Some objects in nature exhibit a repetitive nature. A pendulum swings back and forth. A spinning top rotates over and over. Watch hands rotate too, but at a much slower pace. Planets orbit stars, and stars orbit in galaxies. Technically speaking, we need no new concepts to describe and understand this phenomena. We already have kinematics, dynamics, rotation and fields from the previous four articles. However, physicists are particularly lazy people and would like a more elegant and shorter form to describe repetitive motion. Two particularly helpful schemes will be covered here
The oscillation in the above graph repeats twice every second. Therefore we say its frequency is 2Hz and its period 0.5s. The frequency and period are the inverse of each other. The angular frequency is related to the frequency and period by
\[\omega = 2 \pi f = \frac{2 \pi}{T}\]
And the wave above can be described by the following sin equation
\[y = A \sin(\omega t)\]
Where \(A\) is the amplitude, or largest peak. Here the amplitude is 1. Since the frequency is 2Hz, \(4 \pi\). Thus this wave is perfectly described by \(y = \sin(4 \pi t)\)
However, consider that sin and cosine are related to one another. Specifically they are just shifted with respect to one another. Therefore, a more general solution to the wave shown in the graph could be given by
\[y = A \cos(\omega t + \phi)\]
Which, using trig relations, is the same as
\[y = A \cos(\omega t) \cos(\phi) - A \sin(\omega t) \sin(\phi)\]
If you have a free choice of \(A\) and \(\phi\) you can always find a solution of the following form (remember that \(\phi\) is just a constant)
\[y = C \cos(\omega t) + D \sin(\omega t)\]
This the following equations relate \(A\), \(C\), \(D\) and \(\phi\)
\[C = A \cos(\phi)\]
\[D = -A \sin(\phi)\]
\[A = \sqrt{C^2 + D^2}\]
\[\phi = \tan^{-1}\left(\frac{-D}{C}\right)\]
Now that we know the position of an object let us calculate the velocity and acceleration.
\[u = \frac{dy}{dt} = - \omega A \sin(\omega t + \phi)\]
\[a = \frac{d^2y}{dt^2} = -\omega^2 A \cos(\omega t + \phi) = - \omega^2 y\]
Notice how the acceleration is proportional to the displacement but in the opposite direction? This could be said to be the defining feature of an oscillation.
Also consider then that this means
\[u^2 + \omega^2 y^2 = \omega^2 A^2\]
If we multiplied this through by \(\frac{1}{2}m\) the first term would be the kinetic energy, the second term the potential energy and the final term the total energy. The total energy of an oscillation related to its angular frequency and amplitude. This makes sense intuitively. Imagine a rotating watch hand again. The faster it rotates the more energy we expect it to have. The heavier it is, the more energy we expect it to have.
Let's suppose that some object of mass 0.2kg is free to move in one direction and that movement is given by \(y(t)\). If the force on this object is in the same direction as the movement, it will continue to accelerate away from the start location and never be seen again. However, if the force always points towards some equilibrium point it will oscillate. Let's suppose this force is given by \(F = -ky\) and that \(k = 1.2 \frac{N}{m}\).
Let's do some analysis of this. First, let's look at the momentum
\[\mathbf{F} = \frac{d \mathbf{p}}{dt} = \frac{d (mu)}{dt} = m \frac{d^2 y}{dt^2}\]
\[m \frac{d^2 y}{dt^2} = -ky\]
\[\frac{d^2 y}{dt^2} = -\frac{k}{m}y\]
Compare this with \(\frac{d^2 y}{dt^2} = -\omega^2 y\), therefore
\[\omega^2 = \frac{k}{m}\]
\[\omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{1.2}{0.3}} = 2Hz\]
Let's do a unit analysis. If you have never seen this notation before , square brackets are used to mean "units of". So \([\omega]\) means "units of \(\omega\)". By convention, \([M]\) is mass units, \([L]\) is length units and \([T]\) is time units. Thus, these units could be anything, obviously we would prefer they be in SI! Newtons, \(N\), has units \([N] = [M][L][T]^{-2}\)
\[[k] = \frac{[M][L][T]^{-2}}{[L]} = [M][T]^{-2}\]
\[\left[ \frac{k}{m} \right] = \frac{[M]}{[M][T]^{2}} = [\omega^2] = \frac{1}{[T]^2} \]
So, the units of angular frequency are "per second", which we call a Hertz or Hz. Since \(a = - \omega^2 y\)
\[[a] = [T]^{-2}[L]\]
Which length per time squared, the correct units of acceleration. Consider frequently doing unit analysis on your answers, it will help you know you have got the right answer. I can't count how many times a unit analysis helped me spot a mistake in a long derivation!
