Week 15

Posted on December 8, 2016 by jhubisz

This week, we have focused on boundary value problems. This comes up when you have two regions in which waves obey the usual differential equation for waves, but where the velocity of the waves in the two regions is different:

In region 1, you might have (for a displacement $y_1(x,t)$ )
$\frac{\partial^2 y_1}{\partial x^2} = \frac{1}{v_1^2} \frac{\partial^2 y_1}{\partial t^2}$ ,
and then in region II, (for a displacement $y_2(x,t)$ :
$\frac{\partial^2 y_2}{\partial x^2} = \frac{1}{v_2^2} \frac{\partial^2 y_2}{\partial t^2}$ .

The example we are discussing in the lectures is the example where you have two strings of different mass per unit length, attached to each other (let’s say, at $x=0$ ).

The question at hand is how wave propagation is affected by this “defect” in the medium – a place where the medium changes from one type to another type in a discontinuous manner.

Knowing just the above two wave equations is not quite enough to answer this question. We also need to impose boundary conditions on the displacement functions. The two boundary conditions that we impose (in words) are:

The displacement of the string must be the same on either side of the point where the medium changes (the strings stay tied together)
There must not be a finite force on an infinitesimal mass (otherwise there would be an infinite acceleration)

In equations, these two conditions are:

$y_1 (x=0,t) = y_2(x=0,t)$ $\frac{\partial y_1}{\partial x} (x=0,t) =\frac{\partial y_2}{\partial x} (x=0,t)$

The first equation is easy to understand – the second comes from the fact that the force on a segment of string is proportional to the slope of the string on either side of that segment. If the slopes are not equal, there will be a finite force on a vanishingly small mass.

Applying these conditions gives enough information to solve the wave equation in these joined regions. Note that the conditions themselves may be different for different types of waves! For example, in the text, in one of the HW problems, the second condition is different mathematically (although the physical motivation given in words above is still the same).

Consider, for example, sending in a wave pulse described by the function
$f_1 (t-x/v_1)$

In this scenario, there will be both a wave that is transmitted through the boundary where the media change, and a wave that is reflected back from the discontinuity. The reflected wave is also a pulse, but moves to the left:
$g_1 (t+x/v_1)$ .
There is also the transmitted wave, which moves to the right:
$f_2 (t-x/v_2)$ .
In region 1, the full solution is the superposition of the incident and reflected wave:
$y_1 (x,t) = f_1(t-x/v_1) +g_1(t+x/v_1)$
while the solution is region 2 is just the transmitted wave:
$y_2(x,t) = f_2 (t-x/v_2)$ .

The question at hand can now be phrased as the following: Using the boundary conditions, can we solve for $g_1$ and $f_2$ given $f_1$ ? That is, for an arbitrary incident wave, can we extract both the reflected and transmitted wave? The answer is yes. We have second order equations, and two boundary conditions, which is enough to specify the output wave given our input incident wave.

Week 11

Posted on November 8, 2016 by jhubisz

Today we talked about some specific examples using Fourier Analysis. Much of the discussion related to the following Mathematica Notebook (which I, as always, suggest that you explore in depth!)

https://jhubisz.expressions.syr.edu/phy360/wp-content/uploads/sites/5/2016/11/Week-11.nb

Week 10

Posted on November 4, 2016 by jhubisz

This week we began our study of Fourier Analysis and Fourier Series. Our system of study for this discussion will be a string on which the wave velocity is $v$ , and which has fixed endpoints at x=0 and x=L where x is a position along the length of the string.

Recall that the wave equation is

$\frac{\partial^2 y(x,t)}{\partial x^2} = \frac{1}{v^2} \frac{\partial^2 y(x,t)}{\partial t^2}$

and that we can write any solution to this wave equation as a superposition of normal modes:

$y(x,t) = \sum_{n=1}^{\infty} A_n \cos \left[ \omega_n t - \delta_n \right] \sin \left[ \frac{n \pi x}{L} \right]$

where $\omega_n = \frac{n \pi v}{L}$ .

We would like to go from a set of initial conditions (position and velocity of every segment of the string) to the values of each $A_n$ and $\delta_n$ . We were able to do this for systems with a finite number of degrees of freedom, so we should, in principle, be able to do it in the case of systems with a very large number of degrees of freedom.

Let us first consider just a set of positions at time $t =0$. Let’s say we are given these, and they take the form $y(x,t=0) = f(x)$ , where $f(x)$ is any function with $f(0)=f(L)=0$ . These are the boundary conditions obeyed by our segment of string.

If we now apply this to the expression for y in terms of the normal modes, we find

$y(x,t=0) = f(x) = \sum_{n=1}^{\infty} A_n \cos (\delta_n) \sin \left[ \frac{n \pi x}{L} \right]$

Now for the moment, let us combine the coefficients of each $\sin$ into a set of coefficients $B_n = A_n \cos \delta_n$ , in which case we then have

$f(x) = \sum_{n=1}^{\infty} B_n \sin \left[ \frac{n \pi x}{L} \right]$

Let us now try to get the $B$ coefficients from the function $f(x)$ .

