The solution to SHM Differential Equation is
$y=A\cos wt + iB\sin wt$, upon applying this, where y would be angular displacement, how can an imaginary quantity pop up when applying this to the real world? for example, at time $t=\pi/4$, you get that the angular displacement is $0.5\sqrt2+i0.5\sqrt2$, how can the angular displacement be a complex number? How can you move $i$ radians around an axis?
[Physics] Simple harmonic motion differential equation and imaginary numbers
complex numbersdifferential equationsharmonic-oscillator
Related Solutions
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
$$\theta(t) = \pm\omega\,t+\theta(0)\tag{6}$$
In general they are simply solutions to different differential equations. The simple harmonic oscillator obeys the following differential equation:
$$\frac{d^2 y}{d t^2} = -ky,$$
where $k$ is a constant. The solution to this equation can indeed be written as a sinusoid with a given frequency and phase.
The wave equation is a differential equation given by:
$$\frac{\partial^2 y}{\partial x^2} - \frac{1}{c^2} \frac{\partial^2 y}{\partial t^2} = 0,$$
where the solution is arbitrary function of the form $y(x,t) = F(x-ct) + G(x+ct)$ for any two functions $G$ and $F$. This can be satisfied by a sinusoid, and hence the solution to the SHO differential equation is also a solution to the wave equation. However, not all solutions to the wave equation are simple harmonic oscillators. Fortunately, one can see via Fourier analysis that any function can be decomposed into a sum of simple harmonic oscillators.
Best Answer
These are not the only solutions. You don't have to express solution in this way, you could just say that the solution is cos or sin without the imaginary unit $i$. BUT this solution you just wrote is a valid one. Physics do not have some kind of a monopoly to solutions of differential equations. Physicaly meaningful quantities in this solution you wrote are totally ok, you just ignore the imaginary unit and that is all.
So if you want to talk about the physics you just take the real part of the solution. If you plug in any number into your solution you get a complex number back, in general. But the value or modulus of this number is a real number as you can compute it using Pythagoras theorem.
So to sum up, in physics you are using mathematics to model reality. When you write something like this, using complex numbers, its just because it is easy to do it this way. You have to have in mind that you are taking just the real part because of course, physical quantities can not be imaginary. So if you have your oscillator with the amplitude A and some frequeny $\omega$, you can express it in many ways. The same motion.
In this case, what do you think the amplitude of the motion is? You can not add up real and imaginary number directly. You have to find the amplitude in some other way. And as I said you have to find modulus of your complex number. So complex solution is the most general one and physicist chooses how to describe particular motion. Maybe this would help: http://galileo.phys.virginia.edu/classes/152.mf1i.spring02/ComplexNumbersSHO.htm
So of course you can ask yourself what would be the motion of the point in the complex plane described by your general solution?