Let's assume that this object is at \(y=2m\) at time \(t=0s\) and stationary. Let's calculate the \(C\) and \(D\) coefficients as well as the amplitude. We can use \(u^2 + \omega^2 y^2 = \omega^2 A^2\) but use \(u=0\) so \(y=A\), \(A = 2m\). Now, we can use \(y = C \cos(\omega t) + D \sin(\omega t)\) but our boundary conditions means it becomes \(2 = C\). Finally, \(A = \sqrt{C^2 + D^2}\), so \(D=0\).
We've so far covered the trigonometric representation of an oscillation. Let's cover the other form now: the complex number. \(y = A \cos(\omega t + \phi)\) is exactly equal to
\[y = \Re A e^{i(\omega t + \phi)} \]
Where \(\Re\) means "the real part of"
Let's suppose we have some oscillator that is described by the differential equation
\[m \ddot{x} + r \dot{x} + kx = 0\]
Let's assume a solution to this equation takes the form \(x = \Re (z)\), where \(z\) is some complex number \(Ae^{\alpha t}\) and \(A\) and \(\alpha\) are complex constants. Let's show that this solution holds if
\[(m \alpha^2 + r \alpha + k)z = 0\]
We start with derivatives
\[x = \Re{z} = \Re{A e^{\alpha t}}\]
\[\dot{x} = \Re{A \alpha e^{\alpha t}}\]
\[\ddot{x} = \Re{A \alpha^2 e^{\alpha t}}\]
So,
\[-kx - r \dot{x} = m \ddot{x}\]
\[-k \Re(A e^{\alpha t}) - r \Re(A \alpha e^{\alpha t}) = m \Re(A \alpha^2 e^{\alpha t})\]
Which rearranges to
\[\Re((k + r \alpha + m \alpha ^2)z) = 0\]
And our solution will suffice if \((m \alpha^2 + r \alpha + k)z = 0\) is enforced.
A trivial solution is obviously \(z=0\), but we want to find the more interesting solutions (that's just the kind of people physicists are. At any rate, \(z=0\) indicates that nothing is happening i.e. no movement, no oscillations). So, we need to find the values of \(\alpha\) for which the bracket is 0.
\[m \alpha^2 + r \alpha + k = 0\]
We can use the simple quadratic equation here to help us solve this
\[\alpha = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]
\[a=m, b=r, c=k\]
\[\alpha_1 = \frac{-r + \sqrt{r^2 - 4mk}}{2m}\]
\[\alpha_2 = \frac{-r - \sqrt{r^2 - 4mk}}{2m}\]
If \(r^2 = 4mk\) then the system is critically damped and the solution can be written in the form
\[z = e^\frac{-rt}{2m} (A_1 + A_2 t)\]
Let's show that this solution satisfies the differential equation.