To do this, we use a trick involving integrals of sin functions:

$\frac{2}{L} \int_0^L \sin \left[ \frac{n \pi x}{L} \right] \sin \left[ \frac{m \pi x}{L} \right] dx = \left\{ \begin{array}{ll} 1 & n = m \\ 0 & n \ne m \end{array} \right.$

multiplying both sides of the equation for f(x) by $\frac{2}{L} \sin \left[ \frac{m \pi x}{L} \right]$ and integrating over $x$ from $0$ to $L$, we find

$\frac{2}{L} \int_0^L f(x) \sin \left[ \frac{m \pi x}{L} \right] = \sum_{n=1}^{\infty} B_n \cdot \frac{2}{L} \sin\left[ \frac{n \pi x}{L} \right] \sin \left[ \frac{m \pi x}{L} \right] dx = B_m$

This means that if we know the function f(x), and perform some integrals, then we can get the $B_m$ coefficients!

See this weeks Mathematica code to begin to get an idea of how this works:

https://jhubisz.expressions.syr.edu/phy360/wp-content/uploads/sites/5/2016/11/Week-10.nb

We can then go a bit further. Again, the goal is to get both the $A_n$ and $\delta_n$ . In order to do this, we need not just initial position information, but also initial velocity information. Right now, we have $B_n = A_n \cos \delta_n$ , so we only have fixed a combination of $A_n$ and $\delta_n$ . Let us say we have that velocity information as a given:

$\left. \frac{\partial y(x,t)}{\partial t} \right|_{t=0} = g(x) = \sum_{n=1}^{\infty} A_n \omega_n \sin \delta_n \sin \left[ \frac{n \pi x}{L} \right]$ .

In this case, we can write $C_n = A_n \omega_n \sin \delta_n$ , and thus we can also express the velocity of each point of the string in terms of a sum over sin functions:

$g(x) = \sum_{n=1}^\infty C_n \sin \left[ \frac{n \pi x}{L} \right]$

We can then use the same trick to get:

$C_n = \frac{2}{L} \int_0^L g(x) \sin \left[ \frac{n \pi x}{L} \right]$

So now you have all of the B’s, and all of the C’s (since we presumed we know both initial position, f(x), and initial velocity, g(x)). We then have

$B_n = A_n \cos \delta_n \text{ and } C_n = A_n \omega_n \sin \delta_n$

From this, we can finally get both $A_n$ and $\delta_n$ :

$A_n = \left[ B_n^2 + \left( \frac{C_n}{\omega_n} \right)^2 \right]^{1/2} \text{ and } \delta_n = \arctan \left[ \frac{ C_n/\omega_n }{B_n} \right]$ .

I will soon be uploading a mathematica notebook that goes through an example of a particular set of initial positions and velocities, and go through how to get the A’s and $\delta$ ‘s for a specific example.

Week 9

Posted on November 3, 2016 by jhubisz

During this week, we derived the wave equation by taking the continuum limit of the discrete system (where there were a finite number of masses attached to a string). We started from the equation of motion for a single mass, which we derived last week:

$\ddot{y}_p = - \frac{T}{l m} \left[ 2y_p - y_{p-1} - y_{p+1} \right]$

which we can we write as

$\ddot{y}_p = - \frac{T l}{m} \frac{2 y_p - y_{p-1}-y_{p+1}}{l^2}$

then we use the fact that the second derivative of a function can be approximated by

$f''(x) \approx \frac{f(x+\epsilon) + f(x-\epsilon) - 2 f(x)}{\epsilon^2}$

this allows us to re-write the equation of motion as

$\frac{\partial^2 y}{\partial t^2} = \frac{T}{\mu} \frac{\partial^2 y}{\partial x^2}$

or, in terms of a velocity:

$\frac{\partial^2 y}{\partial x^2} = \frac{1}{v^2} \frac{\partial^2 y}{\partial t^2}$

So we now have the wave equation! This is our new favorite equation, and it is important to note that we obtained it from the behavior of a large number of coupled simple harmonic oscillators – that means that there is nothing intrinsically new here, but we have found a convenient formalism for dealing with systems with a large (essentially infinite) numbers of degrees of freedom.

Let us think about a “case study” of a string fixed at both ends (x = 0 and x = L).

We can now go about solving for the normal modes of oscillation. It is not difficult to check that the following is a solution:

$y_n (x,t) = A_n \cos (\omega_n t - \delta_n) \sin\left(\frac{2 \pi x}{\lambda_n}\right)$

Plugging this into the wave equation, we see that the wave equation is solved if $\frac{\omega_n}{v} = \frac{2 \pi}{\lambda_n}$

Further more, in order for $y_n$ to be zero at x=L, we need $\frac{2 \pi L}{\lambda_n} = n \pi$ , or $\lambda_n = \frac{2 L}{n}$ where n is any positive integer.

