New answers tagged

2

In general, a mixed wave of the form $$ f(x,t)=\cos(x-t)+r\cos(x+t) $$ will not have nodes, at least in the sense of points $x_0$ for which $f(x_0,t)\equiv 0$ for all times $t$. In general, there's relatively little to say beyond what the picture will convey: As you turn $r$ up from 0 to 1, you first start inducing modulations into the amplitude of the ...


6

Dispersion in waves arises from both material property variation with frequency and from the geometry of the fields in question. That wave dispersion will arise from material property variation is obvious. But wave geometry and boundary conditions also matter. Simple example: a conductive waveguide with rectangular cross-section with sidelengths $a$ and $b$...


-2

A physical significance of a wave number would be the speed of a photons oscillation (frequency) times Planck's constant. Which can equate to a force strong enough to knock an electron loose or give you a sunburn.


4

Imagine something oscillating in space and time, for example a plane wave propagating across the axis $x$. This propagation is expressed via the so-called phase $$ \phi(x,t)=\omega \cdot t - k\cdot x = \dfrac{2\pi}{T}\cdot t -\dfrac{2\pi}{\lambda}\cdot x \tag{01} $$ and the magnitude of the plane wave as $$ E(x,t)=A\cos\phi(x,t) \tag{02} $$ As the ...


2

Basically it is whatever you need to multiply a distance by to find a phase difference (in radians). For a traveling wave, the wave number is the amount of phase difference per unit length. For a physical sine wave, it is the ratio between the maximal slope of the wave surface and the amplitude. In other words, it measures how dramatic the local ...


0

Actually both are analogous. When you quantize a single mode which is simply an oscillation associated with a single photon, you adopt the Lorentz model where you imagine your single photon as a mass on a spring, then you translate your quantum mechanical Hamiltonian that includes both position and momentum to phase space where the electric and magnetic ...


0

I think this is very commonly confused topic about this wave-particle thing. Let me clarify. Like the post above indicates, from Maxwell's Equations, we find evidence, in a vacuum, for the electric and magnetic fields to travel together at the speed of light, perpendicular to each other. This is simply the solution of the well known equations which govern ...


1

The difference between a fluid and a solid is the following: Fluid's have zero shear modulus, so they can't carry a shear force, but solids have non-zero shear modulus, so the can carry shear force. Fun little way to visualize this: Let's say we line up a bunch of second graders on rectangular grid. Now we push one of the students along one row. That ...


1

Treating a medium as continuous (which works for our purposes), for sound to travel through the medium the motion of one unit of it has to affect the motion of the next unit, and so on and so on. Sound in a gas or liquid does this by varying pressure: the air pressure next to a speaker suddenly increases, pushing a unit of air outwards towards the adjacent ...


0

According to this website: Sound is transmitted through gases, plasma, and liquids as longitudinal waves, also called compression waves...Through solids, however, it can be transmitted as both longitudinal waves and transverse waves. This means that through most forms of matter (gases, plasma, liquids), such as water, or the air, sound travels as ...


1

In the first case I get points where there is an oscillation in time, while in the second one, as stated, the points "do not oscillate": where there is a minimum, that minimum stays there in time and the same happens for maxima. This is not true. There is identical time dependence, of frequency $\omega$, in both situations; as I said in my answer to your ...


2

Yes, sound waves in a gas, liquid or solid can affect the light passing through it, as the motion of the atoms due to sound waves changes the atomic spacing, and this changes the index of refraction slightly. So the light would be diffracted and some amount of the light would experience a frequency shift up and a frequency shift down by the sound wave ...


6

Any physical phenomenon is potentially capable to cause some change to any other phenomenon, more or less directly. If it was not the case, the physical world could be divided into completely independent realms; there would not be the one single world we call Nature. Practically though, many if not most of the actually existing interactions between systems ...


5

I can answer half your question in that a sound can change the path of light. A change in the density of the air produces a change in the refractive index of the air and so a Schlieren photograph can make this visible. Here is a YouTube video to show a sound wave produced by clapping.


1

The answer has pretty much been given in the comments, but I think a nice pictorial representation might help. The mathematical form of a standing wave is $$y(x) = \sin \left(\frac{2 \pi}{ \lambda} x \right) $$ Here $y(x)$ is the displacement of the string at point $x$. If we plot the waves for the four wavelengths we obtain the following picture I will ...


6

My high school physics teacher was saying that “this is because of interference of sound waves. During the day, there are a lot of sounds and they cancel each other due to interference. But, during the night, there are few sounds and they can reach to our ears without canceling each other”. You need a better high school physics teacher. Temperatures tend ...


-1

If we suppose that the phenomenon you describe is related with wave interference. A wave is a kind of mechanical disturbance in the medium through which it is travelling. A sound wave consists of areas of relatively high and low energy, in the form of relatively high and low pressure. To understand how sound is produced, consider a speaker. The cone or ...


