Physlib.ClassicalMechanics.Pendulum.SimplePendulum.SmallAngle

For a simple gravity pendulum $S$ with mass $m$, rod length $\ell$, and gravitational acceleration $g$, let $I = m\ell^2$ be its moment of inertia. The harmonic oscillator associated with $S$ is defined as the oscillator with mass $m_{\text{osc}} = I$ and spring constant $k = mg\ell$.

For a simple gravity pendulum $S$, the mass $m$ of its associated harmonic oscillator is equal to the pendulum's moment of inertia $I$ about its pivot.

For a simple gravity pendulum $S$ with mass $m$, acceleration due to gravity $g$, and rod length $\ell$, the spring constant $k$ of its associated harmonic oscillator is given by $k = mg\ell$.

For a simple gravity pendulum with mass $m$, length $\ell$, and gravitational acceleration $g$, the angular frequency $\omega_{\text{osc}}$ of its associated harmonic oscillator (defined with mass $I = m\ell^2$ and spring constant $k = mg\ell$) is equal to the frequency of its small oscillations $\omega = \sqrt{g/\ell}$.

For a simple pendulum $S$ with angular frequency $\omega = \sqrt{g/\ell}$, a trajectory $\theta: \text{Time} \to \mathbb{R}^1$ satisfies the linearized equation of motion if, for all times $t$, it satisfies the differential equation: $$\ddot{\theta}(t) + \omega^2 \theta(t) = 0$$ where $\ddot{\theta}(t)$ denotes the second time derivative of the angular displacement $\theta$ at time $t$. This equation represents the small-angle approximation of the pendulum's motion, where the sine of the angle in the exact equation of motion is replaced by the angle itself.

For a simple gravity pendulum $S$, a trajectory $\theta : \text{Time} \to \mathbb{R}^1$ satisfies the linearized equation of motion $\ddot{\theta} + \omega^2 \theta = 0$ if and only if, for all times $t$, it satisfies the rotational form of Newton's second law: $$I \ddot{\theta}(t) = F_{\text{osc}}(\theta(t))$$ where $I$ is the moment of inertia of the pendulum, $\ddot{\theta}(t)$ is the second time derivative of the angular displacement $\theta$ at time $t$, and $F_{\text{osc}}$ is the force of the associated harmonic oscillator.

Consider a simple gravity pendulum $S$ with angular frequency $\omega = \sqrt{g/\ell}$. Let $\theta : \text{Time} \to \mathbb{R}^1$ be a smooth ($C^\infty$) trajectory. Then $\theta$ satisfies the linearized equation of motion $$\ddot{\theta}(t) + \omega^2 \theta(t) = 0$$ for all $t$ if and only if it satisfies the equation of motion of the associated harmonic oscillator (which has mass $I = m\ell^2$ and spring constant $k = mg\ell$), characterized by the vanishing of the variational gradient of its action.

For a simple gravity pendulum $S$, let $\tau$ be the torque function mapping an angle to its corresponding torque. The Fréchet derivative of the torque at the hanging equilibrium (where the angle is $0$), applied to an angular displacement vector $v \in \mathbb{R}^1$, is equal to the force $F_{\text{osc}}(v)$ of the associated harmonic oscillator at that same displacement: $$(D\tau)_0(v) = F_{\text{osc}}(v)$$ Substituting the definitions for the pendulum, this result implies $(D\tau)_0(v) = -m g \ell v$, where $m$ is the mass, $g$ is the gravitational acceleration, and $\ell$ is the length of the rod.

A trajectory $\theta: \text{Time} \to \mathbb{R}^1$ satisfies the linearized equation of motion for a simple pendulum $S$ if and only if, for all times $t$, it satisfies the relation: $$I \ddot{\theta}(t) = (D\tau)_0(\theta(t))$$ where $I$ is the moment of inertia of the pendulum, $\ddot{\theta}(t)$ is the second time derivative of the angular displacement, and $(D\tau)_0$ is the Fréchet derivative of the torque function $\tau$ evaluated at the hanging equilibrium ($\theta = 0$) and applied to the displacement $\theta(t)$.