Thus, for a normal mode solution to the wave equation with the boundary conditions we have chosen (y(x=0) = y(x=L) = 0), we have:

$\lambda_n = \frac{2 L}{n} \text{ and } \omega_n = \frac{n \pi v}{L}$

Now since the wave equation is linear, any superposition of the normal mode solutions is also a solution, and we can write any motion of the string as

$y(x,t) = \sum_{n=1}^{\infty} A_n \cos\left[ \frac{n \pi v t}{L} - \delta_n \right] \sin \left[ \frac{n \pi x}{L} \right]$

(Of course always keep in mind that this only works sufficiently close to equilibrium, where the equation is linear – for large displacements/very jagged string configurations/etc, non-linear terms will be important, and will not be amenable to this type of analysis).

A good question that we will address next week is how to get out the unknown coefficients $A_n$ and $\delta_n$ . This process is referred to as Fourier analysis, in which the sum over normal modes is referred to as a Fourier Series expansion.

As a final note for this week, we note that we can derive the wave equation in a similar way for systems with spatial extent in more than one dimension. A string is 1 D, but we could consider a sheet/membrane, or the surface of water, as examples of media in which 2D waves can travel. Also, sound waves typically are transmitted in 3D spaces (like the classroom, or a shower that you might find yourself singing in).

In 2D, with transverse displacements from equilibrium characterized by a function $z(x,y,t)$ the wave equation is:

$\frac{\partial^2 z}{\partial x^2} + \frac{\partial^2 z}{\partial y^2} = \frac{1}{v^2} \frac{\partial^2 z}{\partial t^2}$

In 3D, where a sound wave is characterized by differences in the density of the ambient gas $\xi (x,y,z,t)$ , we have

$\frac{\partial^2 \xi}{\partial x^2} + \frac{\partial^2 \xi}{\partial y^2} + \frac{\partial^2 \xi}{\partial z^2} = \frac{1}{v^2} \frac{\partial y}{\partial t^2}$

In more than 1D, the patterns of nodes of the normal modes can become quite beautiful, and due to possible degeneracies (i.e. when a 2D medium is a square rather than a rectangle or some other shape), can have unexpected shapes. Our DEMO of the square plate with salt (or looked at with the strobe light) showed you some of this richness. Here is a picture of a guitar back when it is set into motion associated with one of its normal modes:

Normal modes are beautiful things, and in our engineering feats, it is of great value to understand the role that they play so that we can either eliminate them when they are problematic, or accentuate them when they are of value.

This weeks Mathematica notebook is here:

https://jhubisz.expressions.syr.edu/phy360/wp-content/uploads/sites/5/2016/11/Week-9.nb

In it, you should explore the transition from the discretuum to the continuum, and also seek to understand the structure of the normal modes (i.e. their shape, their relative frequencies, and the behavior of superpositions of normal modes).

Week 8

Posted on October 21, 2016 by jhubisz

From the discretuum to the continuum

This week, we spent our efforts on making a transition from systems with a relatively small number of degrees of freedom to those with an essentially infinite number of degrees of freedom. We examined the coupled equations of motion for N masses attached to each other by string which is fixed at either end to walls.

The equations of motion for transverse displacement of the p’th mass was:

$\ddot{y}_p = - \omega_0^2 \left[ 2 y_p - y_{p+1}-y_{p-1} \right]$

where we have $\omega_0^2 = \frac{T}{l m}$ . (Here $l$ is the distance between masses, and $m$ is the mass of each mass).

We found that the spectrum of normal modes is described by amplitudes

$A_p = C \sin \left[ \frac{p n \pi}{N+1} \right]$

and frequencies

$\omega_n^2 = 4 \omega_0^2 \sin^2 \left[ \frac{n \pi}{2 (N+1)}\right]$

With this information, we can write any excitation of the string (sufficiently close to equilibrium) as a superposition over these normal modes! The motion of the p’th mass while the n’th normal mode is excited is:

$y^n_p (t) = C_n \sin \left[ \frac{p n \pi}{N+1} \right] \cos (\omega_n t + \alpha_n)$

finally, the motion of the p’th mass during a completely arbitrary excitation of the string is written as a superposition over all of the normal modes:

$y_p (t) = \sum_n C_n \sin \left[ \frac{p n \pi}{N+1} \right] \cos (\omega_n t + \alpha_n )$

I encourage you all to take a look at this weeks’ mathematica file, which allows you to visualize the normal mode spectrum as you change the number of degrees of freedom, and also as you change the normal mode number, $n$ :

https://jhubisz.expressions.syr.edu/phy360/wp-content/uploads/sites/5/2016/10/Week-8.nb

Week 7

Posted on October 14, 2016 by jhubisz

Forced oscillations of systems with 2 dof:

Recall the demo! We had the usual two pendula with 1kg masses on strings with a weak spring connecting the two masses. The additional ingredient in the demo on Tuesday was that we attached the piston motor so that it would push sinusoidally on one of the masses. As you showed in your homework, this is equivalent to the addition of a sinusoidal driving force acting on one of the two masses. We then slowly scanned through the various different frequencies on the frequency generator that was controlling the motor. We found that at around 0.67 Hz, the symmetric normal mode was excited with large steady state amplitude. We discovered a resonance of the two-body system, and it coincides with a normal mode excitation! We then explored higher frequencies, and we found at at around 0.77 Hz, the second asymmetric normal mode was excited!

