Trumpet players, sound intensity level

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SUMMARY

The sound intensity level of five trumpet players at 75 dB increases to approximately 96 dB when the number of players increases to 30. This calculation is based on the logarithmic relationship between sound intensity and perceived loudness, where an increase in intensity by a factor of 10 results in a perceived loudness increase of about 2 dB. The formula used is Beta = (10 dB) log10(I/I0), with I0 defined as 1.0 x 10^-12 W/m². The intensity ratio of 30 to 5 trumpet players is calculated as 6, leading to the final decibel level.

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  • Familiarity with logarithmic functions and their applications
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  • Basic principles of sound perception and loudness
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Homework Statement



Five trumpet players have a sound intensity level of 75dB. What would be the sound intensity level of 30 such trumpet players? _______dB


Homework Equations



Increasing the sound intensity by a factor of 10 results in an increase in perceived loudness by approx. a factor of 2.

Beta = (10dB)log10(I/I0)
Where I0 = 1.0x10-12 W/m2


The Attempt at a Solution



I tried finding a ratio and applying that to the 2nd situation but my answers were wrong. Can someone please help me ASAP!

Thank you!
 
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10 dB=1 bel, and that's 10^1. 100 dB=10 bel, and that's 10^10. 75=7.5 bel, which is equal to what power of ten?
 
0.75?
3/4?
 
The 30 trumpeters will be 6 times greater intensity than the 5 trumpeters. The ratio in decibels all "drop out" leaving log(6) = 0.78 (about).
30 trumpeters are louder than 6, so to get the new decibels it is 75 / 0.78 = 96 decibels.

Logic check - 6 x intensity is < 10 x, so we expect the difference in decibels to be < 2x.

cheers!
 

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