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9

The energy of an element of a traveling wave is not constant. Halliday-Resnick-Krane is right. For a string of density $\mu$ and tension $T$ the kinetic energy of an element $dx$ is $$dK=\frac 12\mu dx\left(\frac{\partial \xi}{\partial t}\right)^2.$$ For the potential energy we have $$dU=Tdl,$$ where $dl$ is the stretched amount of the string. A small ...


9

A fake derivation We can rather easily compute a horizontal velocity for the string fi we assume that the total velocity vector is everywhere normal to the string (this assumption is not always valid, see below). The following picture then illustrates the computation: Take two infinitesimally separated points $x$ and $x+\mathrm{d}x$ and let the wave ...


8

TL;DR: There is of course a continuous transition between seeing an interference pattern and not seeing one as the width of the source is increased. So it does not really matter if it is $<$ or $\leq$ in the condition, since the relation only represents a qualitative statement about when the contrast in the interference pattern has significantly ...


7

EM waves are a special case of electromagnetic radiation, where typically the source is periodic, or near enough that there is a carrier wave, as with radio and television. Maxwell's equations support a "sourceless" electromagnetic wave, as if it has existed forever. Also see Why do we think of light as a wave? Let's consider two cases of electromagnetic ...


6

From the Wikipedia article on sound: In physics, sound is a vibration that propagates as a typically audible mechanical wave of pressure and displacement. To fully understand how is air vibrating in an open pipe, you have to consider not only the acoustic pressure wave, $$\frac{\partial^2 p}{\partial x^2}=\frac{1}{c^2}\frac{\partial^2 p}{\partial ...


6

Wind instruments work by setting up sanding waves in the air column inside them. Shorter instruments have shorter air columns and thus standing waves with shorter wavelengths resulting in higher pitches.


6

The explanation is not a very full one. As you correctly note, you're taking a limit, so the assumption $\sin\theta \to\theta$ as $\delta z\to0$ becomes exact. So Eq 16-23 contains no approximation. The assumption creeps in subtly when one assumes that the force calculated in Eq 16-23 is at right angles to the $z$ axis. That is, that ...


6

You are absolutely right in everything you said. The momentum is non zero only if the wave has a longitudinal mode, which is in fact the realistic case. Moreover when this is the case, the wave equation is not that simple. Let me try show this. Longitudinal Mode Let us assume that when in equilibrium the string, of density $\mu$, is along with the $x\equiv ...


5

The second derivation is correct, as explained by Diracology. However, the first derivation is 'sort of' correct, in the sense that the location of potential energy can be ambiguous. For example, consider the three following systems. A mass on a stretched spring. A mass sitting on a table. A charged mass next to another charge. These three systems have ...


5

No, the frequency will not change. If the wind is blowing at constant speed and the distance between source and observer remains constant, then the time it takes for a sound wave to get from source to observer will be constant. So the time interval between wave peaks (period T) when they are detected by the observer remains equal to the interval between ...


4

You're wondering why pressure nodes form at an open end of a tube. The answer is, they don't! It's just a reasonably good approximation. Physically, consider the air molecules at the center of the tube. Since they're far away from the edges, there's no way for them to "know" exactly when the tube ends, so the sound wave must "leak out" slightly. The ...


4

There are already good answers here, but I'm afraid that to the best of my knowledge, Diracology's (and indeed Halliday-Resnik-Krane's) expression of the potential energy is not correct. I would like to point to this paper by Lior M. Burko which focusses on the subtleties of the derivation of the kinetic and potential energy of the string as a whole and ...


4

They are not equivalent. A way to see this is to find the normal modes associated to these solutions. Applying the condition $\xi(L,t)=0$ to $$\xi(x,t)=2 A \mathrm{sin}(kx)\mathrm{cos}(\omega t),$$ you get $kL=n\pi$. This gives $\lambda_n=2L/n$ and $f_n=nv/2L$. Which are the correct wavelength and frequency for the rope with both ends fixed. For the ...


4

As a diagram screen $A$ is closer than screen $B$ and so the fringes are closer on screen $A$: Fringe separation $\Delta x = \dfrac {\lambda \; D}{d}$ where $d$ is the slit separation, $\lambda$ is the wavelength of light and $D$ the distance from the slits to the screen.


3

Whether waves == radiation is somewhat a question of semantics, and thus a bit subjective, however... In general wave equations support radiation, but not all solutions to a wave equation are radiative. Evanescent solutions are also called waves ("evanescent waves"), yet typically not considered to be radiation since they do not propagate in three ...


