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When you're driving and you open 1 car window, say the front one, there comes a horrible noise, but when you open another window just the slightest bit, this noise goes away (I'm sure most people know what I'm talking about then I mention this 'noise').

  1. Why does this noise occur?
  2. Why does it go away when another window is slightly opened?

(Not sure about the tag).

Qmechanic
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ODP
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2 Answers2

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The car is behaving like a closed pipe, so you get a resonance set up. There's a Wikipedia article here, but for once the Wikipedia article isn't that great, so there's another better article here. I imagine you (like most of us) will at some point have discovered you can make a sound by blowing across the top of an opened bottle, and it's the same thing happening in your car with the open window acting like the opening in the bottle. Since your car is much bigger than a bottle the resonance frequency is uncomfortably low.

When you open a second window you get an air current flowing through the car and this destroys the resonance.

John Rennie
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In another post of a supposedly identical question today, which was closed while I was writing the following answer, it was claimed that the frequency did not change if the opening of the window was changed. And Helmholtz resonators change frequency with the area of the opening. Therefore I post my answer here:

The excitation of air oscillations in a moving car interior due to the open window is probably due to a vortex shedding phenomenon at the edge of the open window which creates vortices and thus air oscillations with a frequency proportional to the speed of the car. This phenomenon causes the whining or howling that is emitted from objects (e.g. ropes) in strong wind. These oscillation can also couple to the resonances of standing air waves in the car interior which would enhance the effect.

freecharly
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  • I was also about to contribute a similar answer related to the Helmholtz resonance being driven by coupled vortex shedding and I believe this is the key to why the frequency does not change much with the area of the window. Given $f_s = \frac{S \ U}{D}$, $S = 0.2$, $U = 24.5 \ m/s$, $D = 4.9 \ m$, $f_s \simeq 1 \ Hz$, (which seems a bit low) but would explain the noticeably periodic 'woofing' sound of the window. – Halyn Betchkal Feb 12 '18 at 21:03
  • @D.Betchkal - Thank you for your helpful and reassuring comment. I think this is a good estimate, because the characteristic linear dimension $D$ might not necessarily be the total car length. I have experienced the phenomenon myself and think it has a higher frequency than 1Hz but still mostly infrasound. Maybe you can also vote to open the original question: https://physics.stackexchange.com/questions/385962/air-oscillation-at-open-window-of-a-moving-car?noredirect=1#comment866257_385962 – freecharly Feb 12 '18 at 22:06
  • that takes 3000 rep right? I'm a few thousand shy. – Halyn Betchkal Feb 12 '18 at 22:16
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    The frequency of the Helmholtz resonator scales with $\sqrt{V/L_{eff}}$, where the end correction for the finite aperture is $\frac{L}{L+0.3D}$ and the value $D$ is approximately $2h$ where $h$ is the size of the crack of the window (when $h \ll w$). It might be instructive to calculate the expected shift in frequency, and ask whether a casual observer could detect that difference. – Floris Feb 12 '18 at 22:21
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    Doesn't frequency scale with $\sqrt{\frac{1}{V \ L_{eff}}}$? Depending on the value of $L$ (which seems quite mysterious in the case of a car's window) $\Delta f$ is definitely large enough to detect. For $L = 0.1 \ m$, $\Delta f = 359 \ Hz$. Even for an unreasonably large $L=1 \ m$, $\Delta f = 72 Hz$, which is much greater than the Just Noticeable Difference (JND) for the human ear. Treating the car as a simple Helmholtz resonator doesn't appear to be consistent with the observation. – Halyn Betchkal Feb 13 '18 at 01:27
  • @Floris - The resonance frequency of the simplest spherical Helmholtz resonator of diameter $D$ with no neck and just a sound hole of diameter $d$ is given by $$f=const \times v \sqrt \frac {d}{D^3}$$ where $v$ is the sound velocity. See Wikipedia https://en.wikipedia.org/wiki/Acoustic_resonance#Resonance_of_a_sphere_of_air_(vented) Thus the Helmholtz resonator resonant frequency is proportional to $\sqrt d$ and consequently depends significantly on the hole size. – freecharly Feb 13 '18 at 02:06
  • @D.Betchkal - The resonance frequency of the simplest spherical Helmholtz resonator of diameter $D$ with no neck and just a sound hole of diameter $d$ is given by $$f=const \times v \sqrt \frac {d}{D^3}$$ where $v$ is the sound velocity. See Wikipedia https://en.wikipedia.org/wiki/Acoustic_resonance#Resonance_of_a_sphere_of_air_(vented) Thus the Helmholtz resonator resonant frequency is proportional to $\sqrt d$ and consequently depends significantly on the hole size. – freecharly Feb 13 '18 at 02:06
  • @freecharly it is hard to believe that formula - it would mean that as the aperture goes to zero, so does the resonant frequency. Does that seem right to you? – Floris Feb 13 '18 at 03:19
  • @Floris - That's indeed astounding. Maybe, we should consult the references. Actually I am thinking more of the resonant frequencies of the closed box $$f=v\sqrt {(l/L_x)^2+(m/L_y)^2+(n/L_z)^2}$$ as the simples resonant cavity for a simple estimate of the resonant frequencies of the car interior, as long as the window opening is comparatively small. Actually I found an article where the resonant frequencies were measured https://www.linkedin.com/pulse/when-car-turns-resonance-chamber-signal-analysis-case-jose-i-rey – freecharly Feb 13 '18 at 03:59