For a simple gravity pendulum $S$ and initial conditions $IC$ (consisting of an initial angle $\theta_0$ and an initial angular velocity $\dot{\theta}_0$), the **small-angle trajectory** is the function mapping time $t$ to the angle $\theta(t)$ in 1D Euclidean space, defined as the trajectory of the harmonic oscillator associated with $S$: $$\theta(t) = \cos(\omega t) \theta_0 + \frac{\sin(\omega t)}{\omega} \dot{\theta}_0$$ where $\omega = \sqrt{g/\ell}$ is the angular frequency of the small oscillations.

For a simple gravity pendulum $S$ and any initial conditions $IC$ (consisting of an initial angle $\theta_0$ and an initial angular velocity $\dot{\theta}_0$), the small-angle trajectory $\theta(t)$ is a smooth function of time. Specifically, for any $n \in \mathbb{N} \cup \{\infty\}$, the mapping $t \mapsto \theta(t)$ is of class $C^n$ over $\mathbb{R}$.

Let $S$ be a simple gravity pendulum. For any initial conditions $IC$ consisting of an initial angle $\theta_0$ and an initial angular velocity $\dot{\theta}_0$, the small-angle trajectory $\theta(t)$ at time $t = 0$ is equal to the initial angle $\theta_0$: $$\theta(0) = \theta_0$$

For a simple gravity pendulum $S$ and initial conditions $IC$ (consisting of an initial angle $\theta_0$ and an initial angular velocity $\dot{\theta}_0$), the time derivative of the small-angle trajectory $\theta(t)$ evaluated at time $t = 0$ is equal to the initial angular velocity $\dot{\theta}_0$: $$\dot{\theta}(0) = \dot{\theta}_0$$

For a simple gravity pendulum with angular frequency $\omega$ and initial conditions consisting of an initial angle $\theta_0$ and an initial angular velocity $\dot{\theta}_0$, the small-angle trajectory $\theta(t)$ is given by the closed-form expression: $$\theta(t) = \cos(\omega t) \theta_0 + \frac{\sin(\omega t)}{\omega} \dot{\theta}_0$$ where $t$ denotes time.

For a simple gravity pendulum $S$ with angular frequency $\omega$, given initial conditions $IC$ (consisting of an initial angle $\theta_0$ and an initial angular velocity $\dot{\theta}_0$), the small-angle trajectory $\theta(t) = \cos(\omega t) \theta_0 + \frac{\sin(\omega t)}{\omega} \dot{\theta}_0$ satisfies the linearized equation of motion: $$\ddot{\theta}(t) + \omega^2 \theta(t) = 0$$ where $\ddot{\theta}$ denotes the second time derivative of the angular displacement.

Let $S$ be a simple pendulum with angular frequency $\omega = \sqrt{g/\ell}$. For any initial conditions $IC$ (comprising an initial angle $\theta_0$ and an initial angular velocity $\dot{\theta}_0$) and any smooth ($C^\infty$) trajectory $\theta : \text{Time} \to \mathbb{R}^1$, if $\theta$ satisfies the linearized equation of motion $\ddot{\theta} + \omega^2 \theta = 0$ with the initial conditions $\theta(0) = \theta_0$ and $\dot{\theta}(0) = \dot{\theta}_0$, then $\theta$ is identical to the small-angle trajectory $S.\text{smallAngleTrajectory}(IC)$.

Given an initial angle $\theta_0 \in \mathbb{R}$, the function `releasedFromRest` defines the trajectory of a simple pendulum's small-angle motion when released from rest. For a time $t$, the angular displacement $\theta(t)$ is given by: $$\theta(t) = \theta_0 \cos(\omega t)$$ where $\omega = \sqrt{g/\ell}$ is the angular frequency of the pendulum. The result is represented as a one-dimensional vector in $\mathbb{R}^1$ corresponding to the angle.

For a simple pendulum $S$ and an initial angle $\theta_0 \in \mathbb{R}$, the small-angle motion released from rest, $S.\text{releasedFromRest}(\theta_0)$, is identical to the small-angle trajectory $S.\text{smallAngleTrajectory}(IC)$ given the initial conditions $IC$ where the initial angular displacement is $\theta_0$ and the initial angular velocity is $0$.

For any initial angle $\theta_0 \in \mathbb{R}$, let $\theta(t)$ be the small-angle motion of a simple pendulum released from rest. At time $t = 0$, the angular displacement is given by $\theta(0) = \theta_0$. (Note: In the formal statement, the angle is represented as a one-dimensional vector $(\theta_0) \in \mathbb{R}^1$).