3

I would tend to agree that background noise is a factor, but rather than reducing, adding to the sound you are trying to make sense of. So part of that may be how your brain is able to filter the information from the background noise. But at night the temperature is lower and according to this tutorial on sound propagation (which does cite reliable ...


22

You don't. You actually hear the high frequency notes from headphones. The bass really doesn't travel at all well, but the attack noise from the drum or bass guitar is what leaks from headphones. This is why on the tube you hear "tsss tsss tsss tsss" and very little else. From @leftaroundabout's answer on the post that valerio92 linked: Normal ...


3

The classical interference pattern is explained by the equations governing the behavior of light, and energy there is treated as a collective phenomenon, using the Pointing vector Energy transfer in a light beam can be best understood as an emergent phenomenon from the underlying quantum mechanical level. Innumerable photons create the visible interference ...


0

Helmholtz effect occurs when pressure outside the box (i.e. external pressure change). There will be a whistling at certain frequency. The controlling parameters are aperture size and volume of the box. When a loudspeaker inside the box radiates, it just radiates with wave bouncing on the walls and some emits out of the aperture. Resonance occurs at ...


0

Additionally, the fundamental frequency can be obtained by: $$f_0 = v_s/\lambda_0$$ Where $v_s$ is the speed of sound, and $\lambda_0 = 4L$ where $L$ is the length of the air column in the tube. so, $$f_0 = v_s/\lambda_0 = \frac{v_s}{4L}$$ and so, $$f_n = \frac{nf_0}{4L} = \frac{n}{4L}\frac{v_s}{4L} = \frac{nv_s}{16L^2}$$


0

Water in a glass cup can be thought of as a one end closed organ pipe. The more volume of water you add, you effectively increase the length of the water column thereby reducing the length of the air column. The frequency is dependent on the length which can be understood from the following diagram. As you can see the length of the air column changes ...


0

By changing the amount of water in a tube you are changing the effective length of the tube. When you create a sound wave in the air inside the tube, the wave is reflected at the water surface and at the open end. With waves initially traveling in both directions inside the tube, they interfere with themselves and form standing waves. These standing waves ...


0

In a complex number $z=a+i b$,the real part is $a$ and abs is $\sqrt {a^2+b^2}$,. I am not an expert of acoustics but in general, when pressure is given in complex quantity and you want to measure its magnitude then take absolute. If you are comparing two complex numbers then equate real with real and imaginary with imaginary.


1

Difference between real and absolute value in general: Look at count_to_10 's answer. For acoustics and preasure measurement: Absolute pressure - pressure against perfect vacuum. Real pressure: Usually defined as the pressure against a reference-environment. Also called differential pressure. For example the pressure of the air inside a football against the ...


0

In a sense, the size of the universe limits the wavelength of a photon: any photon that has larger wavelength than the size of the universe, cannot exist entirely within this universe. It is not clear that this can ever be tested, however. In high energy (short wavelength) the lack of a limit to thermal radiated light was an important reason for the ...


0

Brian Dodson has posted a brief explanation to this at Quora. Basically it is suggested that "the largest energy that is sustainable as an electromagnetic wave is approximately 1 MeV, or a wavelength of 0.01 Angstrom." https://www.quora.com/Whats-the-longest-possible-wavelength-lowest-possible-frequency-lowest-possible-energy-of-electromagnetic-radiation ...


2

Let's back up. How do you know that a monochromatic wave of frequency $\omega$ doesn't contain any component at $\sqrt{2} \omega$? It's because these two frequencies are not commensurate: if you plot $\sin(\omega t)$ and $\sin(\sqrt{2} \omega t)$ they'll have no clear relation. The peaks of one look like random points in the other. Then it's clear their ...


2

To say that a wave, say with amplitude given by f(t), has a period $T$ means that not only $f(t+T) = f(t)$, but also that $T$ is the smallest value that has this property. Given that $f(t+T) = f(t)$ then it follows that $f(t + nT) = f(T)$ where $n$ is an integer (for example $f(t+2T) = f(t+T+T) = f(t+T) = f(t)$).


2

The assumptions under the statement are that A. the oscillation count in a wave is conserved and B. the passage of time is universal and uniform. Since the frequency of a wave is the count of oscillations measured within a given time interval by a stationary observer, it remains the same anywhere the wave can reach. On the other hand, the wavelength is how ...


0

You can visualize the situation like this. Lets say there is a source and an observer. They are not moving relative to each other. The source is emitting 10 peaks of wave per second (i.e. frequency is 10 Hz) and observer is observing 8 (only) peaks per time. i.e. due to medium property the frequency is changed. In this scenario 10 peaks enter the medium per ...


0

In a polaroid type of polarising material the molecules are aligned in the same direction throughout the sheet when the material is manufactured. Their electric dipoles are therefore also aligned and thus absorb photons whose electric field is parallel with the electric dipoles (to an extent that depends on cos(theta)^2 where theta is the angle between ...


