FREQUENTLY ASKED QUESTIONS

August 27, 2007

Administrative Questions

Can we do more of a broad survey of modern physics rather than go really in depth with 1 or 2 topics? I would rather touch on many topics than go into minute details on 1 or 2 topics.

The course material is indeed more broad than deep. We cover quite a bit, but are prevented from going too deep by the fact that more math is needed to go further in many places.

What do we do with the packet?

These are handouts on statistics and error analysis that will be needed for your lab work. If the material is unfamiliar to you, you should read them. If not, keep the handouts as reference.

Can we have more information about the lab procedures? 5/7 labs?

Everyone does Lab 0, and then you will choose 5 ``regular'' labs out of a possible 7. Each will take two weeks. More details on lab logistics will be given during the first lab period, and information is also available on the course lab web page.

Do we need to bring our text to class?

Not really.

Could we have an recitation section (optional) during the week, led by one or both of the TAs?

As for other upper division classes, you're assumed to by now be proficient enough at studying this kind of material that a formal recitation is not needed. However that is not to say you are on your own! You are very much expected to come get help when needed. The TAs are not required to hold recitations, but they do have two hours of office hours per week; in fact these could turn into ad-hoc recitation-style periods. Plus, we are all available at other times by appointment, in case you don't have an overlap with scheduled office hours. Please do not be shy about coming to ask questions. You can email or phone too.

Let me also suggest getting together in informal study groups. You are encouraged to work with your peers; it's well known that this is enormously helpful for learning (although of course, I don't think I need to remind you that all your assignments must be your own solutions, in your own words).

I'm nervous as to the level of difficulty of the problem sets without a recitation. How are the psets compared to 41/42?

I've never taught 41/42 so I'm not sure how to answer this. But please see above question: you are not on your own, and you're expected to take the initiative to get help from me and the TAs (and each other) when you have trouble with a problem set.

Content Questions

How much of this do I need to understand exactly?

The ``WUN2K'' at the end of class gives a good guide to what you need to know. In general in this course, we will be concerned mostly with physical ideas rather than formal mathematics. In other words, you should understand the physical meaning of the equations. For example, for today's lecture, the key thing to understand is generally how one gets a wave equation (restoring force plus Newton's second law), and what the solution means, i.e. how solutions correspond to wavy behavior in a physical system. The details of the derivations are less important, although you may be interested to look them up.

How does the restoring force connect to waves on a string?

When Newton's second law is applied to a problem with a linear, restoring force, you typically get an equation of motion- a differential equation for the displacement $h(x,t)$ as a function of space and time - that's a wave equation. What it means to be a wave equation is that the solution gives some sort of wavy motion. The simplest example we saw is the case of a spring in one dimension: in that case, we got an equation of motion that led to a sinusoid. The next more complicated example is that of waves on a string: for this, plugging the restoring force into Newton's second law yields the wave equation $\frac{\partial^2 h}{\partial x^2}= \frac{1}{v^2} \frac{\partial^2 h}{\partial t^2}$ (this expression is known as ``$the$'' wave equation.) See link a couple of questions below for the derivation, if you're interested.

Solutions of the wave equation tend to be are wiggly, wavy things, e.g. traveling or standing waves, depending on boundary conditions. A single pulse solution is also possible.

What is a restoring force factor?

This is some factor which is related to the restoring force in the system. Its specific nature depends on the system. In our simple Hooke's law example, it was $k_H$, the Hooke's constant. For the case of a wave on a string, it's the tension $T$ in the spring.

Can you put the derivation up of the wave equation?

Here's an example from the web, for the vibrating string:
www.math.ubc.ca/$\sim$feldman/apps/wave.pdf

An equation like this can be derived for lots of different situations; I hope you've seen a few of them before. It's a typical form that pops up all the time when you apply Newton's 2nd law ($F=ma$) to the case of a material that has Hooke's-Law-like internal forces (an ``elastic medium''). Your freshman physics textbook will definitely have some examples. For instance, see section 14-2 in ``Physics for Scientists and Engineers'' by Fishbane, Gasiorowicz and Thornton.

