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I solved the Lagrangian of a simple pendulum (with help from online examples as this concept is new to me) and ended with the following: $$\ddot{\theta}+\omega^2\theta=0$$ But in the example I was following, the next line written was this:

$$\theta(t) = A\cdot\sin(\omega t + \Phi)$$

And I don't know what steps they took to get there?

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    $\begingroup$ Did you try solving the differential equation? If not, here's a hint: take the ansatz $\theta(t) = Ae^{\lambda t}$, motivated by the fact that successive derivatives of the function yield a multiple of the original function. $\endgroup$ Dec 11, 2020 at 14:23

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Express this 2nd degree equation as \begin{equation} \ddot{\theta}\boldsymbol{+}\overbrace{\left(\rho_1\boldsymbol{+}\rho_2\right)}^{0}\dot{\theta}\boldsymbol{+}\omega^2\theta\boldsymbol{=}0 \tag{01}\label{01} \end{equation} or \begin{align} \left(\ddot{\theta}\boldsymbol{+}\rho_1\dot{\theta}\right)&\boldsymbol{+}\rho_2\left(\dot{\theta}\boldsymbol{+}\dfrac{\omega^2}{\rho_2}\theta\right)\boldsymbol{=}0 \tag{02a}\label{02a}\\ \rho_1&\boldsymbol{+}\rho_2\boldsymbol{=}0 \tag{02b}\label{02b} \end{align} If $\rho_1,\rho_2$ beyond satisfying equation \eqref{02b}, it would be possible to satisfy also \begin{equation} \dfrac{\omega^2}{\rho_2}\boldsymbol{=}\rho_1 \tag{03}\label{03} \end{equation} then equation \eqref{02a} is written \begin{equation} \left(\ddot{\theta}\boldsymbol{+}\rho_1\dot{\theta}\right)\boldsymbol{+}\rho_2\left(\dot{\theta}\boldsymbol{+}\rho_1\theta\right)\boldsymbol{=}0 \tag{04}\label{04} \end{equation} so defining the new variable \begin{equation} \psi\boldsymbol{\equiv}\dot{\theta}\boldsymbol{+}\rho_1\theta \tag{05}\label{05} \end{equation} we have at first the 1st degree equation \begin{align} \dot{\psi}\boldsymbol{+}\rho_2\psi & \boldsymbol{=}0 \tag{06a}\label{06a}\\ \texttt{with}\qquad\rho_1\boldsymbol{+}\rho_2 & \boldsymbol{=}0 \tag{06b}\label{06b}\\ \texttt{and}\qquad\qquad\rho_1\rho_2 & \boldsymbol{=}\omega^2 \tag{06c}\label{06c} \end{align} But then $\rho_1,\rho_2$ are the roots of equation \begin{equation} x^2\boldsymbol{+}\left(\rho_1\boldsymbol{+}\rho_2\right)x\boldsymbol{+}\rho_1\rho_2\boldsymbol{=}0 \tag{07}\label{07} \end{equation} that is of equation \begin{equation} x^2\boldsymbol{+}\omega^2\boldsymbol{=}0 \tag{08}\label{08} \end{equation} Hence \begin{equation} \rho_1\boldsymbol{=+}i\omega\,, \quad \rho_2\boldsymbol{=-}i\omega \tag{09}\label{09} \end{equation} ...continue solving firstly \eqref{06a} with respect to $\psi$ \begin{equation} \dot{\psi}\boldsymbol{-}i\omega\psi \boldsymbol{=}0 \tag{10}\label{10} \end{equation} and with the solution $\psi$ solve secondly \eqref{05} with respect to $\theta$ \begin{equation} \dot{\theta}\boldsymbol{+}i\omega\theta\boldsymbol{=}\psi \tag{11}\label{11} \end{equation}

$=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=$

From \eqref{10} \begin{equation} \psi\left(t\right) \boldsymbol{=}b e^{i\omega t}\,,\quad b\in\mathbb{C} \tag{12}\label{12} \end{equation} If \begin{equation} b \boldsymbol{=}B e^{i\phi}\,,\quad B,\phi\in\mathbb{R} \tag{13}\label{13} \end{equation} then \begin{equation} \psi\left(t\right) \boldsymbol{=}B e^{i\left(\omega t\boldsymbol{+}\phi\right)}\,,\quad B,\phi\in\mathbb{R} \tag{14}\label{14} \end{equation} Inserting this in \eqref{11} \begin{equation} \dot{\theta}\boldsymbol{+}i\omega\theta\boldsymbol{=}Be^{i\left(\omega t\boldsymbol{+}\phi\right)} \tag{15}\label{15} \end{equation} Since $\theta,\dot{\theta},\phi$ and $B \in \mathbb{R}$, equating real and imaginary parts of \eqref{15} we must have \begin{equation} \dot{\theta}\boldsymbol{=}B\cos\left(\omega t\boldsymbol{+}\phi\right)\quad \texttt{and} \quad \theta\boldsymbol{=}B\omega^{\boldsymbol{-}1}\sin\left(\omega t\boldsymbol{+}\phi\right) \tag{16}\label{16} \end{equation} Hence \begin{equation} \theta\left(t\right)\boldsymbol{=}A\sin\left(\omega t\boldsymbol{+}\phi\right)\,, \quad A,\phi \in \mathbb{R} \tag{17}\label{17} \end{equation} The real constants $A,\phi$ are determined from the initial conditions of the pendulum.