We can see this behavior as arising from the equations of motion, as always. The equations of motion are now:

$\vec{F}_A = - m \omega_0^2 x_A - k (x_A - x_B) + F_0 \cos \omega t = m \ddot{x}_A$ $\vec{F}_B = - m \omega_0^2 x_B - k (x_B-x_A) = m \ddot{x}_B$

Note that the motor was only attached to the first pendulum, so there is no driving force acting on mass B.

Now we use the same trick – adding or subtracting the two equations from each other in order to decouple them:

$\ddot{(x_A + x_B)} + \omega_0^2 (x_A +x_B) = F_0/m \cos \omega t$ $\ddot{(x_A - x_B)} + (\omega_0^2 + 2 k/m) (x_A-x_B) = F_0/m \cos \omega t$

So the mathematics leads us to two simple harmonic oscillators, both driven by the same driving force. When $\omega$ coincides with $\omega_0$, you expect to find resonance for the first (symmetric) mode of oscillation. Similarly, when $\omega$ is close to the frequency $\omega_- = \sqrt{ \omega_0^2 + 2 k/m }$ , you expect to be exciting the resonance for the anti-symmetric normal mode. This is exactly what we saw in the demonstration! We also used the frequencies of the normal modes to determine $k$ .

Normal Modes with 2 dof:

We continued our discussion about normal modes of oscillation on Thursday, wrapping up the discussion using a Mathematica demonstration of computing the normal modes of oscillation in a more general case without obvious symmetry. In this case, there is a procedure. The mathematica file that we used is linked here:

https://jhubisz.expressions.syr.edu/phy360/wp-content/uploads/sites/5/2016/10/Week-7.nb

The procedure is to first identify the potential energy function. This demonstration was for the “double pendulum” where one pendulum is attached to the bottom of the first, and both have equal length $l$ and equal mass $m$ . The potential energy function we found was given by (in the small angle approximation):

$U(x_1 ,x_2) = \frac{mg}{2l} \left[ 2 x_1^2 + (x_2-x_1)^2 \right]$

From this one function, we can get everything about the behavior of the system when it is perturbed from equilibrium. First we get the forces on each mass by taking the derivative with respect to the two coordinates, then we apply Newton’s 2nd law. The next step is to assume we have identified a normal mode of oscillation of the system, in which case we have

$x_1 = A r \cos \omega t$ $x_2 = A \cos \omega t$

where $\omega$ is the frequency of the normal mode, and $r$ is the ratio of amplitudes when the normal mode is excited. Recall that for the symmetric pendulum, $r = 1$ for the symmetric normal mode, and $r = -1$ for the antisymmetric normal mode.

Plugging this into the equations of motion, we can re-express the equations as algebraic equations that allow us to determine pairs of $\omega$ and $r$ .

Finally we can plot what the normal modes actually look like, as I did in the mathematica file using the Animate command.

We can also then take arbitrary superpositions of the two normal modes, and adjust the amount of amplitude of each of them.

Week 6

Posted on October 14, 2016 by jhubisz

This was exam week! We spent Thursday after the exam going over solutions, and also wrapping up some discussion about systems with 2 degrees of freedom. I will put material in week 6 for these systems – particularly looking at forced oscillations when there are two degrees of freedom.

Week 5

Posted on September 28, 2016 by jhubisz

Coupled Simple Harmonic Oscillators

In this week, we begin to address systems with two degrees of freedom. We first looked at a demonstration where there are two 1kg masses suspended from strings of about 50cm each in length. The masses are then connected to each other through a spring with a small spring constant (called $k_c$ , for example). In the demonstration we did the following:

First we displaced only one of the masses from equilibrium, and set the system in motion. As with the Wilberforce pendulum, energy “sloshed” from one of the degrees of freedom to the other in manner reminiscent of “beats” that we saw while superposing waves of nearly the same frequency.
Next, we displaced both masses by an equal amount in the same direction, such that the spring connecting the two masses was not stretched or compressed. After letting it go, the system just oscillated back and forth, with the spring never being compressed.
Finally, we displaced each mass by an equal amount, but in the opposite direction. This time the masses again just moved back and forth in an antisymmetric manner, somewhat similar to #2.

Clearly the phenomena associated with two degrees of freedom is very rich, exhibiting beats, and having two special types of excitations. How do these appear in the mathematics?

First, we must write down Newton’s 2nd law for each of the masses:

$\overrightarrow{F}_A = -m \omega_0^2 x_A - k_c (x_A -x_B) = m \ddot{x}_A$

$\overrightarrow{F}_B = -m \omega_0^2 x_B - k_c (x_B -x_A) = m \ddot{x}_B$

The most important aspect of these equations are that they are coupled equations. That is, the equation for $x_A$ depends on $x_B$ and vice-versa. The two degrees of freedom “talk to each other” through the coupling spring. Indeed, if the coupling spring constant is taken down to zero, the communication is lost, and the two degrees of freedom become uncoupled.