1

I'm not really sure that my answer fits with your request of a "uniform" background pressure gradient, but anyway it's a related subject. Have you considered perturbations of a fluid parcel in a hydrostatically balanced atmosphere? Conservation of momentum here relates the pressure gradient to the gravitational acceleration. It can be shown that perturbed ...


2

Keep in mind that frequency is both the wave property that is preserved in a change of medium (both wavelength and velocity change) and the physical property of sound waves that we experience as pitch. So the frequencies you hear in both cases are ones produced by the vocal cords. Nor do we expect the gas environment of the vocal cords to have a large ...


3

To a very rough approximation we can say that frequencies of speech are selected by standing waves in the speakers mouth, larynx etc. If they breath helium the speed of these standing waves increases but their wave length, being constrained by dimensions of their body, remains the same. This results in higher frequency sounds produced. (think $f=v/\lambda$...


1

The formula you cite contains all you need to answer your question. The following is the procedure used to assess the suitability of a link for many microwave communication links defined by many RF link planning standards, indeed pretty much any terrestrial radio link at any frequency these days, since we no longer depend on ionospheric reflexions (used by "...


1

The sound wave is a property of the air, which does not care whether the object producing the sound was moving or not. Therefore in both cases the sound travels at the same velocity. That is the answer your book wanted, and while it is almost precisely correct, there is a slight complication. The point is that if an object in motion emits a sound which ...


1

For a travelling wave there is a phase difference between adjacent points in the medium. When two travelling waves superpose the resultant displacement of the medium is the vector sum of the displacements due to the individual travelling waves. At some points the two travelling waves arrive exactly in phase with one another and that is a position of ...


1

I think the Rayleigh-Taylor instability may be considered to be amplified under a pressure gradient. Due to the difference in densities, there is a pressure gradient across the interface which becomes unstable after a certain time. The instability grows exponentially in time according to an amplitude on the order of: $$a\propto\exp\left(\sqrt{A}\right)$$ ...


5

In general the wavenumber is a vector. That is, $e^{i(\vec{k}\cdot\vec{x}-\omega t)}$ is a solution to the wave equation in 3 (or any number) dimensions. We say this solution is a plane wave propagating in the $\hat{k}$ direction with wavenumber $|\vec{k}|$ or wavelength $\lambda = 2\pi/|\vec{k}|$. So properly the de Broglie relation is $\vec{p} = \hbar \...


0

According to huygens principle each point in the wave front acts as the source of secondary wavelet.By the time, the secondary wavelets from B travel a distance BC, the secondary wavelets from A on the reflecting surface would travel the same distance BC after reflection.Tking A as centre and BC as radius an arc is drawn.From C a tangent CD is drawn to ...


1

Remember that the air at the closed end can't move, so the amplitude at that location is always zero: it is a node for the standing wave. The displacement or amplitude will be maximal at the speaker, it is an antinode. So the length of the pipe determines the wavelength. The distance between node and antinode of a wave is $\lambda/4$, so $l=\lambda/4$. The ...


1

A simple harmonic motion is one where the acceleration (or restoring force) is directly proportional to the displacement and in the opposite direction of the displacement. For a mass $m$ on a spring with spring constant $k$, the differential equation describing the motion becomes: $m\frac{\mathrm{d}^2 x}{\mathrm{d}t^2} = -kx$ That equation has as solution: ...


1

I do not believe you can show it starting by the equation, but it becomes clear when you solve the differential equation that the two quantities are independent: $\omega$ is an arbitrary parameter on the equation of motion, and A is an arbitrary constant that appears in the solution.


1

Yes. Let's consider the phase shifts to be random. Then we can consider each amplitude to be an independent random variable $A_i(t)$. Each one has a variance of $$\text{var}(A_i) = \langle A_i^2 \rangle - \langle A_i \rangle^2 = \langle A_i^2 \rangle \propto I_i$$ because the average amplitude should be zero, and intensity is proportional to amplitude ...


0

You cannot prove that because it is not generally true, or because it depends on your definition of "equal loudness". Consider the case where the sound sources are inside a flaring pipe with a standing wave (say, a modified trumpet). A standing wave is typically drawn as a standing wave of the air displacement in the pipe (illustrated), but note that the ...


1

I remember working this out in the opposite direction, hoping to get a paradox: that the amplitude of the sound produced by a choir of 100 singers is only 10 times the amplitude of the sounds produced by one of them. Each individual sound can be represented as a vector in 2-dimensional space, with length 1 (since all are equally loud) and random phase. The ...


4

They do cancel each other when the peaks of one coincide with the troughs of the other, but since they move in opposite directions, this only happens at specific moments in time (every half period). At other moments (one fourth period later), the peaks of both waves will coincide (and so will their troughs). In the figure above, the red and green wave move ...


0

The reason the waves do not cancel each other out at every position is that to do that the waves from the two sources must be exactly out of phase at every position and that is impossible. Suppose that you have two sources $A$ and $B$ emitting waves of exactly the same frequency and wavelength at each other. Further suppose that the waves are emitted from ...



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