1
$\begingroup$

Lets say that there are two waves. They are both identical in every way except that one is traveling through a gas and another through a liquid. As we know they are traveling at different speeds and that if you hear the gas wave while standing in a room it sounds different than if you are in a pool and hear the wave through the water.

My question is how are they heard differently in a quantitative way? Is there any way that I could have a recording of a sound through one medium and determine roughly what it would sound like through a different medium? At the very least I am assuming that there is a way to determine how a very simple wave changes with the medium.

Thanks in advance, and if it is not clear feel free to ask for any clarification!

$\endgroup$
  • 1
    $\begingroup$ $v=\lambda f$, so if the speed is different, one or both of the wavelength or frequency must be as well... $\endgroup$ – KBriggs Jun 1 '16 at 20:34
2
$\begingroup$

Assume that the sound you are interested in is heard at 1000 Hz in air, where the sound is traveling at 343 m/s. If the device generating the sound (some type of waterproof speaker) is placed underwater, it will still generate a 1000 Hz sound, because the frequency is determined by the device that generates the sound, not by the speed of the sound in a particular medium. This means that the pitch of the sound will be perceived as being the same in air or in water, where sound travels considerably faster than in air (e.g., approximately 1500 m/s).

Obviously, something changed between the air and water scenarios. Since sound velocity is equal to frequency multiplied by wavelength, the 1000 Hz sound will have a wavelength that is considerably longer in water than in air. This has practical implications. There is normally a small difference in timing between when sound reaches each of your ears. In air, your ears are far enough apart that this difference is large enough for you to locate the approximate direction that the sound is coming from. However, in water the timing difference is MUCH smaller due to the much higher speed of sound in water, meaning that even though the pitch of the sound is unaffected by the speed difference, you don't get the information regarding the direction of the sound. Thus, under water, you easily hear the 1000 Hz tone, but you can't tell where it is coming from.

$\endgroup$
1
$\begingroup$

Or in short, it is harder to make liquid vibrate than the air because they have different qualities. If you test a subwoofer right beside you, or far from you, you can measure how long it takes for the bass to reach your ears. If you try doing it in water, let's say you have a subwoofer built for, and to submerged underwater, it will be engineered to have the same pressure as the water where you will be playing it. Since it is engineered to generate the same vibration in water, I think you will hear the same thing you heard when you were not underwater, as long as there are no air bubbles inside your ear, and if your ear canal was just like a straight pipe. In order for you to hear the same sounds, you just need your ear drums (tympanic membrane) to be hit by waves the same way. Apparently, making liquid water molecules vibrate the same way as the air is difficult because they are liquid. At least when no wave is generated yet. And no liquid molecules can move faster than the gas molecules because that is a fact. Generating a pressure wave in water is slower than in air. You hear sounds differently underwater because the waves traveling the air doesn't have the sufficient energy to make water vibrate the same way.

"Sound is a pressure wave, but this wave behaves slightly differently through air as compared to water. Water is denser than air, so it takes more energy to generate a wave, but once a wave has started, it will travel faster than it would do in air."

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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