71

But that still means the ear is only moving at one frequency at any given time No, it doesn't mean that at all. It means the eardrum is moving with a waveform that is a superposition of all the frequencies in the sound-wave it is receiving. Then, within the inner ear, hair cells detect the different frequencies separately. It is entirely possible for ...


20

A photon cannot lose all of its energy by Compton scattering, as that would violate conservation of four-momentum. Imagine a photon with four-momentum $(p,\vec p)$ gives all of its energy (and thus all its momentum) to an electron with four-momentum $(m,0)$, in $c=1$ units. Then by conservation of four-momentum, the new four-momentum of the electron would be ...


14

Given any photon, there is a frame of reference where it has any energy you like. Arbitrarily high or low. There may be a wavelength so long and an energy so low that you cannot detect it. But it is still a photon. If it has an arbitrarily low energy and momentum and it scatters off a particle at rest in your frame, it will not deflect the particle very much....


13

the human ear separates out and detects all the frequencies within its range individually (in parallel) in real time, and sends that decomposition to the brain along a bundle of nerves. The eardrum responds to the instantaneous sum of all the different frequencies impinging upon it. That complex amplitude sum looks like a crazy squiggle which moves the ...


9

It may help your thinking to distinguish the term "photon" from the term "light pulse". The concept "photon" comes from quantum physics, and it refers to the degree of excitation of of the electromagnetic field at some given frequency. This means that a photon cannot change frequency, by definition. But in a process such as ...


6

What happens to a photon when it loses all its energy? A photon is by definition a quantum mechanical particle of mass zero, spin one, and energy equal to $hν$ where $ν$ is the frequency of the classical electromagnetic wave that a lot of photons of that energy will build up. It follows special relativity rules, i.e. it is described by a four-vector and ...


5

So too much frequency and you do lose the ability to decipher it and it starts to just sound like nosie? If that happens, it's because of how your brain interprets the signals that it receives from your ears, and not because of the physics of how your ears work. If the total sound pressure level is not so great as to damage your hearing, then your ears will ...


3

Here is the Feynman diagram for Compton scattering: A Feynman diagram is the pictorial representation of of the integral needed to calculate an interaction crossection, and depending on how you read the axis it describes the corresponding interaction. If the x axis is taken as time, it describes the scattering of an electron on a photon, and a new photon ...


3

As you mentioned, all the different waves get added together in a Fourier series. The hair cells in the inner ear are essentially performing a Fourier analysis of this combined wave, splitting it back into its component frequencies. The amplitudes of each frequency are then sent to the brain, which performs higher level analysis to recognize specific types ...


2

"Radians" aren't really units. E.g. $\frac{\pi}{2}$ (or $90^{\circ}$) is just a dimensionless number. That's why in physics/math we prefer "radians" to degrees. $\mathrm{radians/second}$ is thus really $\mathrm{s^{-1}}$, in $\text{SI}$ units. Your expression: $$x=\left(1+\omega_{0} t\right)e^{-\omega_{0} t}$$ cannot refer to a distance, ...


2

can wave frequency in fact influence its velocity in a medium? Yes. This phenomenon is called dispersion and media in which dispersion occurs are called dispersive. Dispersion can lead to all sorts of interesting effects, including most familiarly the production of rainbows.


2

This is a partial answer because I am not going to discuss the many different ways it is possible to change photons' frequency. I want to emphasize an issue related to the way the question has been formulated. I think it may be useful to appreciate the meaning of changing the frequency of a photon, whatever method is used. Photons are definitely neither like ...


2

When a photon interacts with an atom, three things can happen: elastic scattering, like mirror reflection, this is Rayleigh scattering, the photon keeps its energy and phase, and changes angle https://en.wikipedia.org/wiki/Rayleigh_scattering inelastic scattering (your example, Compton scattering), the photon gives part of its energy to the absorbing ...


1

If you have a general oscillator with restoring forces and damping forces: $$F_{rest}=-kx,\text{ } F_{damp}=-b\frac{dx}{dt}$$ Such that the equation of motion would be: $$m \frac{d^2x}{dt^2}=-b\frac{dx}{dt}-kx$$ $$m \frac{d^2x}{dt^2}+b\frac{dx}{dt}+kx=0$$ This equation has solution of the form : $$Acos(\omega x + \theta)e^{t/\tau}$$ Where you can ...


1

It depends on what you mean by "at a time". If you mean an exact instant of time, no frequencies can be detected because frequencies are distinguished by how they change over time. However, research has shown that times under around 100 ms are perceived by the human brain as "at the same time". For a 20 kHz signal, that means that there ...


1

The short version: that's not how the ear works. The eardrum resonates in sympathy with all vibrations in the air, within its physical constraints. It transmits those vibrations through three small bones (named the "hammer, anvil and stirrup" because of their shapes) into a fluid-filled, spiral-shaped chamber called the cochlea. The cochlea is is ...


1

The human ear doesn't hear one frequency at a time. A way to physically see what the ear is experiencing is to take a pool of water. Now make some waves. Splash around. See how the entire tub gets full of up and down motion all over the place? In the air, similar things happen; but instead of "up and down" those waves correspond to changes in ...


1

"increasing power doesn't change s.p.". This statement is misleading. It should read "increasing the power of the same wavelength of monochromatic light doesn't change the stopping potential. The standard explanation is that by increasing the power in this way, we are not changing the energies of the individual photons, but simply increasing ...


1

$\omega=2\pi f$ so angular frequency depends only on frequency. And frequency depends only on the source that produces it, that is, it is not a function of time or position. So angular frequency is constant.


1

I think it’s boundary conditions—an assumption that the wave is continuous: If something is wiggling at a frequency when it hits an interface, it wiggles at that frequency on “both sides” of that interface.


1

What justifies is that people did it, and found that the reflected signal was at the same frequency, every time (almost). Science is, at the heart, an empirical process. What we find is that the way sound operates and interacts with mediums is defined by a wave equation: $\frac{\partial^2u}{\partial t^2}=c^2\frac{\partial^2u}{\partial x^2} $ Whether you ...


1

Buy a bandpass filter, which transmits around 315nm. A bandpass filter is a "glass plate", which absorb "all" wavelength except some within a certain wavelength range. Look at the transmission spectrum of the bandpass filer and check whether or not this is acceptable.


1

Use a prism to disperse the light by wavelength, and then just put the sensor where the UV part of the spectrum falls, or physically block out the wavelengths that you don't want.


1

Your doubt is common; we should take $2X$ for the change in the potential energy, but potential energy is given by change in the position of C.G. of body, and since it is a uniform, cross-section its C.G. is half of its length i.e. $.5*2X$. If you are not satisfied, then integrate the potential energy. You might remember the work done problem as well, where ...


1

Great thread! But there's more to it with harmonicas (and I don't know the full details). You can see this because different holes bend more than others...and the amount of possible bend is related to the pitch of the other reed at that hole. Hole 2: Draw/inhale, can bend 2 semitones. Hole 2 blow/exhale is 3 semitones down. Holes 4, 6: Draw can bend 1 ...


1

Actually.... Regardless of the volume of the "bell" changing due to the vortex created, (its a very small factor), from what I understand, its actually the movement of the tea, or coffee itself. That movement changes the frequency of the sound wave created as it travels through the liquid. I'm not sure if its technically the same as the Doppler ...


1

If you have light that is reasonably monochromatic, then you can shift its frequency using two types of related devices: acousto-optic modulators (AOMs) and electro-optic modulators (EOMs). Strictly speaking, these fit within the categories of ptomato's answer as nonlinear optical processes, but they are distinct enough that they are typically considered on ...


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