(Try also googling for ``wave equation vibrating string'' or the like).

In the wave equation why did $v^2$ show up in the denominator on the right hand side of $\frac{\partial^2 h}{\partial x^2}= \frac{1}{v^2} \frac{\partial^2 h}{\partial t^2}$?

If you follow the derivation linked above, you can see where it comes from.
To get a feeling for it: we have $v^2 = \frac{\partial^2 h}{\partial t^2}/\frac{\partial^2 h}{\partial x^2}$. The expression $\frac{\partial^2 h}{\partial t^2}$ is acceleration of a particular piece of string; if the ratio of $\frac{\partial^2 h}{\partial t^2}$ to the ``space acceleration'' $\frac{\partial^2 h}{\partial x^2}$ (rate of rate of change of displacement with respect to space) is large, then the pattern moves quickly.

How did the wave equation $\frac{\partial^2 h}{\partial x^2}= \frac{1}{v^2} \frac{\partial^2 h}{\partial t^2}$ generalize $x= -\frac{m}{k} \ddot{x}$, and how does it incorporate a restoring force?

I wouldn't say that the wave equation exactly ``generalizes'' the Hooke's law equation. It's more that it's another example of a similar, slightly more complex situation, for which displacement is a function of both $x$ and $t$ (rather than the Hooke's law case for which displacement is a function of time only). The specific restoring force can be tension in a string (see derivations referred to above), or pressure in a gas, and comes in to the $v$ constant in the equation. For instance for a string $v=\sqrt{T/\mu}$, where $T$ is tension and $\mu$ is mass density; for a gas (sound wave solution), $v=\sqrt{P/\rho}$, where $P$ is pressure and $\rho$ is gas density.

It is important to note that the general wave equation with $v$ constant only describes waves in ideal elastic media. Real media do not have perfectly linear restoring force function.

That is indeed true. In physics we typically start with idealized situations, which are often a very good approximation and yield a lot of insight (although perhaps not in the ``Consider a spherical cow'' example).

How did you get the velocity of the harmonic wave?

The $v$ of the wave equations appears in $\sin{(k(x-vt))}$. In writing this as $\sin{(kx-\omega t)}$ we have have defined $\omega$ to be $kv$. So the $v$ of the wave equation is $v=\omega/k$. But since the wave repeats in time for $\omega T = 2\pi$, $\omega= 2 \pi/ T$, and since the wave repeats in space for $k \lambda = 2 \pi$, $k= 2 \pi / \lambda$. Plugging these in, we get $v = \lambda / T$. The velocity of the wave is the distance it travels from peak to peak over the time it takes to go from peak to peak-in other words $\lambda/T$. So the $v$ in the wave equation is the velocity of the wave.

What is the distinction between the velocity of the traveling wave solution and the velocity of any given point on the wave (i.e. velocity as first derivative of $h$)?

The velocity of the traveling wave is the velocity with which the entire pattern moves along (see above question). The velocity of a given point on the wave is a different thing: it's the speed of that little piece of string (or whatever medium). This is given by $\frac{\partial h}{\partial t}$.

What does $g(x-vt)$ mean?

$g$ is any function you can dream up; call it $g(y)$ (it has to be well enough behaved, e.g. twice-differentiable). It just has to be a function of $y=x-vt$ in order for it to work as a solution of the wave equation (note that $v$ can be negative). For instance, take $g(y) = y^2$. $h(x,t)=g(x-vt) = (x-vt)^2$ solves the general wave equation. Similarly $h(x,t)=1/(x-vt)$ works, and $h(x,t)=\sin(x-vt)$ works, and so on.

(Note that any old function $g$ won't work for specific physical situations, for which boundary conditions determine what solutions are allowed!)

How did you find the general solution $h(x,t)=g(x-vt)$?

Here the wave equation's general solution was sort of pulled out of a hat for the lecture, because the solution takes a fair amount of time and would sidetrack the story. Here is some more info for those interested:

The wave DE is asking you: ``I'm a function of $x$ and $t$... differentiate me twice with respect to time, divide by $1/v^2$ and you get my 2nd derivative with respect to space... what function am I?'' And you have to come up with such a function that also respects the boundary conditions.

