I have a physical pendulum that, for small oscillations, can be modeled with the simple harmonic motion approach. In determining the motion equation, I need to figure out the amplitude: I know that the pendulum has length $L$ and is released from an initial angle of $\theta = 15.0°$. My intuition suggest me that the amplitude can be calculated as $A = L\,\sin\,\theta$, but the solutions book actually uses the formula $A = r\,\theta$ (where in this case $r = L$). But the last formula is used to calculate the arc of circumference, so I wonder if it is the right way to do it and why.
[Physics] Pendulum and simple harmonic motion
harmonic-oscillatoroscillators
Related Solutions
Your calculation of $\alpha$ is correct. You got two $\alpha$, as you should for second‐order homogeneous linear differential equations. The general solution of the equation is then a linear combination:
$$\theta = c_1 e^{-\alpha_1 t} + c_2 e^{-\alpha_2 t}$$
So from the first boundary condition you get: $$c_1+c_2 = \theta_0$$
And the second boundary condition gives: $$c_1 \alpha_1 + c_2 \alpha_2 = 0$$
Solving for $c_{1,2}$ you get: $$c_1 = \frac{\alpha_2\theta_0}{\alpha_2 - \alpha_1}$$ $$c_2 = \frac{\alpha_1\theta_0}{\alpha_1 - \alpha_2}$$
Regarding your questions. This should solve your first problem with the second boundary condition.
Why are there two solutions for alpha?
Because it is a second order linear ODE with constant coefficients.
Solving differential equations
Technically you did not assume a single solution. What you did was an Ansatz, i.e. you tried whether $A e^{-\alpha t}$ is a solution. You found that it is a solution for two different choices of $\alpha$. That means you found two solutions ($\theta_{1,2}$ to the differential equation). Because it is a linear homogeneous ODE, any linear combination of these solutions is also a solution. Note that this does include either $\theta_1$ or $\theta_2$ only solutions as one of the coefficients in the linear combination could be zero. Now you just need to proof (or ask a mathematician), that these are all the solutions of the equation (which they are), i.e any solution to this ODE can be written as linear combination of $\theta_{1,2}$.
Physical interpretation
The velocity is (for $\alpha_1\neq\alpha_2$):
$$\dot{\theta} = \theta_0 \frac{\alpha_1\alpha_2}{\alpha_1-\alpha_2}\left(e^{-\alpha_1 t} - e^{-\alpha_2 t}\right)$$
In the overdamped limit ($\frac{b}{2m}\gg\sqrt{\frac{g}{l}}$), $\alpha_1\gg\alpha_2$ and from the structure of the velocity in the equation above you see that for small times ($t\ll\alpha_1^{-1}$), the first exponential ($e^{-\alpha_1 t}$) governs the motion (acceleration), while in the opposite limit of large times ($t\gg\alpha_2^{-1}$), the second exponential governs the motion (deceleration).
I think you're worrying too much. This is the correct approach (I'm going to be slightly flippant, so don't take this first paragraph too seriously on a first reading :) ):
- Step 1: Understand the meaning of the Picard-Lindelöf Theorem;
- Step 2: Understand that, by assigning state variables to all but the highest order derivative, you can rework $\ddot x +\omega^2\,x=0$ into a vector version of the standard form $\dot{\mathbf{u}} = f(\mathbf{u})$ addressed by the PL theorem and that, in this case, the $f(\mathbf{u})$ fulfills the conditions of the PL theorem (it is Lipschitz continuous)
- Step 3: Choose your favorite method for finding a solution to the DE and boundary conditions - tricks you learn in differential equations 101, trial and error stuffing guesses in and seeing what happens ..... anything! .... and then GO FOR IT!
Okay, that's a bit flippant, but the point is that you know from basic theoretical considerations there must be a solution and, however you solve the equation, if you can find a solution that fits the equation and boundary conditions, you simply must have the correct and only solution no matter how you deduce it.
In particular, the above theoretical considerations hold whether the variables are real or complex, so if you find a solution using complex variables and they fit the real boundary conditions, then the solution must be the same as the one that is to be found by sticking with real variable notation. Indeed, one can define the notions of $\sin$ and $\cos$ through the solutions of $\ddot x +\omega^2\,x=0$ and they have to be equivalent to complex exponential solutions through the PL theorem considerations above. You can then think of this enforced equivalence as the reason for your own beautifully worded insight that you have worked out for yourself:
"So using sin/cos and even is essentially equivalent so long as you allow for complex constants to provide a conversion factor between the two."