\[Z = A_1 e^{-rt/2m} + A_2 t e^{-rt/2m}\]
This section will get quite messy, so I will redefine \(e^{-rt/2m} = E\). You're always allowed to group certain symbols together and define them as some new symbol, but remember that this new \(E\) stands for something else! This is especially important in the differentiations we are about to do
\[Z = A_1 E + A_2 t E\]
\[\dot{Z} = -A_1 \frac{r}{2m} E + A_2 E - A_2 \frac{r}{2m} t E\]
\[\ddot{Z} = A_1 \frac{r^2}{4m^2}E - A_2 \frac{r}{2m}E - A_2 \frac{r}{2m}E + A_2 \frac{r^2}{4m^2} t E\]
Now we need to substitute these three equations into \(m \alpha^2 + r \alpha + k = 0\) and check that everything cancels
\[ A_1 \frac{r^2}{4m} E - A_2 \frac{r}{2} E - A_2 \frac{r}{2} E + A_2 \frac{r^2}{4m} t E - A_1 \frac{r^2}{2m} E + A_2 r E - A_2 \frac{r^2}{4m} t E + A_1 E k + A_2 ktE = 0 \]
\[ -A_1 \frac{r^2}{4m}E - A_2 \frac{r^2}{4m} t E + A_1 E k + A_2 ktE = 0 \]
But \(r^2 = 4mk\)
\[ -A_1 k E + A_1 k E - A_2 r E + A_2 r E - A_2 k t E + A_2 ktE = 0 \]
Which does indeed totally cancel out.
If \(r^2 - 4mk > 0\), then the system is overdamped. Both \(\alpha_1\) and \(\alpha_2\) are real and negative and the solution can be written in the form
\[z = A_1 e^{\alpha_1 t} + A_2 e^{alpha_2 t}\]
Try demonstrating that this solution works, in the same way I did for the critically damped solution.
If \(r^2 - 4mk < 0\) then we have an underdamped situation.
Try showing that the solutions are now
\[ -\frac{r}{2m} \pm i \sqrt{\frac{k}{m} - \frac{r^2}{4m^2}} \]
Let's dig a little more into how waves work - we'll go over the wave equation itself. Let's start with a wave in the form
\[ y = A \cos(\omega t - kx + \phi) = \Re A e^{i\phi} e^{i(\omega t - kx)} \]
Let's first consider \(\frac{\partial y}{\partial x}\)
\[ \frac{\partial y}{\partial x} = Ak \sin(\omega t - kx + \phi) = - \Re i k A e^{i\phi} e^{i(\omega t - kx)} \]
In this section I will do every operation twice, once for the trig representation and another for the imaginary representation
So this is telling us about the behaviour of the wave when we move along its length. Imagine a wave on a string, and you freeze time. Then you move along the x axis, \(\frac{dy}{dx}\) is telling you about this. In other words, you're getting the momentum of each tiny piece of rope.
Now let's take the second derivative with respect to \(x\)
\[ \frac{\partial^2 y}{\partial x^2} = -k^2 A \cos(\omega t -kx + \phi) = -k^2 y \]
\[ \frac{\partial^2 y}{\partial x^2} = -k^2 \Re Ae^{i\phi} e^{i(\omega t -kx)} = -k^2 y \]
So the force acting on each little piece of the rope is inversely proportional to its height. This makes sense, when a part of the rope goes above \(y=0\) it needs a restoring force back to \(y=0\) and vice versa. We've come across \(k\) before, in springs. In many ways this constant is analogous. It certainly is a constant that describes the same kind of restoring force.
Now, let's do another partial differentiation - this time \(\frac{\partial y}{\partial t}\)
\[ \frac{\partial y}{\partial t} = - \omega A \sin(\omega t - kx + \phi) = \omega i \Re A e^{i\phi} e^{i(\omega t - kx)} \]
Now, this situation is like taking a tiny piece of the rope, ignoring the rest of the rope, and then watching it move up and down in time. So, \(\frac{\partial y}{\partial t}\) is telling you about the velocity of each tiny piece of rope.
Consider now the second derivative with respect to time
\[ \frac{\partial^2 y}{\partial t^2} = -\omega^2 A \cos(\omega t - kx + \phi) = -\omega^2 y \]
\[ \frac{\partial^2 y}{\partial t^2} = -\omega^2 \Re A e^{i\phi} e^{i(\omega t -kx)} = -\omega^2 y \]
So this is telling us about the acceleration of each tiny piece of rope. Again, this shouldn't be a surprise to us by now - we've come across acceleration as related to the angular frequency.