3

That is because of the termination impedances at the pipe end and mouth. The above described relationship for fundamental frequency is given for zero termination impedances (ideally open pipe) which is not the real case. The simplest way to account these impedances is to introduce corresponding length corrections, so for the $f_0$: $$ f_0 = ...


3

The clue here is " how far apart". The question is asking for distance which must be in terms of the wave's wavelength. Phase measures fractions of wavelength. And you are given information of the wave's speed and the periodic time in which it propagates (frequency). The fundamental "distance = rate * time" applies in terms of the wave speed, wavelength and ...


3

This is a subtle and somewhat complicated question, but I think the basic answer is ``no''. 1) The relativistic Boltzmann equation is $$ p^\mu\partial_\mu f = C[f] $$ which has the same structure as the non-relativistic Boltzmann equation. This equation can be used to derive relativistic Fokker-Planck equations. One example is the Landau collision term, ...


3

For small enough amplitudes, the speed of sound is independent of how loud the sound is. It is also true that for a wide range of frequencies, the speed of sound doesn't vary with the pitch. When you move to large amplitudes (the assumptions of linear material are challenged) and high frequencies (when the wavelength of the sound is comparable to the spacing ...


3

In Wikipedia the caustic is defined as follows. In optics, a caustic is the envelope of light rays reflected or refracted by a curved surface or object, or the projection of that envelope of rays on another surface. You can think of the envelope of a family of curves as a curve that is a tangent to each of them. Here is a diagram on page 60 of "A Treatise ...


3

Right off the bat in eq (16-23) it's assumed that the restoring force is linear in the displacement. That's only true for small displacements.


2

It depends: 1) On whether you mean mass density or particle density, 2) on the temperature and type of gas. The main observation is that speed of sound is approximately independent of particle density, and mostly a function of temperature. 1) The speed of sound (squared) is given by the compressibility at constant entropy per particle $$ c_s^2 = \left. ...


2

No, it is not true that sound travels faster in denser media. In fact it travels slower. In the adiabatic approximation we assume that the portions of the gas vibrate so fast that it is not able to exchange heat with the surroundings. The the longitudinal displacements can be shown to satisfy the wave equation $$\frac{\partial^2 u(x,t)}{\partial ...


2

Yes, it is possible to have two different waves of different frequencies on a string (Imagine driving the string with a square wave; it consists of multiple frequencies with multiple amplitudes; See Fourier transform). However, the velocity of waves on a thin string are constant and depend only on the tension of the string and the density. Thus, you cannot ...


2

At the same amplitude, higher frequency waves make the rope move faster (because every point has to move the same distance in less time). That means there is more kinetic energy in the wave at higher frequencies. You notice this because it is more tiring to move the rope quickly (that also has to do with the fact that loss mechanisms are stronger at higher ...


2

The answer is "it depends". If you have a very thin and light "rod" (for example, a piece of fishing wire), the material from which that is made has a shear modulus, and it is in principle possible to have a piece of fishing wire vibrate without being held in tension. However, if you pull both ends tight, like the string of a guitar, then waves will travel ...


2

If the source were truly point-like, yes. But nothing is truly point-like, so no: your formula for $I(r)$ is modified for short distances (where the inner structure of the source becomes relevant). This means that, if your source is a set of speakers, then $I(r)\propto R^{-2}$ is only valid for $R\gg\ell$, where $\ell$ is, say, the radius of the diaphragm of ...


2

$T= k f^2$ and $T + \Delta T= k(f + \Delta f)^2 \Rightarrow 1 + \frac{\Delta T}{T} = (1+ \frac{\Delta f}{f})^2 = 1+ \frac {2 \Delta f}{f} + \left ( \frac{\Delta f}{f}\right )^2$ $$\frac{\Delta T}{T} = \frac {2 \Delta f}{f} + \left ( \frac{\Delta f}{f}\right )^2$$ So your methods differ by the last term


2

When working with waves the wavelength and frequency (pitch) are inversely related. Sound waves have the relation frequency times wavelength equal the speed of sound. Wind instruments are using the longitudinal dimension of air in the instrument as a medium like a "string". In both the string and wind instruments, shorter wavelengths in the excited medium ...


2

In general, the term cannot be neglected. It's entirely possible that the term has to stick around when advection is significant. But, since you are asking how it goes away, I'm guessing you are in a situation where it does, in fact, go away. Like any other order of magnitude analysis, you have to non-dimensionalize your equation. This means you have to ...



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