Here is d'Alembert's general solution to the wave equation, for the more math-inclined:
http://mathworld.wolfram.com/dAlembertsSolution.html

Lots of solutions (an infinite number, of course) of the general form, $g(x-vt)$, are possible and different ones are appropriate for different boundary conditions and situations. To get the specific solutions we covered, traveling and standing waves, you don't really ``solve'' the equation the way you would some algebraic equation to solve for $x$; the answers are figured out from knowledge of what functions have specific properties. These traveling and standing wave functions have specific properties that apply for specific physical conditions.

How does $h(x,t)=g(x-vt)$ work for all $g$?

You can see it for yourself by plugging into the wave equation. Write $h(x,t)=g(x-vt)=g(y)$, where $y=x-vt$. Then, employing the Chain Rule of derivatives, $\frac{\partial h}{\partial x} = \frac{\partial g}{\partial x}= \frac{\partial g}{\partial y} \frac{\partial y}{\partial x} = \frac{\partial g}{\partial y}$, and $\frac{\partial^2 h}{\partial x^2}= \frac{\partial^2 g}{\partial y^2}$. We also have, $\frac{\partial h}{\partial t} = \frac{\partial g}{\partial y} \frac{\partial y}{\partial t} = - v \frac{\partial g}{\partial y}$. Then the $LHS$ of the equation is $\frac{\partial^2 h}{\partial x^2}= \frac{\partial^2 g}{\partial y^2}$, and the $RHS$ is $\frac{1}{v^2}\frac{\partial^2 h}{\partial t^2} = \frac{(-v)^2}{v^2} \frac{\partial^2 g}{\partial y^2}= \frac{\partial^2 g}{\partial y^2}$, so $LHS=RHS$ no matter what $g$ is.

You mentioned that $h(x,t)=g(x-vt)$ is the most general solution to the wave equation. However what about $f(x+vt)$? Isn't this an independent solution, assuming $v$ is constat?

For $h(x,t)=g(x-vt)$, $v$ is a constant which can be positive or negative. So $f(x+vt)$ is indeed a function of this form (just redefining $v$ as the negative of itself). $g(x-vt)$ represents a wave traveling in the $+x$ direction, while $f(x+vt)$ represents a wave traveling in the $-x$ direction.

Where do the solutions of the differential equations come from?

I didn't derive them: I selected them. Any function of the form $g(x-vt)$ will be a solution to the general wave equation. I wrote down two types of common solutions which have this form, the traveling wave and the standing wave. These two types of solutions have different properties appropriate to describe different situations.

Given a 2nd order, linear, 2 variable partial differential equation, what must the boundary conditions look like in order to get quantized solutions?

I do not know the formal math answer to this off the top of my head (the question probably needs to be defined carefully). Standing wave solutions to wave equations with finite boundary conditions will typically have quantized solutions; we'll see lots of examples.

What was the difference between the time and space waves and which part affected the phase for each? Why did it change?

I'm assuming you're asking about the traveling wave. For the harmonic wave (traveling wave), $H(x,t) = H \sin(kx- \omega t)$, first consider a point in space, $x=x_0$. You are looking at what happens to that point in space (say, a point on a string) as a function of time. The function describing displacement as a function of time is a sinusoid, $H(x,t) = H \sin(kx_0- \omega t)$, where here $kx_0$ is a constant. A constant inside the sinusoid acts as a phase: it tells you how much the sinusoidal pattern is shifted along the $t$ axis. In other words it tells you when the wiggle starts.

Now consider a snapshot in time, $t=t_0$. You are now looking at the space pattern at a particular time, like a frame in a movie. Now the function describing the space pattern is $H(x,t) = H (kx- \omega t_0)$. In this case, $\omega t_0$ acts as a constant phase: it tells you how much the wave is shifted in space. At a different time, $t=t_1$, the phase is different: now it's $\omega t_1$. The wave is shifted along. In other words, the wave pattern changes in time: it moves. It's a traveling wave!

