RootsBooster Posted January 18, 2013 Share Posted January 18, 2013 I'm normally quite clued up on physics but my son asked me a question last night that I wasn't sure about. To put it simply, if there was a buzzer that emitted 20db, would it double to 40db if there were 2 buzzers? Link to comment Share on other sites More sharing options...
Gormenghast Posted January 18, 2013 Share Posted January 18, 2013 I'm normally quite clued up on physics but my son asked me a question last night that I wasn't sure about. To put it simply, if there was a buzzer that emitted 20db, would it double to 40db if there were 2 buzzers? The dB level of sound sources can't simply be added together. Doubling the number of 20dB sources would give a total of 23dB as far as I remember from a course I did a few years ago. Link to comment Share on other sites More sharing options...
stereolab Posted January 18, 2013 Share Posted January 18, 2013 I agree with the poster above. http://www.howstuffworks.com/question124.htm On the decibel scale, the smallest audible sound (near total silence) is 0 dB. A sound 10 times more powerful is 10 dB. A sound 100 times more powerful than near total silence is 20 dB. A sound 1,000 times more powerful than near total silence is 30 dB. Based on this I believe you would need 1000 times more buzzers than you started with to get from 20 to 40 dB. Link to comment Share on other sites More sharing options...
metalman Posted January 18, 2013 Share Posted January 18, 2013 Tis indeed logarithmic and it would give you 23 dB. It's all in the fount of wisdom. Link to comment Share on other sites More sharing options...
Annie Bynnol Posted January 18, 2013 Share Posted January 18, 2013 The answer could be silence. Or any amplitude between zero and double. Sound waves can cancel/add each other out by destructive/constructive interference. Link to comment Share on other sites More sharing options...
Obelix Posted January 18, 2013 Share Posted January 18, 2013 I'm normally quite clued up on physics but my son asked me a question last night that I wasn't sure about. To put it simply, if there was a buzzer that emitted 20db, would it double to 40db if there were 2 buzzers? As the others have said above, it would in theory increase to 23dB (if you want to be really picky it's 23.01dB plus change) There are other effects to consider though for very loud sounds as there is an ultimate limit to loudness in a gas because you can form a vacuum. You also strictly should relate it to a reference as a dB is merely a scaling factor. The standard reference pressure for listening is 20 microPa at 1kHz which is the onset of hearing in most people. ---------- Post added 18-01-2013 at 09:42 ---------- I agree with the poster above. http://www.howstuffworks.com/question124.htm On the decibel scale, the smallest audible sound (near total silence) is 0 dB. A sound 10 times more powerful is 10 dB. A sound 100 times more powerful than near total silence is 20 dB. A sound 1,000 times more powerful than near total silence is 30 dB. Based on this I believe you would need 1000 times more buzzers than you started with to get from 20 to 40 dB. You need a 100, you are increasing it by 20dB which is a factor of 10^2 = 100 Link to comment Share on other sites More sharing options...
NightFlight Posted January 18, 2013 Share Posted January 18, 2013 ---------- Post added 18-01-2013 at 09:42 ---------- You need a 100, you are increasing it by 20dB which is a factor of 10^2 = 100 I disagree on method, if not entirely different answer you need an increase of 20dB. Each doubling of the number of the speakers gives you a 3dB rise, so you need 20/3 = 6.666 doublings to get the desired output. 2^6.666 = 101.59 speakers needed. (or 102 in sensible numbers) Link to comment Share on other sites More sharing options...
Obelix Posted January 18, 2013 Share Posted January 18, 2013 Because that's how the Bel is defined. An increase of one Bel is defined as log(10)(Increased level/base level). A decibel is simply one tenth of a Bel. Flipped the other way, if you are going from a known increase of 20dB then that is 20/10 = 2 Bel, which is a power increase of 10^2 over the base power - ergo 100 times. Double the speakers and you get *almost* an increase of 3dB but not quite - your increase is a little more than 3dB and that's why you have the excess of 1.59 in your example above. Link to comment Share on other sites More sharing options...
NightFlight Posted January 18, 2013 Share Posted January 18, 2013 Of course. I have vague recollections of seeing that Log(10)(increased/base) from my IoA diploma a few years back. Wish I'd done something with that but seemed to graduate right at the time jobs dried up. Perhaps time to dig out the Little Red Book and get looking again Link to comment Share on other sites More sharing options...
Ghozer Posted January 18, 2013 Share Posted January 18, 2013 if the buzzers were exactly the same frequency, volume, and of opposite phase, they would likely cancel each other out, leaving silence! Link to comment Share on other sites More sharing options...
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