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  • $\begingroup$ What is the motivation for( 03 ) equation satisfication Frobenius? How to show that no other equation relating p1 and p2 would be possible to solve that new differential equation? $\endgroup$
    – Orion_Pax
    Mar 12, 2022 at 22:26
  • $\begingroup$ @Orion_Pax : The constant factors $\;\rho_1,\rho_2\;$ are used in a trick to convert a 2nd order differential equation to a 1srt order. The uniqueness of the solution \eqref{17} is not proved through the uniqueness of the pair $\;\left(\rho_1,\rho_2\right)$. $\endgroup$
    – Frobenius
    Mar 12, 2022 at 22:42
  • $\begingroup$ That means we still can have solutions to theta(double dot) = -w^2theta , which till now no one knows of ? Till now just the exponential was worked out using the trick method right ? $\endgroup$
    – Orion_Pax
    Mar 12, 2022 at 23:32
  • $\begingroup$ @Orion_Pax : In my previous comment I didn't write that there exist other solutions. The solution \eqref{17} is unique. $\endgroup$
    – Frobenius
    Mar 12, 2022 at 23:36
  • $\begingroup$ Understood . Is it because a first order differential will have only one solution ? $\endgroup$
    – Orion_Pax
    Mar 13, 2022 at 0:25
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This differential equation can be solved quite easily because we are aware of the relation $$\alpha=\omega\frac{d\omega}{d\theta}$$which is the relation between angular acceleration $\alpha$, angular velocity $\omega$ and angular displacement $\theta$ in a simple pendulum ,i.e, circular motion.(Note: $\ddot{\theta}=\alpha)$

Now,$$\alpha=-k^{2}\theta$$(or)(Note: k is the angular frequency which was taken as $\omega$ in the question) $$\omega\frac{d\omega}{d\theta}=-k^{2}\theta$$integrating on both sides, $$\int_{\omega_0}^\omega \omega d\omega=\int_0^\theta-k^{2}\theta d\theta$$ $${\omega}^{2}-{\omega_0}^{2}=-k^{2}{\theta}^{2}$$ $${\omega_0}^{2}-{\omega}^{2}=k^{2}{\theta}^{2}$$ Here, $\omega_0$ is the angular speed at the mean position ,i.e, the lowermost position which is taken as mean position and $\omega$ is the angular speed with angular displacement $\theta$ at an insant t. (Note: The max angular dislplacement is A which is the amplitude.). Also, ${\omega_0}^{2}=k^2A^{2}$ which can easily be derived by energy conservation between the mean position and amplitude position.

So,$$\omega=\sqrt{{\omega_0}^{2}-k^2{\theta}^{2}}$$ $$\omega=\sqrt{k^{2}A^{2}-k^2{\theta}^{2}}$$ $$\frac{d\theta}{dt}=k\sqrt{A^{2}-{\theta}^{2}}$$ Integrating on both sides, $$\int_0^\theta \frac{d\theta}{\sqrt{A^{2}-{\theta}^{2}}}=\int_0^tkdt$$ $$\sin^{-1}\left({\frac{\theta}{A}}\right)=kt$$ Which is, $$\theta=A\sin(kt)$$ If there exists a phase constant $\phi$ we can write the angular displacement $\theta$ as a function of time as,(Now, we rewrite k as $\omega$ foe angular frequency to get the equation asked in the quesion) $$\theta(t)=A\sin(\omega t + \phi)$$

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This is one of the standard differential equations, and its solution can indeed be used like that.

If you want to write a formal proof, then the simplest route is to:

  1. Show that the functions $\theta_1(t) = \sin(\omega t)$ and $\theta_2(t) = \cos(\omega t)$ are solutions of the differential equation, and that they are linearly independent of each other.
  2. Argue, from the theorem of existence and uniqueness of solutions of linear differential equations, that the equation is of second order in time and therefore admits exactly two linearly-independent solutions, so if $\theta(t)$ is a solution then it must be a linear combination of $\theta_1(t)$ and $\theta_2(t)$, i.e., $$\theta(t) = a\sin(\omega t) + b\cos(\omega t).\tag{$*$}$$
  3. Use the standard trigonometric identities to rephrase $(*)$ into the form in your example.

If you want more details, see any textbook on differential equations.

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As I recall, solving a differential equation often involves taking a guess and seeing if you can get it to work. In this case, you need a function (of θ) with a second derivative which is proportional to θ. A sine (or cosine) serves nicely (with the constants adjusted to match the initial conditions). By the way, with a simple pendulum, your θ should be a sin(θ) which leads to a much more complex solution. Your equation is valid only for small angles where sin(θ) is nearly equal to θ (in radians).

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  • $\begingroup$ Note: as mentioned in Emilio's answer, the $A \sin\omega t + B \cos\omega t$ that is obtained as an answer may be converted to $C\sin(\omega t + \phi)$ using standard trigonometric manipulations, where $C$ and $\phi$ depend on $A$ and $B$. $\endgroup$ Dec 11, 2020 at 16:49