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I think I did not get this completely, in particular how is the existance of the privileged frame translated in a different physical phenomenon in the cases in which the observer or the source move? Imagine a case where one of the two is at rest with respect to the medium and the other is moving away at $c$. If the emitter is moving, the receiver ...


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The only thing which matters is relative positions or velocities of two bodies. If you have only two bodies in a vacuum, then it doesn't matter which one is moving - what matters is their relative velocity. If you now have three bodies, say A, B, C then still what matters is their relative velocities. "A is moving 10 km/h while B and C are staying" is not ...


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No, the frequency will not change. If the wind is blowing at constant speed and the distance between source and observer remains constant, then the time it takes for a sound wave to get from source to observer will be constant. So the time interval between wave peaks (period T) when they are detected by the observer remains equal to the interval between ...


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Your errors are 4 Hz, plus and minus. If that was 1 sigma (standard deviation), that'd be 4 in 40000 approx, or 10^(-4). Check the specs for your counter, that may be somewhere around its accuracy. It could also be a bias in the counter, since plus or minus numbers are symmetric. Errors can be due also to the temperature measured (though 1 degree error ...


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The error for both directions is very similar - approximately 4 Hz - and with a consistent sign (measured frequency is 4 Hz closer to the actual source frequency than the "theoretical value"). From this I infer that most likely your velocity measurement is very poor, or the alignment of velocity vector and sound vector is very bad. If your microphone is not ...


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Frequency can be viewed as rate-of-change of phase, so in the formula there are 2 position at which the mobile device observes the transmitted wave field. There is one position at $t=0$ and another a $t=\Delta t$, with respective phases $\phi=\phi_0$ and $\phi=\phi_0+\Delta \phi$, giving an additional rate-of-change of phase $\frac{\Delta\phi}{\Delta t}$ ...


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This question raises several concerns. From a purely signal processing point-of-view, where the aforementioned pulse is not a photon, but rather a time-domain pulse of a certain duration (here less than 10^-100 seconds)--then NO: you cannot have pulses shorter than the period of wave in question--as the wave an be defined. Consider sound: when a 440Hz A note ...


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Yet another way. Consider a wave of a form $f(kx-\omega t)$ in the emitting frame, being observed as some other wave $f^{\prime} (k^{\prime} x^{\prime}-\omega^{\prime} t^{\prime})$ in a frame moving at $v$ along the x-axis. We actually don't care what $f^{\prime}$ is, only the argument of the function matters. So use the Lorentz transforms to change $x$ and ...


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Neither origins nor emitters nor receivers are required: consider a photon with momentum $\vec{p}$: it has 4-veclocity $p_{\mu}=(||p||/c, \vec{p})$. To compute what another observer moving at $\vec{v}$ sees, do a Lorentz transform: $p'_{\mu} = \Lambda_{\mu}^{\ \nu}p_{\nu}$. From this you'll get the relativistic (time-dilated) Doppler effect and stellar ...


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The motion does affect the observed continuous spectrum. The flux at each wavelength is doppler shifted to a new wavelength. The result is of course a new continuum, so a doppler shifted continuous spectrum is rather hard to distinguish from one that is not. In particular, unless the continuum has some feature or break in its slope, then it can be ...


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That's correct. Imagine a sound emitter and receiver traveling at twice the speed of sound in the same direction with the emitter behind the receiver. Even though the relative velocity is zero, the sound will never reach the receiver because sound doesn't travel fast enough. At this point, the frequency calculated from Doppler shift formulas would have ...



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