For a simple pendulum $S$ and any initial angle $\theta_0 \in \mathbb{R}$, the small-angle motion released from rest at $\theta_0$ is genuinely released from rest: its angular velocity at time $t = 0$ vanishes. Mathematically, this is expressed as $\partial_t (S.\text{releasedFromRest } \theta_0) (0) = 0$.

For a simple pendulum with angular frequency $\omega = \sqrt{g/\ell}$, the trajectory $\theta(t)$ resulting from releasing the pendulum from rest at an initial angle $\theta_0$ (given by $\theta(t) = \theta_0 \cos(\omega t)$) satisfies the linearized equation of motion: $$\ddot{\theta}(t) + \omega^2 \theta(t) = 0$$ where $\ddot{\theta}$ denotes the second time derivative of the angular displacement.

The small-angle period of a simple pendulum $S$ is defined as the period of its associated harmonic oscillator. It is given by the formula $T = 2\pi \sqrt{\frac{\ell}{g}}$, where $\ell$ is the length of the pendulum rod and $g$ is the acceleration due to gravity.

The small-angle period $T$ of a simple pendulum $S$ is given by the ratio of $2\pi$ to its angular frequency $\omega$: $$T = \frac{2\pi}{\omega}$$ where $\omega$ is the angular frequency of the pendulum's small oscillations (defined as $\sqrt{g/\ell}$).

For a simple pendulum $S$, the small-angle period $T$ is given by the formula $$T = 2\pi \sqrt{\frac{\ell}{g}}$$ where $\ell$ is the length of the pendulum rod, $g$ is the acceleration due to gravity, and $\pi$ is the mathematical constant pi.

For a simple pendulum $S$, the small-angle period $T$ is strictly positive ($T > 0$).

Given a simple pendulum $S$ with rod length $\ell$ and gravitational acceleration $g$, let $\theta(t)$ be the small-angle trajectory corresponding to the initial conditions $IC = (\theta_0, \dot{\theta}_0)$. Then the trajectory $\theta(t)$ is periodic with the small-angle period $T = 2\pi \sqrt{\frac{\ell}{g}}$, such that for all $t$, $\theta(t + T) = \theta(t)$.

For any initial angle $\theta_0 \in \mathbb{R}$, the trajectory $\theta(t)$ of a simple pendulum's small-angle motion released from rest is periodic with the small-angle period $T = 2\pi \sqrt{\ell/g}$. That is, $\theta(t + T) = \theta(t)$ for all time $t$. This implies that after each time interval of $T$, the motion returns to the initial angle $\theta_0$ with zero angular velocity.

Consider a simple gravity pendulum with mass $m$, rod length $\ell$, moment of inertia $I = m\ell^2$, and gravitational acceleration $g$. Given initial conditions $IC$ consisting of an initial angle $\theta_0$ and an initial angular velocity $\dot{\theta}_0$, let $\theta(t)$ be the small-angle trajectory. The total energy $E(t)$ of the associated harmonic oscillator along this trajectory is constant for all time $t$ and is given by the sum of the initial rotational kinetic energy and the linearized potential energy: $$E(t) = \frac{1}{2} \left( I \|\dot{\theta}_0\|^2 + m g \ell \|\theta_0\|^2 \right)$$

For a simple gravity pendulum with mass $m$, rod length $\ell$, and gravitational acceleration $g$, let $x \in \mathbb{R}^1$ be the angular displacement such that $\theta = x_0$. The difference between the torque $\tau(x)$ of the pendulum and the force $F_{\text{osc}}(x)$ of the associated harmonic oscillator is given by $$\tau(x) - F_{\text{osc}}(x) = m g \ell (\theta - \sin \theta) \mathbf{e}_0,$$ where $\mathbf{e}_0$ is the unit basis vector of the one-dimensional Euclidean space. This result expresses the exact term discarded by the small-angle linearization of the pendulum's dynamics.

