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Is there any reason for why a small speaker cone cannot produce low end sound at a comparable volume to higher frequencies?

I can understand how a larger speaker would be in contact with more air and so would produce a louder sound, but why specifically is this important for the low end?

As an aside, I think this phenomenon can also be seen elsewhere, such as the size increase of the violin family (violin: small/high pitched, double bass: enormous/low pitched). If you tuned the strings of a violin to double bass pitch I doubt you’d hear much!

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  • $\begingroup$ It's mostly about impedance matching, i.e.making it possible for the supplied energy to get transferred to a pressure wave in the air rather than just wobble the speaker. $\endgroup$ – Andrew Steane Mar 17 '19 at 0:22
  • $\begingroup$ Lower pitch sounds have longer wavelengths. $\endgroup$ – Gert Mar 17 '19 at 0:26
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    $\begingroup$ Same question on music.SE: music.stackexchange.com/questions/77453/… $\endgroup$ – user4552 Mar 17 '19 at 0:39
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    $\begingroup$ @StephenG: Headphones and in-ear phones seem to quite capable of producing the relative volume-frequency performance you're suggesting is a problem. Actually that's mostly a psycho-acoustical illusion. Your ear hears the higher harmonics of the bass, and your ear-brain system perceives the pitch based on the period, which correspondings to the frequency of the fundamental, even though the fundamental can't be reproduced. $\endgroup$ – user4552 Mar 17 '19 at 14:59
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    $\begingroup$ @David I can't attest to it in the case of ear-buds, but there is a reasonably common demonstration in which the present reminds the audience that the range of human hearing is roughly 20-20k Hz, and then fires up a big woofer with a signal generator dialed in to about 30 Hz. You don't hear a thing even when it's turned up to eleven. Then the presenter mixes in a little 60 Hz (or 90 or 120 Hz) signal and suddenly you hear both. It's ... spooky. $\endgroup$ – dmckee --- ex-moderator kitten Mar 19 '19 at 1:14
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To radiate sound effectively, you require the radiating object to present a good impedance match to the air surrounding it. In so doing, you will maximize the radiation resistance of the sound source. In practical terms this means the physical dimensions of the object should be approximately the same as the wavelength of the radiated sound. For example, to radiate well at 1000Hz (wavelength ~1 foot), the radiator should be about 1 foot in size- so a 12" loudspeaker will be well-coupled to the air surrounding it at 1000Hz; a 6" speaker will be well-coupled at 2000Hz, 3" at 4000Hz, 1.5" at 8000Hz, and so on. This is why the tweeters in your hi-fi speaker system have a cone diameter of about 1", the midrange will be 4", and the bass will be 12"- or as big as you can fit into the enclosure of your choice.

This rule extends to low frequencies too, but here the practicality of the scaling law breaks down: To radiate well at 100Hz requires a speaker 10 feet in diameter, and a pair of these will not fit in your living room. Instead, you make up for the poor impedance match with more cone excursion: You drive the speaker cone farther back and forth so as to pump the same volume of air that the big speaker would, only with a smaller speaker.

This is an inefficient way to get good low-frequency response, and requires the power amp that is driving the woofer cone to feed a lot more juice to it- but at the present time, power amplification is inexpensive compared to the cost of (for example) an 18" diameter loudspeaker, which needs to be mounted in an enclosure the size of a refrigerator in order to engage the air properly.

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    $\begingroup$ There's nothing incorrect here, but this doesn't explain the physics, it just asserts a rule of thumb. $\endgroup$ – user4552 Mar 17 '19 at 13:50
  • $\begingroup$ I will edit.... $\endgroup$ – niels nielsen Mar 17 '19 at 17:53
  • $\begingroup$ I still don't see any physical argument here, just the assertion that the result is true. $\endgroup$ – user4552 Mar 18 '19 at 1:57
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    $\begingroup$ @BenCrowell, why not post your own answer? then I'll delete mine. $\endgroup$ – niels nielsen Mar 18 '19 at 2:29
  • $\begingroup$ I don't have a complete answer. My comment at the top describes my best understanding, but I don't understand case #1 of the three cases I listed, and that's the most relevant one here. I'm planning to offer a bounty. $\endgroup$ – user4552 Mar 18 '19 at 13:25
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I wonder whether the impedance-matching explanation would turn out to be equivalent to one based on diffraction.