For the standing waves, how did you derive $\phi=0$ everywhere?

For the case of the case of the standing wave on a string fixed at both ends, we must have $h=0$ at $x=0$ for all values of time $t$. Since we have $h(x,t)=H\sin(kx+\phi)\cos(\omega t + \delta)$, the sine part has to be zero for $x=0$. So we must have $h=H\sin(\phi)=0$. That requires $\phi=0$.

For the standing waves, where did the $kL=n\pi$ come from?

The string is fixed at $x=0$ and $x=L$, so its displacement $h$ is zero there. So $h(x,t) = H \sin{(kx+\phi)}\cos{(\omega t + \delta)}$ must be zero for all values of $t$ at $x=0$ and $x=L$. Since in general the cosine term isn't zero, we must have the sine term equal to zero for $x=0$ and $x=L$. Take $x=0$: $\sin{(0+\phi)} = 0$, so we must have $\phi=0$ to satisfy that boundary condition. Now take $x=L$: plugging that in, we find we must have $\sin{kL}=0$ to satisfy the $x=L$ boundary condition. This means that $kL=n\pi$.

Can you explain how standing waves are generated from interference?

A standing wave is a superposition of traveling waves in opposite directions (see also below). I don't think I said that today; we'll get to that in a bit when we talk about interference next class.

What's the difference between traveling and standing waves?

A traveling wave moves in time along $x$. In contrast, a standing wave does not travel along, but rather wiggles up and down ``in place''. To see how the standing wave equation, with space and time separated out, matches the physical situation: think of a snapshot in time, say $t=t_0$: as a function of $x$ you see a sinusoid. But unlike for the harmonic wave $H\sin(kx-\omega t)$ case, a different snapshot in $t$ ($t=t_1$) does not correspond to a different phase ($\omega t_0$ vs $\omega t_1$) and a shift of the wave along the $x$ axis. Rather, the displacement at each $x$ is multiplied by the new time sinusoid value. If you imagine watching each $x$ spot in time, you will see it thrashing up and down (sinusoidally). This is exactly what a standing wave does.

Examples of standing waves everyday life are a vibrating string, or a column of gas (pressure waves creating sound). A standing wave can be created from two traveling harmonic waves moving in opposite directions (you can show this yourself using a few trig identities). This can happen when one wave is reflected from a boundary and interferes with (is added to) itself after reflection: see http://physics.usask.ca/$\sim$hirose/ep225/animation/standing1/anim-stwave1.htm. We'll discuss superposition and interference next class.

Can you explain the space/time separation of the standing wave?

Basically, when the time and space parts are separated out, as for $h(x,t) = H \sin{(kx+\phi)}\cos{(\omega t + \delta)}$, rather than intermingled in the same sinusoid argument, the phase of the space wiggle does not change with time. So the standing wave doesn't march along the $x$ axis. See above question also.

How does the standing wave equation have the form $g(x-vt)$?

You can manipulate the standing wave equation with trig identities to get it in this form. The easiest way to see it is perhaps as follows: a standing wave is actually a superposition of two traveling waves with the same amplitude, traveling in opposite directions:

$$h= h_1(x,t)+h_2(x,t)$$

where

$$h_1 = A \sin{(kx-\omega t)},  h_2 = A \sin{(kx+\omega t)}$$

Using the trig identity

$$\sin\theta_1 + \sin\theta_2 = 2 \sin{\frac{1}{2}(\theta_1 + \theta_2)} \cos{\frac{1}{2}(\theta_1 - \theta_2)}$$

you can show that

$$h= 2A \sin (kx) cos (\omega t)$$

This can be shown more generally with phases, too, using a bit more algebra.

In the standing wave equation, $h(x,t)=H\sin(kx+\phi)\cos(\omega t + \delta)$, how do you know to separate space and time wiggle to get a standing wave as opposed to a traveling wave? Where did this solution come from?

I'm not sure this kind of solution is exactly ``derived'': it's written down as a particular function we happen to know about that has the desired properties (sometimes called ``solution by inspection''). It's sort of like having a job to do and already having the right tool for it, so you don't have to go out and manufacture one.