$ is equal to the Euler-Lagrange operator $\frac{\partial L}{\partial \theta} - \frac{d}{dt} \left( \frac{\partial L}{\partial \dot{\theta}} \right)$ applied to the pendulum's Lagrangian $L$.

For a simple pendulum with moment of inertia $I$ and torque function $\tau$, let $\theta: \text{Time} \to \mathbb{R}^1$ be an infinitely differentiable function representing the angular position over time. The variational derivative of the action $\frac{\delta \mathcal{S}}{\delta \theta}$ is given by the difference between the torque and the product of the moment of inertia and the angular acceleration: $$\left( \frac{\delta \mathcal{S}}{\delta \theta} \right)(t) = \tau(\theta(t)) - I \ddot{\theta}(t)$$ where $\ddot{\theta}$ denotes the second time derivative of $\theta$.

For a simple pendulum $S$ with moment of inertia $I$ and gravitational torque $\tau(\theta) = -m g \ell \sin \theta$, let $\theta: \text{Time} \to \mathbb{R}^1$ be an infinitely differentiable ($C^\infty$) trajectory of the pendulum's angle. Then $\theta$ satisfies the equation of motion $$I \ddot{\theta} = \tau(\theta)$$ if and only if the variational derivative of the action $\frac{\delta \mathcal{S}}{\delta \theta}$ vanishes: $$\frac{\delta \mathcal{S}}{\delta \theta} = 0.$$ This expresses that the smooth motions of the pendulum are precisely the critical points of its action.

For a simple gravity pendulum $S$, let $\theta : \text{Time} \to \mathbb{R}^1$ be a function representing the angular position over time. Then $\theta$ is a solution to the equation of motion if and only if $\theta$ is infinitely differentiable ($\theta \in C^\infty$) and the variational derivative of the action $\frac{\delta \mathcal{S}}{\delta \theta}$ is identically zero.

For a simple pendulum, let $\theta : \text{Time} \to \mathbb{R}^1$ be a smooth ($C^\infty$) trajectory representing the angular position. If $\theta$ satisfies the equation of motion $I \ddot{\theta} = \tau(\theta)$, then the time derivative of the total energy $E$ along this trajectory is zero: $$\partial_t (E(\theta)) = 0$$ where $E(\theta)$ is the sum of the kinetic and potential energies of the pendulum.

For a simple gravity pendulum $S$, let $\theta : \text{Time} \to \mathbb{R}^1$ be a smooth trajectory representing the angle of the bob over time. If $\theta$ satisfies the equation of motion $I \ddot{\theta} = \tau(\theta)$, then for any time $t$, the total energy $E(t)$ is equal to the initial energy $E(0)$.

For a simple pendulum $S$, let $\theta: \text{Time} \to \mathbb{R}^1$ be a smooth solution to the equation of motion. Then for any time $t$, the total energy $E$ of the pendulum at time $t$ is equal to its initial value at time $t = 0$: $$E(t) = E(0)$$ where $E(t)$ is defined as the sum of the kinetic and potential energies of the pendulum at time $t$.