Why Are Room Modes Making Your Room Sound Terrible?!

[@pecacartun]

Speed of sound also depends of the temperature in the room, but assuming 68°F /20°C it's closer to around 1126 ft/s / 343.2 m/s. And excact formula is: 331.3 + 0.606 * °C (Temperature). Amroc use exactly 343 m/s, which is about 1125.328 ft/s.

[@JackofAllMinistries]

Thanks for the info! Knew I was in the ballpark, but makes sense it was a metric thing.

[@sawabhacks8050]

this is the best tut on KZbin.on this topic. Thank you thank you thank you.

[@JackofAllMinistries]

Means a lot because I feel like there are some pretty good ones out there. I appreciate you saying it!

[@brycegardner6171]

Just wanted to comment on something that maybe confusing in you room mode description. When you draw your modes, you can draw them as acoustic velocity or acoustic pressure. I think you said this: at the wall the acoustic velocity is zero and the pressure is maximum. Typically we think of the modes as acoustic pressure (as that is what we hear). If you want to visualize what the mode looks like, you need the draw your modes starting at the peak of the wave (1/4 wavelength from where you started your mode shape drawing). Then you would find the pressure nodes in the room (where you don't hear that mode). The way that you drew you modes is more of a representation of the acoustic velocity: at the walls is zero (or very small as the air must move the wall) and it peaks 1/4 wave away from the wall. I think you mentioned that putting some fuzz there would be the most effective (slightly off the wall). This is because acoustic absorbing materials tend to impede the movement of the air and work best where there is most movement (highest velocity) which is typically about 1/4 wavelength away from a hard surface. I just wanted to mention this because you were talking about the pressure nodes in the room, but drew the shape of the velocity nodes. It just may be confusing because you said that the pressure was high against the wall but drew a diagram where it looked like it was zero on the wall (because you drew the velocity shape). Hopefully, I haven't added to the confusion. Although writing this long of a reply on KZbin is almost guaranteed to not be useful.

[@JackofAllMinistries]

I definitely see your point. I literally lost sleep over that one after I uploaded the video, wondering if that would confuse people. Honestly, it was confusing me as I went back and watched it to check. I thought about taking it down and remaking it more clearly, but I felt that since it wasn't "wrong" I could leave it, even if it's a bit confusing. I still felt like it was helpful and decided to leave it. Nice catch

[@brycegardner6171]

@@JackofAllMinistries It is a pretty subtle thing. I thought it you go looking for pressure nodes or anti-nodes in a room mode, it would leave you looking in the wrong place.

[@JackofAllMinistries]

@@brycegardner6171 Definitely appreciate the help!

[@tonofquestions]

Thank you for the video - one quick question from the 3:18 time frame. Not sure if I'm misunderstanding??? When you show the wave reflecting from the surface on the left... you have it return along the same wave (below the 0 point)... and then you say that the waves double in amplitude. It seems to me that the wave would reflect or "bounce" off of the surface going above the 0 point and thus provide a subtraction to the wave... causing a cancelation effect rather than a doubling one. Am I off base? (no pun base / bass haha)

[@JackofAllMinistries]

I’m a little confused by the question, but I have a hunch I can clarify the explanation… the problem with using this type of visual to explain the wave is that we picture the waves taking the shape of the illustration. In reality, it is pressure that is changing in a wave, not a shape. So when it returns it’s pressurized air and excited particles that are actually reflecting. Does that make sense? Does that help?

[@Rene_Christensen]

If you have a hard wall, the reflection coefficient is 1, and so the amplitude of the reflected way is equal to that of the forward propagating wave, so combined pressure is doubled. The issue comes when trying to visualize this, because a mode is a steady-state phenomenon, so you not see this forward and backward propagation, as the field has had enough time to build. When you just have a standing wave, you will measure a high (combined) pressure at the wall.

[@everyonesalama4447]

very very helpful thanks so much

[@JackofAllMinistries]

Glad I could help!

[@theproduceryoudidntcounton1826]

Thank you so much

[@JackofAllMinistries]

My pleasure!

[@6.5inthemix]

Thank u so much.

[@JackofAllMinistries]

You’re so welcome!

[@songsongsingasong]

Super useful thank you!!!!

[@JackofAllMinistries]

You're welcome!!

[@jdwpraise]

Aren’t you adorable

[@JackofAllMinistries]

I like to think so!!