The sound of a level dog barking is 83 dB. The sound of a thunder clap is 102 dB. How many times louder is the thunder clap than the dog?
I know the answer is 19.
Math help!!?
2018-05-09 9:28 am
回答 (11)
2018-05-09 9:37 am
Decibels are a logarithmic measure...
db increase is 19 (102 - 83)
And a 20 db difference is about 4 times as loud
read https://music.eku.edu/sites/music.eku.edu/files/ekuhealthandsafety.pdf
db increase is 19 (102 - 83)
And a 20 db difference is about 4 times as loud
read https://music.eku.edu/sites/music.eku.edu/files/ekuhealthandsafety.pdf
2018-05-10 3:58 am
The sound level of a dog barking is 83 db.
Thunder is louder.
Doubling of sound intensity (acoustic energy) belongs to a calculated level change of +3 dB. +10 dB is the level of twice the perceived volume or twice as loud (loudness) in psychoacoustics − mostly sensed.
Twice the sound pressure is a Sound Pressure Level of +6 dB
Thunder is louder.
Doubling of sound intensity (acoustic energy) belongs to a calculated level change of +3 dB. +10 dB is the level of twice the perceived volume or twice as loud (loudness) in psychoacoustics − mostly sensed.
Twice the sound pressure is a Sound Pressure Level of +6 dB
2018-05-09 3:41 pm
thanks to John Napier :
a)
in terms of sound intensity :
thunder dB = dog dB+10log (n)
102-83 = 19 = 10*Log(n)
1.9 = Log (n)
n = 10^1.9 = 79.43
b)
in terms of sound pressure :
thunder dB = dog dB+20log (n)
102-83 = 19 = 20*Log(n)
0.95 = Log (n)
n = 10^0.95 = 8.91
a)
in terms of sound intensity :
thunder dB = dog dB+10log (n)
102-83 = 19 = 10*Log(n)
1.9 = Log (n)
n = 10^1.9 = 79.43
b)
in terms of sound pressure :
thunder dB = dog dB+20log (n)
102-83 = 19 = 20*Log(n)
0.95 = Log (n)
n = 10^0.95 = 8.91
2018-05-09 11:30 am
There is a 19 dB difference.
From http://www.sengpielaudio.com/calculator-levelchange.htm
Ratio doubling means:
− a power level of +3 dB, or a sound intensity level of +3 dB
− an electric voltage level of +6 dB, or a sound pressure level of +6 dB
− a loudness level of about +10 dB
So the sound intensity is increased by a factor of 79,
The sound pressure level is increased by a factor of 8.9
And the perceived loudness increases by a factor of about 3.7.
From http://www.sengpielaudio.com/calculator-levelchange.htm
Ratio doubling means:
− a power level of +3 dB, or a sound intensity level of +3 dB
− an electric voltage level of +6 dB, or a sound pressure level of +6 dB
− a loudness level of about +10 dB
So the sound intensity is increased by a factor of 79,
The sound pressure level is increased by a factor of 8.9
And the perceived loudness increases by a factor of about 3.7.
2018-05-10 2:41 am
It is not that simple decibels are on a logarithmic scale.
http://www.animations.physics.unsw.edu.au/jw/dB.htm
http://www.animations.physics.unsw.edu.au/jw/dB.htm
2018-05-09 3:39 pm
firstly you can see 83 to 103 dB is 20 deci Bell = 2 Bell
Which would be a factor of 10 ^ 2 ie 100 times greater.
Which means that the answer that you KNOW is incorrect.
The sound is just a bit under 100 times greater.
Far, far more than 19
To do it formally the ratio is 10 ^ (n/10) where n is the increase in dB
in this case 19 dB
so the sound of the thunder is 10 ^ (19/10) = 79 times greater.
Which would be a factor of 10 ^ 2 ie 100 times greater.
Which means that the answer that you KNOW is incorrect.
The sound is just a bit under 100 times greater.
Far, far more than 19
To do it formally the ratio is 10 ^ (n/10) where n is the increase in dB
in this case 19 dB
so the sound of the thunder is 10 ^ (19/10) = 79 times greater.
2018-05-09 9:41 am
Each increase of 3 dB is a doubling of the sound level.
2018-05-09 9:35 am
1,473 dB
2018-05-09 9:33 am
If this is a percentage, an easy way to remember how to do them is to remember that 10 is 10% of 100, so how do you get 10% from 100 and 10? That way you don't have to ask such basic questions.
2018-05-09 9:31 am
102/83 = 1.2289
~ 22% louded
~ 22% louded
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