Sound Pressure Level?

Question

The sound pressure level in a factory is recorded as 69dB. What will be new SPL if the sound pressure is decreased by 10%. Express your answer to the nearest dB.

If someone could write out and explain the formula used for this so I am able to use it again in the future that would be great!

Many thanks

Answer 1

Given:

Sound pressure level (SPL) Lp = 69 dB.



Reference sound pressure po = 20 µPa = 2×10^−5 Pa (Threshold of hearing)

Reference sound pressure level Lpo = 0 dB-SPL (Threshold of hearing level)



Get sound pressure p in Pa when entering sound pressure level Lp = 69 dB:

Sound pressure p = po×10^(Lp/20) Pa (= N/m²) = 2×10^−5×10^(69/20) Pa

= 0.056368 Pa. (100%)



10 percent of this sound pressure is: p = 0.0056368 Pa.

Total sound pressure when decreased by 10% sound pressure:

0.056368 Pa minus 0.0056368 Pa = 0.050731 Pa (90% sound pressure).



Get sound pressure level Lp in dB when entering sound pressure p = 0.050731 Pa:

Sound pressure level Lp = 20×log (p / po) dB = 20×log (0.050731 / 2×10^−5) dB

= 68.085 dB (90% sound pressure).



The answer to the nearest dB = 68 dB. That is 1 dB less than 69 dB.



Cheers ebs



PS: The answer of Ronald B is wrong.

You cannot take 10 percent of a decibel value.

Answer 2

As the sound pressure is 69dB a reduction 10% will be 10% of 69 db = 6.9 db= 62.1dB or 62dB

Answer 3

There are 2 meanings for decibels (dB). One is absolute sound intensity or rigidity, and the various is a ratio of two parts of ability or 2 amplitudes. For sound intensity I, the component in dB = 10*log10(I/Iref), the situation Iref = = 0 dB ability intensity = 1E-12 w/m^2 For sound rigidity P, the component in dB = 20*log10(P/Pref), the situation Pref = 0 dB rigidity = 2E-5 Pa For any ratio of ability P, the ratio in dB = 10*log10(P1/P2) For any ratio of amplitudes A, the ratio in dB = 20*log10(A1/A2)