It is not obvious at this point how to approach finding the solutions to this equation, but you can check that if you add (and subtract) the two equations from each other you get equations that look exactly like the original (uncoupled) SHO equations:

$\ddot{x}_+ = - \omega_0^2 x_+$

$\ddot{x}_- = - (\omega_0^2 + 2 k_c/m) x_-$ .

Here we have defined $x_+ = x_A + x_B$ and $x_- = x_A - x_B$ .

We know these equations! So we know their solutions! We have:

$x_+ (t) = A_+ \cos(\omega_+ t + \alpha_+ )$

$x_-(t) = A_- \cos(\omega_- t + \alpha_- )$ .

There are now 4 coefficients that are determined only after inputting a particular initial condition for the positions and velocities of the two masses.

These solutions can now be used to get back the results for $x_A \text{ and } x_B$ :

$x_A = \frac{1}{2} \left[ A_+ \cos (\omega_+ t + \alpha_+ ) + A_- \cos (\omega_- t + \alpha_- ) \right]$ $x_B = \frac{1}{2} \left[ A_+ \cos (\omega_+ t + \alpha_+ ) - A_- \cos (\omega_- t + \alpha_- ) \right]$

To explore this solution, let’s consider a particular initial condition where at t=0, we have $x_A = A$ and $x_B = 0$ and both velocities are zero. This corresponds to $\alpha_+ = \alpha_- = 0$ and $A_+ = A_- = A$ . Then we have

$x_A = \frac{A}{2} \left[ \cos (\omega_+ t) + \cos(\omega_- t) \right]$

$x_B = \frac{A}{2} \left[ \cos (\omega_+ t) - \cos(\omega_- t) \right]$

Superposing these gives the beat formula! Using $\bar{\omega} = \frac{\omega_+ + \omega_-}{2}$ and $\Delta{\omega} = \frac{\omega_- - \omega_+}{2}$ , we have:

$x_A = A \cos(\bar{\omega} t) \cos\left( \frac{\Delta \omega}{2} t \right)$

$x_B = A \sin(\bar{\omega} t) \sin\left( \frac{\Delta \omega}{2} t \right)$

So we see that, in the equations, the motion/energy “slosh” between the two degrees of freedom. Here is the mathematica notebook that you can use to explore the behavior of this very interesting system!

https://jhubisz.expressions.syr.edu/phy360/wp-content/uploads/sites/5/2016/09/Week-5.nb

Week 4

Posted on September 23, 2016 by jhubisz

Forced Damped Simple Harmonic Motion

This week we began by reviewing the simple harmonic oscillator including dissipative damping via a viscous force. I encourage you to read last week’s blog posts to see the material associated with this, as well as this weeks mathematica notebook:

https://jhubisz.expressions.syr.edu/phy360/wp-content/uploads/sites/5/2016/09/Week-4.nb

The new material for this week is the forced damped simple harmonic oscillator. For this case, we are now continuously injecting energy into and out of the system via some sort of periodic driving mechanism.

The demonstration associated with this is the mass on spring in water, where the spring is hung from a motor and piston that drives the entire mechanism up and down sinusoidally. Experimentally, we verified that at high frequencies, there is not a great deal of steady-state response (by steady-state, we mean the motion that persists a long time after the driving force has turned on). Similar behavior was noted for low frequencies. If, however, we drove the system at a frequency that is close to the natural frequency (the frequency of oscillation if there were no damping or driving) that the response is significant (the mass oscillates with a large steady state amplitude).

The phenomenon of large response at a given characteristic frequency is called resonance. You are all likely familiar with resonance from using swings as children, and driving your legs at the right frequency leads to a more exciting (larger amplitude) ride.

How does the mathematics work out for this scenario? There is now one additional force, the force that is an external input into the system. This force is what it is, independent of the displacement of the oscillator from equilibrium:

$\vec{F}_\text{drive} = F_0 \cos \omega t$ .

Note that here $\omega$ is not a characteristic of the oscillator itself, but rather of the external input. This force is not dependent on x(t).

The total force then (let’s say the system is a mass on a spring with spring constant k, and there is a viscous damping coefficient b) is given by:

$\vec{F} = F_0 \cos \omega t - k x(t) - b \dot{x}(t)$

Applying Newton’s 2nd law, we have an equation of motion given by

$\vec{F} = F_0 \cos \omega t - k x(t) - b \dot{x}(t) = m \ddot{x}(t)$

We rearrange the equation of motion, dividing by $m$ , and we have:

$\ddot{x}(t) + \gamma \dot{x}(t) + \omega_0^2 x(t) = \frac{F_0}{m} \cos \omega t$

where $\gamma = b/m$ and $\omega_0^2 = k/m$ .

Note the important difference from the equations we have looked at before – there is now a term in the equation that does not depend on x(t) at all. To solve this equation, let us presume that the long-time behavior of the system (a long time after the driving force is turned on) is given by oscillation with the same frequency as the driving force. Let us express this in the complex plane, however, when the equation of motion is expressed as

$\ddot{z}(t) + \gamma \dot{z}(t) + \omega_0^2 z(t) = \frac{F_0}{m} e^{i \omega t}$ .

