harmonic-oscillatorharmonicshomework-and-exercisesoscillators
Im trying to get my head around SMH out of curiosity because it seems simple yet I'm not getting the concept behind some ideas.
For a SMH equation :
$$ x=a \sin(\omega t+\phi) $$
we know that :
$$ x'=-\omega \cos(\omega t + \phi) $$
$$ x''=-\omega^2 \sin(\omega t + \phi) $$
Also :
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 :) ):
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
$$\theta(t) = \pm\omega\,t+\theta(0)\tag{6}$$
Note that in the diagram below the velocity leads the displacement by $\dfrac \pi 2$ which is the same as the velocity lagging the displacement by $\dfrac{3\pi}{2}$.
So when mentioning phase it is important to state which two quantities are being compared, eg $A$ and $B$, and then whether there is a lead or lag between them, eg $A$ leads/lags $B$.
$A$ leading $B$ by $\phi$ is the same as $A$ lagging $B$ by $2\pi -\phi$.
Best Answer
The easiest way to determine maximum and minumums of a function is to set the derivative equal to zero. Thus, in this case, setting the equation for acceleration equal to zero and solving for the variables of interest will give you what you want. Thus, in this case we have the equation for position:
$$ x = a \sin(\omega t + \phi) $$
One way to see the maximum and minimum is to plot the function. I find that playing around with the constants (in this case $a, \omega,$ and $\phi$) helps you see what effect they have.
You can see that $a$ is an amplitude that determines how large the swings are from the origin. The termn $\omega$ tells you something about how quickly the position moves back and forth, thus it is known as the "angular frequency." Finally, $\phi$ tells you where you are when $t=0$ (or any other time for that matter) and is called the "phase."
For the position function, it is obvious that the value is maximized when the value of the sine function equals $1$ and, from what you learned in trig class, this happens when $\omega t + \phi = \pm \pi/2$ depending on the sign of $a$. Since $\omega$ and $\phi$ are constants set by the conditions and $t$ is the variable, the position is maximum when $t = (\pm \pi/2 - \phi)/\omega$, again depending on the sign of $a$.
ps- sorry for the curt answer the first time, but Wikipedia has a good article on this topic and there are tons of resources out there for it.