2
$\begingroup$

I've read other answers about how vinyl records reproduce sound, but they don't quite address a main thing I'm curious about.

Play Middle C on a piano or a Clarinet, its the same note that can be played at the same volume because the main note is the same. You can tell the instruments apart because there is a different timbre in the instruments due to the components of the other harmonics present in the sound that differ between the instruments. You can break this down using Fourier Analysis.

Part of the reason for these differences is the natural frequency of the materials involved. Strike metal with a small hammer, you typically get a higher pitched sound than if you strike a similar volume of wood.

A vinyl record has its own natural frequency, yet it can sound just like instruments neither of which sound like each other. Somehow the natural frequency of the vinyl doesn't matter. What's going on there?

$\endgroup$
4
  • 4
    $\begingroup$ You are playng with words here. When you listen to sound recorded on a vinil disc you don't listen to the sound produced by mechanical excitation of the body of the disk. The vinil does not "sound", it just store the information about the recorded sound. $\endgroup$
    – nasu
    Jun 2 at 17:42
  • $\begingroup$ You're right, but I think this kicks the can down the road. How is any part of the apparatus having natural frequencies distinct from the instruments each very different from each other yet able to reproduce sounds from those instruments? I guess ultimately anything that vibrates the air the right way will reproduce those sounds. In that case the most important part of the system is the vibrating membrane responsible for agitating the air. I'm guessing somehow that membrane is able to reproduce the specific sounds of the instruments. $\endgroup$
    – R. Romero
    Jun 2 at 18:08
  • 2
    $\begingroup$ Any object can vibrate at any frequency, no matter what are it's resonant frequencies. The relevant vibrating object here is the pick-up tip but the motion of this tip is not a free vibration but forced vibration, determined by the pattern etched in the vinil. Free vibrations are relevant here only for the designer who make sure they are overdamped and so play no role. $\endgroup$
    – nasu
    Jun 2 at 21:16
  • 2
    $\begingroup$ To complement @nasu's comments, you can think of a loudspeaker. The cone (and the extremely complicated systems comprised of all the parts of the loudspeaker, from cabin to coil) has its resonances, yet it can (and in some cases very well) reproduce a very large spectrum of frequencies. As nasu again stated, any mechanical system can be excited at any frequency. The resulting amplitude (of displacement, velocity, acceleration or any other derivative) does depend on the material and geometry (among other things) though. $\endgroup$
    – ZaellixA
    Jun 2 at 23:22

2 Answers 2

7
$\begingroup$

A little less handwavy:

Hold a vinyl record up to your ear in a quiet room. With your free hand, tap the center of the record and listen carefully for any "bong" noise. If the disc has any resonant "drumhead" frequencies in the audio range, you will notice them by this test.

Now note that a disc on a rubber-backed turntable is well-damped for the fundamental "drumhead" resonance which greatly reduces the chance that the resonance can be excited while the disc is being played. Note also that the turntable is spring-isolated from its base, so the disc cannot be resonated by any vibrations coming up through its base either.

In any case, let us know the results of your experiments here! And have fun.

$\endgroup$
4
$\begingroup$

Handwaving:

The vinyl is, on the relevant scales, very stiff and light, so the force of the needle doesn't excite it much, and the resonances are beyond human hearing.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.