Then our guess that the response is characterized by the same frequency is equivalent to the mathematical statement

$z(t) = Z_0 e^{i \omega t}$ .

Here $Z_0$ is an as yet undetermined constant. If we plug this guess into the equation of motion, we obtain an equation for $Z_0$ :

$Z_0 = \frac{ F_0/m}{(\omega_0^2-\omega^2) + i \omega \gamma}$

Now this looks like a relatively complicated formula, with both real and complex parts, but let us unpack it a bit. First, let us recall that any complex number can be expressed as an amplitude and a phase:

$Z_0 = A e^{i \alpha}$

We can find $A$ first:

$A = \frac{F_0/m}{\sqrt{\omega_0^2-\omega^2)^2+\omega^2 \gamma^2}}$

You can do some geometry in the complex plane to convince yourself that

$\alpha = \arctan \left[ \frac{ - \gamma \omega}{\omega_0^2-\omega^2} \right]$

Be careful that the signs in the numerator and denominator have meaning, as the arctan function is double valued. You must figure out which quadrant the point is in based on the signs of the numerator and denominator separately.

So what does all this mean in terms of the physics??

We are characterizing a response (the motion of the mass) due to a stimulus (the applied driving force). As we have found a solution, we see that for a given stimulus (an $F_0$ and an $\omega$ , there is a response with amplitude $A$ that leads the stimulus in phase $\alpha$ . To get a sense of what this means, please take a look at this weeks mathematica file, in which you can vary the stimulus frequency and watch the stimulus and response change with respect to each other in the complex plane.

Note that the complete solution can be written as

$z = C_+ e^{\rho_+ t} + C_- e^{\rho_- t} + Z_0 e^{i \omega t}$ ,

where the first two parts of the solution are damped, and thus not relevant for the behavior of the system at times that are large in comparison with the damping times. These are frequently referred to as transients of the system.

Week 3

Posted on September 13, 2016 by jhubisz

More degrees of freedom (uncoupled)

Examples of SHO’s (torsion balance, rigid rod, complicated pendula)

Damped SHO

You can use the following mathematica file to play with Lissajous figures and also the damped simple harmonic oscillator: https://jhubisz.expressions.syr.edu/phy360/wp-content/uploads/sites/5/2016/09/Week-3-1.nb

Multiple Degrees of Freedom

The first thing we discuss this week is a revisit of the idea of having systems with more than one degree of freedom. Recall the video we watched last week (you can download it in the “course documents” section on the right hand side of the page), where there were multiple bowling ball pendula with varying lengths that were chosen to create great visual effects when you view their superposition (combined oscillation given particular initial conditions – their effects don’t actually add together visually). In the beginning of the first lecture this week, we look at the simplified case where there are two masses on springs, and then we view the electrical analog using two function generators whose frequencies we can manipulate and also view utilizing an oscilloscope.

You can use this weeks mathematica file to play with dual oscillators and make your own Lissajous figures.

We will soon see (in a couple of weeks) that coupled oscillators are extraordinarily interesting in terms of the phenomena that can occur. As an example, I showed in class the Wilburten pendulum, where a mass can both spin and bounce on a spring, and where the energy slowly moves between the rotational degree of freedom and the translational (vertical) degree of freedom.

Also, you might enjoy the following video, which emphasizes the role in coupling between metronome degrees of freedom, where the metronomes become “phase locked” due to the coupling: https://www.youtube.com/watch?v=5v5eBf2KwF8

Damping:

Secondly, we began to discuss the phenomenon of damping. Damping is present in nearly all systems (certainly all that you are familiar with in day-to-day life). Damping is the result of energy in a system being lost to its environment through various different channels (sound wave energy, friction/viscous stirring, heat, etc…). In many systems, damping occurs through a velocity dependent force term. If the velocity is small, the force is approximately

$\overrightarrow{F}_\text{damp} \approx - b \dot{x}(t)$

*Note – again, the vector arrow on top of the force indicates the importance of the sign – i.e. that the force is in the opposite direction as the velocity. The value of $b$ is determined by physical properties of the system such as air density, temperature, or water density, etc.

In this case, Newton’s 2nd law is:

$\overrightarrow{F} = -k x(t) - b \dot{x}(t) = m \ddot{x}(t)$

and we can write the new equation that includes the damping term as

$\ddot{x}(t)+\gamma \dot{x}(t)+\omega_0^2 x(t) = 0$ , where $\gamma = b/m$ and $\omega_0^2 = k/m$ .

Before we delve into the mathematics of this equation, let us remember the phenomena that occur as a result. We performed a demo in lecture where we placed a mass on a spring, as before, but then placed the mass into a narrow glass container holding water. The water must move around when the mass is moving, but there is a viscous force resisting the motion of a mass that is of the form discussed above. The motion that the mass underwent was oscillation, but with an amplitude that decreased quickly with time. The energy of motion and compression of the spring dissipated into motion of the water (which eventually manifests as an increased temperature of the water).