Drop the word "essentially" and you've got it all sorted!
Actually, let's go back to the Step 2 in my "tongue in cheek" (but altogether theoretically sound) answer as it shows us how to unite all of these approaches and bring in physics nicely. Break the equation up into a coupled pair of first order equations by writing:
$$\dot{x} = \omega\,v;\, \dot{v} = -\omega\,x$$
and now we can write things succinctly as a matrix equation:
$$\dot{X} = -i\,\omega \, X;\quad i\stackrel{def}{=}\left(\begin{array}{cc}0&-1\\1&0\end{array}\right)\text{ and } X = \left(\begin{array}{c}x\\v\end{array}\right)\tag{1}$$
whose unique solution is the matrix equation $X = \exp(-i\,\omega\,t)\,X(0)$. Here $\exp$ is the matrix exponential. Note also, that as a real co-efficient matrix, $i^2=-\mathrm{id}$. Now, you may know that one perfectly good way to represent complex numbers is the following: the field $(\mathbb{C},\,+,\,\ast)$ is isomorphic to the commutative field of matrices of the form:
$$\left(\begin{array}{cc}x&-y\\y&x\end{array}\right);\quad x,\,y\in\mathbb{R}\tag{2}$$
together with matrix multiplication and addition. For matrices of this special form, matrix multiplication is commutative (although of course it is not generally so) and the isomorphism is exhibited by the bijection
$$z\in\mathbb{C}\;\leftrightarrow\,\left(\begin{array}{cc}\mathrm{Re}(z)&-\mathrm{Im}(z)\\\mathrm{Im}(z)&\mathrm{Re}(z)\end{array}\right)\tag{3}$$
So if, now, we let $Z$ be a $2\times2$ matrix of this form, then we we can solve (1) by mapping the state vector $X = \left(\begin{array}{c}x\\v\end{array}\right)$ bijectively to the $2\times 2$ matrix $Z = \left(\begin{array}{cc}x&-v\\v&x\end{array}\right)$, solving the equation $\dot{Z} = -i\,\omega\,Z$, i.e. $Z(t) = \exp(-i\,\omega\,t)\,Z(0)$, where $Z(0)$ is the $2\times 2$ matrix of the form (2) with the correct values of $x(0)$ and $v(0)$ that fulfill the boundary conditions, and then taking only the first column of the resulting $2\times 2$ matrix solution $Z(t)$ to get $X(t)$.
This is precisely equivalent to the complex notation method you have been using, as I hope you will see if you explore the above a little. The phase angles are encoded by the phase of the $2\times2$ matrix $Z$, thought of as a complex number by the isomorphism described above.
Moreover, there is some lovely physics here. Consider the square norm of the state vector $X$; it is $E = \frac{1}{2}\,\langle X,\,X\rangle = \frac{1}{2}(x^2 + v^2)$ and you can immediate deduce from (1) that
$$\dot{E} = \langle X,\,\dot{X}\rangle = X^T\,\dot{X} = -\omega\,X^T \,i\, X = 0\tag{4}$$
This has two interpretations. Firstly, $E$ is the total energy of the system, partitioned into potential energy $\frac{1}{2}\,x^2$ and kinetic $\frac{1}{2}\,v^2$. Secondly, (4) shows that the state vector, written as Cartesian components, follows the circle $x^2+v^2=2\,E$ and indeed this motion is uniform circular motion of $\omega$ radians per unit time. So that simple harmonic motion is the motion of any Cartesian component of uniform circular motion.
You could also solve the problem by beginning with (1), deducing (4) and then make the substition
$$x=\sqrt{2\,E}\,\cos(\theta(t));\quad\, v=\sqrt{2\,E}\,\sin(\theta(t))\tag{5}$$
which is validated by the conservation law $x^2+v^2=2\,E$ with $\dot{E}=0$. Then substitute $x$ back into the original SHM equation to deduce that
$$\theta(t) = \pm\omega\,t+\theta(0)\tag{6}$$
Best Answer
Why would you consider the amplitude to be equal to the horizontal displacement? There's also a vertical displacement, which is arguably more important, without vertical displacement there would be no oscillation, since the potential energy would remain the same. Or take for example a torsion pendulum:
You wouldn't describe the amplitude as $r\sin{\theta}$, because that would give an amplitude of zero when the pendulum swings 360 degrees. It makes more sense to say that the amplitude is equal to the length of the path the pendulum travels, i.e. the length of the arc.