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90

What you cannot see by drawing the picture is the velocity of the individual points of the string. Even if the string is flat at the moment of "cancellation", the string is still moving in that instant. It doesn't stop moving just because it looked flat for one instant. Your "extra" or "hidden" energy here is plain old kinetic energy. Mathematically, the ...


48

ACuriousMind's excellent description was missing a picture. Here it is: This clearly shows that for the wave moving to the right, the front is moving up and the rear is moving down. For the opposite wave traveling to the left, the front (now on the left) is moving down and the rear is moving up. Summing them, you get a straight line with significant ...


25

Just to complement the other excellent answers, here's an animation showing what two wave pulses with opposite amplitude passing through each other actually look like: You can clearly see that, at the instant when the string is momentarily flat, it's not stationary but rather moving quite rapidly, and thus will not stay flat for long. (Obviously, the ...


15

Your question seems to imply that you think that, over a deep-ocean tsunami where the wave is travelling at speeds in excess of the speed of sound in air, the water in the wave is travelling faster than that speed of sound. This is not the case. The wave is travelling faster than the speed of sound in air, but the wavelength of the wave is so long that the ...


14

Dispersion of sound in air, with constant temperature and pressure, is very slight, increasing for very short wavelengths, and for very loud noises. Why? Because the rapid sequence of weak compression/decompression steps as the sound propagates are adiabatic, or energy-conserving, for the normal ranges of sound. This leaves the local pressure, temperature ...


8

One must distinguish the underlying quantum mechanical framework from the emergent classical mechanics and electrodynamics framework when discussing waves. In classical mechanics wave equations are solutions of differential equations which depend on the $(x,y,z,t)$ variables of massive ideal particles which are derivable from differential equations . These ...


8

As the energy of the electrons in that case is much greater than their mass, you can consider the approximation $E \sim pc$. So the formulas are equivalent.


7

Often, when dealing with high-energy (relativistic) particles the rest mass of the particle can be neglected when performing calculations. Use your expression for $p$ from relativistic considerations, plug in the numbers and see the negligible change when you include and neglect to include the mass of the electron. A good tip for when you enter into higher ...


5

Take a look at Griffiths Introduction to Electrodynamics, particularly the section called "The Frequency Dependence of Permittivity". Dispersion can arise from the constraints, or bound nature, of the constituent particles in a given medium. For the example of optical dispersion in a dielectric medium, we could picture the electrons as bound, damped ...


4

In a strict, linear view of the world, waves simply transfer energy through space or media. But the world is almost never linear, and if one looks closer mass will be set in motion or transported with the wave energy. Examples? Ocean surface waves. Although energy is transported across the surface, one can trace local elliptical transport of water molecules. ...


4

Let's take the argument of the function i.e, $kx-\omega t$. The argument of the function should remain constant,(equivalently the phase must remain constant)for a particular section of the wave. \begin{equation} kx-\omega t=\lambda \end{equation} where $\lambda$ is a constant. Differentiating both sides we get, \begin{equation} k\frac{dx}{dt}=\omega ...


4

The waves, the particles are associated with, are purely of mathematical construct. The wave or more formally the wavefunction assigns probability amplitude at each spatial coordinate at a specific time; the square of which gives the probability of finding the particle at that coordinate at that specific time. These waves don't transport energy, charge or ...


4

At least two reasons: The instantaneous energy density tells you nothing about the total energy content of the wave, so it's of little utility. Experiments measure the average energy.


4

This wikipedia article is written like the perfect answer to your question: https://en.wikipedia.org/wiki/Hull_speed The answers it provides are: Question 1: When the wavelength of the bow wave is equal to the length of the ship, the bow wave interferes constructively with the stern wave, causing taller waves in the wake and thus more energy radiated in ...


4

The electric field strength is related to the power of the laser by the Poynting vector. This is given by: $$ \mathbf{S} = \mathbf{E} \times \mathbf{H} $$ and the magnitude of $\mathbf{S}$ is the power. Assuming we can treat your laser as a plane wave (which seems reasonable) then $\mathbf{E}$ and $\mathbf{H}$ are at right angles so the power is simply: ...


3

Well spotted. The average intensity of the whole fringe system must be $1+4=5\;\mu$W. However there will be places where the intensity is a maximum, $1+4+2\sqrt{1\times4} = 9\;\mu$W and places where the intensity is a minimum, $1+4-2\sqrt{1\times4} = 1\;\mu$W. Your chosen position is one which is nearer a maximum than a minimum. Later Here is a ...