We can solve this equation generally using complex notation, where we use $z(t)$ in place of $x(t)$ , and we suppose there are solutions like $z(t) = C e^{\rho t}$ . We don’t know what $\rho$ is yet, but we can plug into the equation above, and we see that the differential equation turns into an algebraic equation for $\rho$ :

$\rho^2 + \gamma \rho + \omega_0^2\rho =0$ ,

which is solved for two values of $\rho$ , as expected from the quadratic formula:

$\rho_\pm = - \gamma/2 \pm i \sqrt{ \omega_0^2 - \gamma^2/4 }$

This means that there are two types of solutions to the motion, each with their own coefficient:

$z(t) = C_+ e^{\rho_+ t} + C_- e^{\rho_- t}$ where the two $C$ ‘s are determined by initial conditions, and the $\rho$ ‘s are determined by the physical properties of the system.

There are three possibilities for the physical behavior of the system, and they depend on the ratio $Q = \omega_0/\gamma$ . The value of $Q$ for a system is referred to often as the “Quality factor.” In terms of $Q$ , you can rewrite the $\rho$ ‘s as:

$\rho_\pm = \omega_0 (-1/2/Q \pm i \sqrt{1-1/4/Q^2})$

The three possibilities correspond to whether or not the quantity in the square root in $\rho_\pm$ is positive, zero, or negative.

The three possibilities are

Under-damped (or sub-critical damping). In this case, $Q > 1/2$ and the quantity in the square root is positive. The system has oscillatory solutions, with frequency $\omega' = \omega_0 \sqrt{1-1/4/Q^2}$ . The amplitude of the oscillations decays exponentially. After taking the real part of the complex valued solution, the oscillation is characterized by: $x(t) = A e^{- \omega_0 t/2/Q}\cos (\omega' t + \alpha)$ .
Over-damped (or super-critical damping). In this case, $Q < 1/2$ . In this case, there are no oscillatory solutions, only two damped exponentials. The solution has a fast-damped piece (sometimes referred to as a transient) and a slower damped piece. At small times after excitation, the fast damped piece plays a role, but the slow damped solution is the only remaining one at long times after the system has been displaced from equilibrium. The two exponentials have decay coefficients given by $\lambda_\text{slow} = \gamma/2 - \sqrt{1-4 \omega_0^2/\gamma}$ and $\lambda_\text{fast} = \gamma/2 + \sqrt{1-4 \omega_0^2/\gamma}$ (note that both $\lambda$ ‘s are positive), and the solution can be written as $x(t) = A e^{-\lambda_\text{slow} t}+B e^{-\lambda_\text{fast} t}$ .
Critically damped In this special case, we have $Q = 1/2$ . At first glance, this situation looks strange, in that $\rho_+ = \rho_- = -\gamma/2$ , and you might think that there is only one solution then: $x(t) = A e^{- \gamma/2 t}$ . In fact, there are still two solutions, and we can write $x(t) = (A + B t) e^{-\gamma t/2}$ . You can verify that this solution works by plugging it into the original equation of motion. You can also obtain this solution by a Taylor expansion of the over or underdamped scenarios as I suggested in class, so thought it looks quite different in form, it is not drastically different in terms of the phenomena that occur at this critical point.

Week 2

Posted on September 6, 2016 by jhubisz

Complex Exponentials and Simple Harmonic Motion

The Superposition Principle

In lecture on Tuesday, we covered the use of complex exponentials to represent simple harmonic motion. The reason we can use this notation is due to the following three facts:

The projection of 2 dimensional rotation at constant angular velocity onto one axis is identical to simple harmonic motion
the complex plane carries the same amount of information as the 2D plane
Euler’s formula provides a very concise packaging of this 2d motion into a single formula (just one expression rather than separate ones for the x and y component, for example)

This notation using the complex plane will provide useful for the following scenarios (as examples):

Superposition of multiple simple harmonic oscillators (avoiding trig)
Studying the behavior of 1D systems that include damping/dissipation and/or driving forces (rather inconvenient using trig) (Note that this is also the case in engineering when you have RLC circuits).
it is the fundamental tool in quantum mechanics, where the complex phases then have real physical meaning, so you better get used to it!! 😉

Here is a link to the mathematica file we are using this week:

https://jhubisz.expressions.syr.edu/phy360/wp-content/uploads/sites/5/2016/09/Week-2.nb

As we discussed briefly in class on Tuesday, in response to a question from Emily Syracuse, there is a very interesting and reason for why we can use this complex notation. Recall that in introducing complex notation we bring in extra information, taking 1D motion and turning it into 2D circular motion in the complex plane (the real and imaginary axes are the 2 dimensions). In the end, we can always throw this extra information (the imaginary part) away to get at the physics (the projection of the circular motion onto one axis).

The reason for why this works is due to the fact that the equation of motion for a simple harmonic oscillator is linear.

Recall the simple harmonic oscillator equation is

$\ddot{x}(t) = - \omega^2 x(t)$

There are no terms like $x^2(t)$ or $\dot{x}^2(t)$ , or other higher powers of $x(t)$ and its derivatives.

