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Your interpretations all make two basic mistakes. You assume the recorded data was mathematically accurate, and the FFT algorithm used somehow produces "exact" results. Some of the "broad spectrum" at low frequencies is most likely just environmental background noise. The signal to noise ratio compared with the peak amplitude is around 40 ...


5

What is the energy spectrum of all photons in the observable universe? All photons means: photons from the cosmic microwave background. It is the best black body radiation spectrum known at very low temperature. 2)photons radiated by stars in galaxies. The approximate black body spectrum of stars is in the thousands of kelvin, similar to our sun, the ...


5

Two quantum mechanical systems which are defined on the same Hilbert space (e.g. the free particle on a line and the harmonic oscillator, which are both defined on $L^2(\mathbb R)$) are distinguished only by which operator is chosen to be the Hamiltonian. It follows immediately that any procedure which gives you information about the set of energy ...


3

These are the definitions of the two spectral radiances. $B(\lambda, T)$ is defined by stating that the intensity emitted between wavelengths $\lambda_1$ and $\lambda_2$ is $$I = \int_{\lambda_1}^{\lambda_2} B(\lambda, T)\ d\lambda, $$ and similarly for $B(\nu, T)$. Or, in physicist language, the intensity in a small interval $d\lambda$ is $B(\lambda, T)\ d\...


3

The amount of radiance in an interval of the electromagnetic spectrum is the same regardless of whether that spectral interval is described in terms of a range of wavelengths or a range of frequencies. For example, if you asked how much radiance is in green light, it’s the same whether you define green as being 500-565 nm or 530-600 THz. These are the same ...


3

"Decimated" is another word for "downsampling," or in other words resampling the original ("undecimated") data with a lower sampling rate. So the decimated data will have a lower sampling rate than the undecimated data. Edit As pointed out in the comments, it is a very good idea to low pass filter the data before downsampling, ...


2

The frequency is, as always, the number of cycles per second of the oscillations. It is related to the spatial wavelength by $f = \frac{c}{2 \pi} |\vec{k}|$, where $c$ is the speed of propagation of free waves in the medium and $$ \vec{k} = (k_x, k_y, k_z) = \left( \frac{2 \pi}{\lambda_x}, \frac{2 \pi}{\lambda_y}, \frac{2 \pi}{\lambda_z} \right). $$ and so ...


2

"Dispersion occurs when pure plane waves of different wavelengths have different propagation velocities, so that a wave packet of mixed wavelengths tends to spread out in space. The speed of a plane wave, v, is a function of the wave's wavelength $\lambda$" The differential equation of a wave is: $$\frac{\partial^2 A}{\partial t^2}=v(\lambda)^2\,\...


2

If you have a point and you are standing at the point, when a wave passes through you will see $f$ number of waves (peaks and throughs) pass through the point in some time, say $t$. The wavelength tells you the distance between two consecutive peaks (or throughs, depending on how you want to define wavelength). Throughout this answer, we assume that you can ...


2

We don’t assume $\omega^2=k/m$: this follows from the solution. The differential equation you have can be written as $$ \ddot{x}=-\frac{k}{m}x\, . \tag{1} $$ Now, what functions have the property that their 2nd derivative is a negative multiple of the original function, as in (1)? The trig functions have this property to try $x(t)=A\cos(\omega t)+ B\sin(\...


2

The solution of your differential equation is: $$x(t)=A\,\sin\left(\sqrt {\frac{k}{m}}\,t+\varphi\right)$$ just for convenience we defined $$\omega:=\sqrt {\frac{k}{m}}\quad \text{unit}\quad \frac{1}{\text{s}}$$ and name it "angular frequency" with: $$\omega=2\,\pi\,f$$ you get the frequency $f\quad $ ( Unit [Hz]) of your sine wave which ...


1

They characterize the response in one of the circuits triggered by the signal in the other. E.g., if $F(\omega)$ has only one component: $$F(\omega) = (f_1(\omega),0),$$ Then the responses in the two circuits will be given by $$x_1(\omega) = \Theta_{11}(\omega)f_1(\omega),\\ x_2(\omega) = \Theta_{21}(\omega)f_1(\omega),$$ where $\Theta(\omega)$ is the matrix ...


1

Inductors have a DC resistance, as others have mentioned. But simply assuming that $\omega L = <V>/<I>$, where $<V >$ and $<I>$ are the RMS values read from a multimeter, ignores the phase factor introduced by the resistive ($R$) and reactive ($\omega L$) parts of the impedence. This phase factor introduces a trigonometric factor into ...


1

If the coil has a finite DC resistance it will present what looks like inductance at zero frequency. this will cause your line to miss the origin by an amount that equals the DC resistance of the coil. I recommend that you use frequency increments that are spaced thusly: 500-1000-1500-2000-3000-4000-5000-6000-7000-8000-9000-10,000 and see what the linear fit ...


1

A photon, by itself, doesn't have a frequency or an energy because it has no rest frame. It just "is". When you pick a rest frame, then (ignoring polarization), it is described by a 4-wave vector: $$k^{\mu}=(\omega/c, \vec k) $$ so that the frequency is $f = \omega/2\pi$, the direction is $\hat k$, and the wavelength is $\lambda = 2\pi/k$. That's ...


1

if $\omega$ is the frequency of an ac signal, then the inductive and capacitive reactances are given by $X_C = \frac{1}{C\omega}$ $X_L = L\omega$ You know that at resonant frequency, $\omega_0$, both $X_L$ and $X_C$ are equal. Now if we increase $\omega$, $X_C$ decreases and $X_L$ Increases from the same value. Opposite is true when we decrease $\omega$ from ...


1

My understanding is that we are talking here about the molecules tapping on the eardrum, i.e. the shot noise - similar to the noise of the rain drops. It is a white noise, i.e. its spectrum has the same amplitude at all frequencies. On a deeper level however there are at least two characteristic timescales that would limit the width of the spectrum: the ...


1

You ask about which definition is most useful. This is the right way to ask the question. The answer is as follows. For $\beta \ll \omega_0$ you can see that $\omega_r \approx \omega_0$. Under these conditions we have that both definitions of $Q$ are approximately equal, that is, $Q_r \approx Q_0$. We also have, under this condition, that $Q_r, Q_0 \gg 1$. ...


1

As you will have found, there is more than one way of defining what is meant by the resonance frequency. That for which the amplitude is a maximum, and that for which the peak power dissipation is a maximum are two favourite ones. For amplitude resonance, $\omega_{res}^2 =\omega_0^2-2\beta^2$, whereas for 'power resonance', $\omega_{res} =\omega_0$. Power ...


1

You're talking about waves. In the real world, waves are created by something that moves back and forth for awhile, and then it quits. That is, things are mostly wavelets. When we talk about waves, we are talking about something that keeps oscillating back and forth long enough that it makes sense to think about it continuing, and we can spend at least part ...


1

Oh this is a fun question. It took me a few readings to “get” it though. The basic relation: $v(f, t)$ So if we were to actually reconstruct what is happening from first principles, we would probably instead say something like this: “Dispersion is when the speed of a wave through some medium depends on its frequency, $v = v(f)$ where $f$ is the frequency of ...


1

$v(\lambda)=\lambda f(\lambda)$ is usually written $V=\lambda F$, so $F=V/\lambda$. So your statement "The function f(λ) tells us that the frequency of a wave depend solely on its wavelength" is wrong. Frequency depends on the propagation speed and the wavelength. If the wave is a E/M wave then $V=c$, a constant, and the statement would be true.


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