3

In short, yes, it will be louder. In the simplest case, if you were able to duplicate the exact signal everywhere in space, you would actually get 4 times the intensity - sound waves add linearly, but intensity adds quadratically. For two uncorrelated sources (if you played different white noise signals through each speaker) you would only get a factor ...


3

The first equation correctly states that $$ v_g = \frac{c}{n}\left(1 + \frac{\lambda}{n}\frac{dn}{d\lambda}\right). $$ But if you look at the wikipedia page that you linked to, you'll see that the second equation should read $$ v_g = \frac{c}{n}\left(1 - \frac{\lambda_0}{n}\frac{dn}{d\lambda_0}\right)^{-1}, $$ where $\lambda_0$ is the wavelength in vacuum, ...


3

The momentum of an electron, which is not travelling at very high velocity will not have any relativistic effects. So, its momentum is given by $$p=m_0v$$ where $m_0$ is the rest mass of electron ($9.1\times 10^{-31}~\rm kg$) and $v$ is the velocity. But to observe phenomena like diffraction (which observed with radiations like X-rays), the energy ...


3

The 4-momentum vector is given by ${\bf p}=(\frac{E}{c},p^{1},p^{2},p^{3})$. Now taking the scalar product with itself we have, \begin{equation} {\bf{p.p}}=E^2-(pc)^2=m_{0}^2c^4 \end{equation} Now for extremely relativistic case , we can use the condition that $E\gg m_0c^2$, thus this yields $p=\frac{E}{c}$.


3

This is a common misconception about what boundary conditions do and how they do it (for example here). You discussed two types of boundary conditions, Neumann and Dirichlet. In Neumann boundary conditions, we impose that the derivative of the variable normal to the boundary is specified, generally to be zero. With Dirichlet, we impose the value that the ...


3

Mechanical waves are those that need a specific medium for their propagation. Of course, the medium has to be physical. There are two types of mechanical waves: transverse and longitudinal. But, first of all, electromagnetic waves that show transverse property are not mechanical waves since they don't essentially need any medium for their propagation. The ...


2

Those two equations tell you 1) The speed of a wave on a string depends on only tension and density of the medium, not the frequency of the source. 2) IF the frequency of the source if $f$, you can find the wavelength by $\lambda = v/f$. High frequency sources produce shorter wavelengths, and vice versa. You're NOT free to choose both the wavelength and ...


2

The tension of the string is a constant, if there is no vibration on the string. A wave is produced on the string when you give an unbalanced force on the string which varies the original tension of the string. The velocity of the wave now depends on the value of the tension. The given equation is valid only for small amplitude vibrations. The tension is ...


2

When deriving the wave equation we assume the horizontal component of the tension in the string is constant and equal to $T$ (the tension when the string is at rest). To calculate the tension in the string let's start with the wave then zoom in to a small segment of it. If we take a segment small enough that we can consider it as a straight line, then the ...


2

Suppose an x-axis which points to the right. A wave produced by a source travels to the right. The displacement due to the wave at position $x$ at a time $t$ is $y = f(kx \pm \omega t)$ where $f(kx \pm \omega t)$ is a function which satisfies the wave equation. It might help your visualisation if you think of that displacement corresponding to a peak. ...


2

Do u really think that velocity is constant?i think in this equation nothing is constant.if tension increases per unit mass decreases and it may make change in velocity or not if the ratio remains the same.further tension depends on some variables such as intermolecular force,elasticity etc


2

I don't fully comprehend why the positive (and therefore quicker) parts appear to be retarding and eventually making a N-like sawtooth wave? I wrote a more detailed answer at http://physics.stackexchange.com/a/139436/59023, but the basic idea is that the larger amplitude parts of the wave have a higher phase velocity than the lower amplitude parts. ...


2

There are two types of radiation that are emitted from an accelerating charged particle: synchrotron radiation (if the acceleration causes circular motion) and bremsstrahlung radiation (if the acceleration is from speeding up or slowing down). The total power emitted from synchrotron radiation is given by: $$P = \frac{q^2\gamma^4a^2}{6\pi\epsilon_0c^3}.$$ ...


2

For a general answer concerning these types of systems, we can take the simple gravity pendulum as a representative example. The equation of motion is generally non-linear, but can be approximated to be linear at small angles (small amplitudes). See also the small angle approximation section in the linked wikipedia article. For progressive waves you will ...



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