The SHO equation is linear because all potential energies near equilibrium are quadratic in displacement, meaning the force

$\overrightarrow{F} = - \overrightarrow{\nabla} U$

is linear.

Equations like this – linear equations – obey a very important principle that is called the Superposition Principle.

The Superposition Principle states that if we have two solutions to an equation, the sum of the solutions is also a solution (or the difference, or any arbitrary linear combination of the two solutions). You may be familiar with this principle from electromagnetism, and there is it also a result of the equations (Maxwell’s equations, in this case) being linear.

It is worth revisiting the “big picture” at this point:

All systems with one degree of freedom near equilibrium have a potential energy which is quadratic in the displacement from equilibrium
Thus the force is linear, and application Newton’s 2nd law yields a linear equation of motion
Thus if $x_1(t)$ and $x_2(t)$ are both solutions, then so is $x_1(t) + x_2(t)$ or $a x_1(t) +b x_2(t)$ where $a$ and $b$ are any number at all (including imaginary numbers).

As we will see, it is very convenient to write solutions to the equation of motion for the SHO in the complex plane, even if the physical situation at hand refers only to real quantities, and so long as both the real and the imaginary part are both solutions, then this will work. Circular motion in the complex plane is just the sum of two solutions, one is $\cos \omega t$ , the other is $\sin \omega t$ , and the coefficients $a$ and $b$ are $1$ and $i$ respectively.

An important consequence of the superposition principle is that of interference. Recall now the demonstration where we look at what happens when 2 circular water waves are generated at 2 points some distance from each other. When the crests coincide, you get a larger crest. When a crest lies on top of a trough, you get no motion at all. That is, the net effect of the two waves is the sum of the effects of each one individually.

Week 1

Posted on August 30, 2016 by jhubisz

Ubiquity of Vibrations and Waves

Degrees of Freedom

Simple Harmonic Motion (and relation to 1D systems near equilibrium)

Waves are everywhere – everything waves (Howard Georgi)

Waves and vibration are ubiquitous – nearly all systems under some conditions undergo oscillation of some form. The electric guitar demo in this lecture showed a chain of events that involved wave and oscillation phenomenon at many levels:

neuro-electronic pulses
vibration of guitar string
vibration of electromagnetic field
pulses sent through electronics of guitar down to amplifier
amplifier vibrates speaker panel using electromagnetic coils
sound waves created by speaker travel through air
your eardrums vibrate
new electronic pulse sent through your brain

Many forms of oscillation are present in this “simple” demonstration!

We introduced the “period” (the time which a repeating phenomenon takes to repeat itself). This is often referred to as “T”

Degrees of freedom – it is crucial in systems to identify the moving parts, and the ways that they can move. The number of degrees of freedom is the number of coordinates necessary to specify what a system looks like in a stationary picture. For a point particle in a normal room, there are 3 degrees of freedom (the x,y and z coordinates that locate the particle). For a extended rigid body (like a bowling ball), there are 6 degrees of freedom: The x,y, and z coordinates along with 3 angles describing the orientation of the object. A mass on a spring that can only move in 1 direction counts as a single degree of freedom.

Simple Harmonic Motion – Simple harmonic motion is pure sinusoidal motion. This kind of motion is what a mass on an ideal spring undergoes when you displace it from equilibrium and let it go.

Mathematica Demo – go to the following site to get Mathematica:

http://its.syr.edu/mobile/licenses/Mathematica.html

The mathematica file we used in class is here (use shift-click to download rather than opening in the browser):

https://jhubisz.expressions.syr.edu/phy360/wp-content/uploads/sites/5/2016/08/Week-1.nb

SHM is ubiquitous – As we discussed, all systems sufficiently near an equilibrium point undergo simple harmonic motion when they are displaced from equilibrium. The reason for this is that all functions (like the potential energy) are quadratic in the displacement from equilibrium, and so the force, which is the negative of the derivative of the potential energy, is linear. See the full lecture notes for details.

EXAMPLE DEMO:

Pendulum – potential energy is sinusoidal, but near minimum is is quadratic, and force is linear. Thus for small oscillations, the pendulum is just like a mass on a spring, and undergoes simple harmonic motion. This is not the case for large oscillations! The motion is then periodic, but NOT sinusoidal.

Motivations

Posted on August 30, 2016 by jhubisz

The ubiquity of vibratory and wave phenomena in our world is reflected in the depth to which such solutions of classical mechanics and electromagnetic equations have been studied, and in the breadth of the language and tools we have formed around them. In Physics 360, you will learn the techniques essential for efficiently solving real-world problems in which oscillation of some form is a dominant component.

As you progress through your physics, engineering, or more generally your STEM curriculum, you will find that the tools you learned in Physics 360 are a foundation for your future exploration. Our studies of materials such as crystals and protein structure involve scattering light or electron beams through them, and interpreting the resulting patterns in terms of their microscopic structure. The probabilities associated with outcomes of experiments that push into the quantum regime are calculated via solutions to a particular wave equation. The fundamental particles of our world are currently best understood in terms of quantized waves in fields that permeate our universe.

Regardless of your future career trajectory, a deep knowledge of wave phenomena is of extraordinary value.