The diffraction - based explanation is that if the wavelength of the sound is considerably greater than the diameter of the speaker (e.g. for 100 Hz, $\lambda \approx 3 \text{m}$), then if the speaker cone goes forward and creates a compression in front of it, the rarefaction created at the back diffracts round to the front (and vice versa) so compression and rarefaction tend to cancel each other out.

Where diffraction is hindered, for example by mounting speakers behind large wooden boards (with holes in the middle of the boards in front of the cones), or in cabinets, the speakers have a far better bass response than unmounted speakers.

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  • $\begingroup$ could you clarify what you mean by "C and R tend to cancel each other out" (I assume you are making the electric circuit analogy here)? $\endgroup$ – wcc Mar 17 '19 at 19:29
  • $\begingroup$ A misunderstanding caused by my laziness. I've corrected my answer. There will, of course, be $some$ circuit analogy! $\endgroup$ – Philip Wood Mar 17 '19 at 19:35
  • $\begingroup$ it seems to me that diffraction describes the real physics behind the inefficiency of producing long wavelength sound with small speakers, while the term impedance mismatch facilitates thinking quantitatively about the extra work needed to produce long wavelength sound (hyperphysics.phy-astr.gsu.edu/hbase/Audio/spk2.html#c3) $\endgroup$ – wcc Mar 17 '19 at 19:46
  • $\begingroup$ This doesn't seem very plausible to me. The air inside the speaker's enclosure isn't coupled to the outside. $\endgroup$ – user4552 Mar 17 '19 at 20:37
  • $\begingroup$ @BenCrowell, I think the description in this post applies to speakers without enclosures, and enclosures actually help mitigate the problem described (as described in the HyperPhysics link I put above) $\endgroup$ – wcc Mar 17 '19 at 20:52
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The question asks about both a violin and a loudspeaker. It would make sense to imagine that these were closely analogous, and I initially thought they would be, but after researching this a little more online, I think the physics is actually completely different.

Violin et al.

I think this analysis applies to any member of the violin family, including viola, cello, and double bass. It should also apply to a guitar and other stringed instruments that have a soundboard.

I wrote up a relatively nontechnical answer to this on music.SE, so anyone who wants a less densely technical discussion could take a look at that. The instrument has a resonance of air breathing in and out through the f-holes, and also a set of complicated resonances of the soundboard, the ones that we commonly see visualized as Chladni patterns. The important point about the soundboard resonances is that the ones that are actually excited by the instrument all seem (based on the sources I've seen) to be dipoles or higher-order multipoles. These don't radiate much power when the wavelength is long compared to the size of the soundboard, because there is strong cancellation between parts of the soundboard oscillating with opposite phases.

Speaker

There are lots of possible speaker designs out there, but a common one for good audio quality is a sealed enclosure with a speaker cone sitting on a suspension and driven electromagnetically. The driver (cone+suspension, not sure if I'm getting the terminology right) has a spring constant $k_s$ and a quality factor $Q$, with typically $Q\approx 1$ for good-quality audio. In addition to the spring constant of the suspension $k_s$, we have a spring constant due to the spring of the air inside the enclosure, $k_a\propto 1/V$, where $V$ is the volume. These spring constants add, so no matter what, you can't have $k<k_a$. Therefore if you make the volume of the speaker too low, you get a big $k$ and a high resonant frequency. At low frequencies, the roll-off of the Lorentzian response of this resonator is 12 db/octave, and this ensures that if the volume is small, the response at low frequencies will be weak.

Of course this doesn't mean that no other type of design can ever evade this physical restriction. IIRC there are all kinds of interesting designs out there, including things like ribbons that work through electric rather than magnetic fields. However, my impression is that if you want realistic bass with good audio quality (good transient response and other criteria), usually the best engineering trade-off at most price ranges today is still something governed by the physics I've described above.

Even for this design, it's not totally obvious to me why you can't do things like increasing the mass of the driver to make the resonant frequency lower. Maybe that requires impractically large magnet coils?

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