For a simple pendulum with mass $m$, rod length $\ell$, and gravitational acceleration $g$, let $\tau(x)$ be the torque of the pendulum at an angular position $x \in \mathbb{R}^1$, and let $F_{\text{HO}}(x)$ be the force of the associated harmonic oscillator at the same position. The norm of the difference between the pendulum's torque and the force of its associated harmonic oscillator satisfies the cubic bound $$\|\tau(x) - F_{\text{HO}}(x)\| \leq \frac{m g \ell \|x\|^3}{6}.$$

For a simple pendulum with mass $m$, length $\ell$, and gravitational acceleration $g$, let $\tau(\theta)$ denote the torque at an angle $\theta$. The difference between the torque and its linear approximation $-m g \ell \theta$ is bounded by $$|\tau(\theta) + m g \ell \theta| \le \frac{m g \ell |\theta|^3}{6}.$$ This result demonstrates that the error of the small-angle approximation for the pendulum's torque is cubic in the angle.

For a simple gravity pendulum $S$ and a smooth ($C^\infty$) trajectory $\theta: \text{Time} \to \mathbb{R}^1$, let $\frac{\delta \mathcal{S}_P}{\delta \theta}$ and $\frac{\delta \mathcal{S}_{HO}}{\delta \theta}$ be the variational gradients of the action functionals for the pendulum and its associated harmonic oscillator, respectively. The difference between these variational gradients at any time $t$ is equal to the difference between the pendulum's torque $\tau(\theta(t))$ and the associated harmonic oscillator's force $F_{HO}(\theta(t))$: $$\frac{\delta \mathcal{S}_P}{\delta \theta}(t) - \frac{\delta \mathcal{S}_{HO}}{\delta \theta}(t) = \tau(\theta(t)) - F_{HO}(\theta(t)).$$ This equality holds because the inertial terms in the Euler-Lagrange equations for both systems are identical and thus cancel out.

For a simple pendulum with mass $m$, rod length $\ell$, and gravitational acceleration $g$, let $\theta: \text{Time} \to \mathbb{R}^1$ be a smooth ($C^\infty$) trajectory representing the angular position. Let $\frac{\delta \mathcal{S}_{\text{pend}}}{\delta \theta}(t)$ and $\frac{\delta \mathcal{S}_{\text{osc}}}{\delta \theta}(t)$ denote the variational gradients of the action functionals for the pendulum and its associated harmonic oscillator (defined with mass $I = m\ell^2$ and spring constant $k = mg\ell$) respectively. Then, at any time $t$, the norm of the difference between these variational gradients is bounded by the cubic term: $$\left\| \frac{\delta \mathcal{S}_{\text{pend}}}{\delta \theta}(t) - \frac{\delta \mathcal{S}_{\text{osc}}}{\delta \theta}(t) \right\| \leq \frac{m g \ell \|\theta(t)\|^3}{6}.$$ This implies that the two actions share the same critical-point equation to cubic accuracy in the angle $\theta$.

Consider a simple gravity pendulum $S$ with mass $m$, rod length $\ell$, and moment of inertia $I$. Let $\theta: \text{Time} \to \mathbb{R}^1$ be a trajectory that satisfies the linearized equation of motion $\ddot{\theta} + \omega^2 \theta = 0$, where $\omega = \sqrt{g/\ell}$ is the angular frequency. Then at any time $t$, the norm of the residual of the pendulum's actual equation of motion satisfies: $$\|I \ddot{\theta}(t) - \tau(\theta(t))\| \leq \frac{mg\ell \|\theta(t)\|^3}{6}$$ where $\tau(\theta(t))$ is the torque of the pendulum at angular displacement $\theta(t)$, and $\ddot{\theta}(t)$ denotes the second time derivative of the trajectory.

Consider a simple gravity pendulum $S$ with mass $m$, rod length $\ell$, and gravitational acceleration $g$. Let $\theta : \text{Time} \to \mathbb{R}^1$ be a smooth ($C^\infty$) trajectory representing the lift of the pendulum's angle. If $\theta$ satisfies the linearized equation of motion $\ddot{\theta} + \omega^2 \theta = 0$ (where $\omega$ is the natural frequency of small oscillations), then at any time $t$, the norm of the variational gradient of the pendulum's action, denoted by $\frac{\delta \mathcal{S}}{\delta \theta}(t)$, is bounded by a cubic term in the angle: $$\left\| \frac{\delta \mathcal{S}}{\delta \theta}(t) \right\| \leq \frac{mg\ell \|\theta(t)\|^3}{6}$$