- #1
Chaos' lil bro Order
- 683
- 2
Here is an interesting application of sound resonance. I was sitting in my computer room when I noticed that if I sat forward in my chair and settled in a particular spot, the high pitched 'buzz' coming from my computer sounded extra loud. I concluded that I was listening to a resonance created by this high pitched 'buzz' being reflected by my room's walls. The extra loud sound (or frequency resonance) that I heard was only particularly noticeable in my left ear, so we will ignore my right ear for now. The area of resonance had a planar area of about 5x5cm(paralell to the ground) and I didn't test for z-axis (height) variations. In other words, if my left ear was anywhere within this 5x5cm resonance zone, I heard the extra loud 'buzz'.
Let me setup of the parameters of my computer's spacing with respect to the dimensions of my room now:
Room: a square room with 4 walls, each wall is 4 meters long
Computer location in the Room: in a corner of the room, located 0.5 meters from one wall and 1 meter from the other wall. The computer's height from the ground is equal to the height of my ear's from the ground.
So I guess my question would be can you calculate the frequency of my computer's 'buzz' given what I told you about the resonance zone I found and the dimensions of my room?
I'm guessing the answer has to do with the Wavelength = speed of sound/ frequency formula. And perhaps also to do with how sound behaves in a pipe with two closed ends (like my room's walls close off my room). If memory serves me correctly and I don't think it does here, the formula is something like f = 2L (that doesn't look right does it).
PS. Physics 101 says that if waves constructively interefere with one another so that their respective peaks overlap, a resonance is formed where the waves overlap to form a larger amplitude. But I must admit that I always thought of resonance as a sharp peak that is either ON or OFF, but not somewhere in the middle. If I want to get picky, the resonance zone that was 5x5cm had an even smaller region inside of it 1x1cm which was where the absolute peak sound amplitude was heard. This smaller region was uniform in sound amplitude and I could not distinguish any minute movement within it (aka. it all sounded the same).
Let me setup of the parameters of my computer's spacing with respect to the dimensions of my room now:
Room: a square room with 4 walls, each wall is 4 meters long
Computer location in the Room: in a corner of the room, located 0.5 meters from one wall and 1 meter from the other wall. The computer's height from the ground is equal to the height of my ear's from the ground.
So I guess my question would be can you calculate the frequency of my computer's 'buzz' given what I told you about the resonance zone I found and the dimensions of my room?
I'm guessing the answer has to do with the Wavelength = speed of sound/ frequency formula. And perhaps also to do with how sound behaves in a pipe with two closed ends (like my room's walls close off my room). If memory serves me correctly and I don't think it does here, the formula is something like f = 2L (that doesn't look right does it).
PS. Physics 101 says that if waves constructively interefere with one another so that their respective peaks overlap, a resonance is formed where the waves overlap to form a larger amplitude. But I must admit that I always thought of resonance as a sharp peak that is either ON or OFF, but not somewhere in the middle. If I want to get picky, the resonance zone that was 5x5cm had an even smaller region inside of it 1x1cm which was where the absolute peak sound amplitude was heard. This smaller region was uniform in sound amplitude and I could not distinguish any minute movement within it (aka. it all sounded the same).