# Tag Info

2

For general audio programming or playback, 96kHz or 192hHz is simply useless. Indeed, the Nyquist theorem tells you that a signal can be exactly reproduced given that the sample rate is greater than the highest frequency contained in the original signal. The "excuse" of the slope of analog filter required after digital to analog conversion is no longer ...

1

because the higher the sampling rate is the sloppier the (annalogue) filtering preceding the sampler can be to reduce the aliased noise/interference

5

If you consider your photon as a point object, it cannot bend its own path. It will always travel on the ridge it creates, speaking in terms of curvature of space. The other idea is possible. Two photons having a momentum, attract each other, trapping each other, like a positronium (typical example for this behavior). In the model of relativity this is ...

2

They control the spectrum with regulations: Below is the diagram of frequency allocations in the united states. When people think of radio, they typically think of only the FM or AM parts of the spectrum. However, any frequency of electromagnetic radiation can be used to communicate with others. There is nothing that a government can do that could ...

0

They have both frequency and wavelength (or wave vector $\mathbf{k}$), but the frequency and the wavelength are not related as in a real photon. Instead, they are independent integration variables of a Fourier integral representing a near field. For example, you can represent a "moving" Coulomb field $V(\mathbf{r}(t))$ as a Fourier integral over $\omega$ and ...

0

Virtual photons are not on the mass shell, that means that their mass can be different than zero in the calculations using the Feynman diagrams. They are not observable in any sense other than as the result of the calculations that need them. They appear because in an interaction quantum numbers are conserved, and these photons are used to carry the quantum ...

1

The speed of sound should apply to $v$ because the sound waves are travelling through the air after it leaves the organ pipe. The speed of sound is approximated by the following formula: $$v = 331.3 + 0.606T$$ Where $T$ is the temperature in degrees Celsius, and $v$ is the velocity in meters per second. In your case, suppose you're at room temperature ...

1

It is best to first simplify the situation and think about why is the frequency of the wave the same as the source. Considering the case of a string attached to an oscillator, if the frequency of the oscillator and the wave were different, there would need to be a discontinuity in the string. The fact that the string is attached to the oscillator means ...

2

According to Maxwell's theory of electromagnetism, a light pulse (or generic electromagnetic wave) carries momentum, which can be transferred to an absorbing surface hit by the pulse. This momentum transfer is known under the name 'radiation pressure'. Despite carrying momentum, light carries no mass. Yet a light pulse does carry energy. For a light pulse ...

-3

The total energy of a photon, the carrier of the electromagnetic force, is given by $E=hf$ where $h$ is Planck's constant and $f$ is the frequency of the light. So yes, if two EM waves have the same energy, they will have the same frequency and wavelength, meaning they have the same colour. Photons have no rest mass, but they do have a relativistic ...

1

No, frequency does not have an impact on intensity. If you treat sound waves from a Planck view, the energy of each quant (phonon) is neglectably small: such low-frequency waves behave totally as classical waves, which are made up of zillions of coherent quanta. Changing the precise number (for a given frequency) changes the intensity, obviously. Changing ...

3

Assuming you're talking about propagation through free space, the beam will be diffracted by the aperture you pass it through (3cm in this case) and that will cause the beam to diverge. The far field angular divergence, $\theta$, is approximately given by the equation for the Airy disk: $$\sin\theta \approx 1.22 \frac{\lambda}{d} \tag{1}$$ where $d$ is ...

0

Your two proposed approaches are the ones usually followed by researchers in acoustics. There is a third one that would be an experimental approach. For example you could place a microphone inside the tube and move it to a large number of positions to construct a sound pressure map. The first approach you mentioned is comparable to a lumped elements model ...

0

The very definition of a frequency is the rate at which something occurs over time. Therefore by shortening the time interval in which something occurs you are increasing its frequency, and by increasing the time interval you decrease the frequency. Hence, Frequency is inversely proportional to time.

1

First, the context is a function of time that is periodic which means that it is repetitive with repetition period $T$. $$g(t) = g(t + T)$$ So, if one sampled the function every $T$ seconds, one would get the same value each time. Now, we have the period of time $T$ which tells how long it takes for the signal to go through one cycle. The inverse ...

3

To answer the question "same wavelength, different frequencies, which arrives first?": naturally, the one with biggest speed, which is proportional to frequency AND wavelength according to the formula: $$v = \lambda f$$ So, for the same wavelength $\lambda$, the one with bigger frequency $f$ will have bigger speed $v$, thus arriving earlier. For the ...

1

Here is the bookwork answer. Consider the boundary between two media. Draw a rectangular loop of side $\delta x$ and $\delta y$. Have an E-field either side of the boundary that is parallel to the boundary. The E-field is $E_1$ in medium 1 and $E_2$ in medium 2. Now use the integral form of Faraday's law.  \oint {\bf E} \cdot d{\bf l} = - \int ...

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