2006-11-19 17:07:03 · 4 answers · asked by Anonymous in Science & Mathematics ➔ Physics
db = 10* log (I2/I1)
40 = 10*log (I2/I1)
4 = log (I2;I1)
I2/I1 = 10^4; the anwer is 10,000 times as intense.
NOTE in answer to the responses below, if the statement says "intensity" that means the POWER level of the sound, not the pressure amplitude. (See the reference). Therefore the factor of 10 is correct.
Quote: "When referring to measurements of power or intensity it is:
10log(I1/I2)"
Also, 0db is an arbitrary reference level, and does not mean zero intensity or ampltude. In acoustics, 0db refers to a pressure amplitude of 10 micropascals.
In electrical circuits you have to state whether you are referring to power db or voltage db (BOTH are used). For example, in rating the output of a power amplifier, the ratings are given in power db, but for a voltage amplifier, they are in voltage db.
EDIT AGAIN, final statement: it is true that if you measure sound with a SPL meter, the db is 20log(I1/I2). That is because an SPL meter is a Sound Pressure Level meter.
2006-11-19 17:10:55 · answer #1 · answered by gp4rts 7 · 0⤊ 0⤋
Gp4rts gives you a good cite, but didnt apply it properly. Note that when you are dealing with SOUND (or electrical circuits) every 20 db = 10x amplitude. So a 40 db sound will be only 10*10 or 100 times as loud...
Added later - Gp4rts has a point about the question asking "how intense" is the sound as opposed to asking "how powerful" is the sound. Normally though when you are measuring sound (out of speakers for example) the 20 multiplier is correct...
2006-11-20 01:29:23 · answer #2 · answered by heartsensei 4 · 0⤊ 0⤋
the above answer is almost correct, the formulas are right, but at 0db there is no sound and to find how much more intense it is you would have to divide by zero and is therefore undefined.
2006-11-20 01:16:03 · answer #3 · answered by frmme2u2me2u 2 · 0⤊ 0⤋
40dB=10^(40/10)=10^4=10 000 times more intense.
2006-11-20 01:50:45 · answer #4 · answered by zee_prime 6 · 0⤊ 0⤋