Soundwave Intensity & Sound Level

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SUMMARY

The discussion focuses on calculating sound intensity and sound level from a given soundwave amplitude of 5 x 10^-5 N/m^2. The sound intensity is derived using the formula Sound Intensity = Power/Area, leading to a calculated intensity of 6.3 x 10^-12 W/m^2. The sound level is determined using the formula L (dB) = 20*Log(P/Pref), resulting in a sound level of approximately 8 dB. The user acknowledges a misunderstanding of the formulas involved, which has implications for their test preparation.

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Homework Statement


The amplitude of a soundwave is 5 X 10 ^-5 N/m^2
a) What is the intesity of the sound energy from this sound wave?
b) What is the sound level?

The Attempt at a Solution



Well I believe that Sound intensity is Power/amplitude. However, i only have the amplitude so.. does that mean that my problem is impossible?
 
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L (dB) = 20*Log(P/Pref) = 10*Log(I/Iref)

According to the units, you are given the pressure (50 uN/m^2 = 50 uPa). Sound Pressure Level is 20*Log(P/Pref) where Pref is 20uPa. You get around 8 dB.

Sound Intensity = Sound Power / ***Area***. Now this 8 dB is just a ratio and is also equal to 10*Log(I/Iref) where Iref is 10^-12 W/m^2. You get 6.3*10^-12 W/m^2.
 
OH. Thanks for your solution. I realize that I made a complete blunder now. So it was my formula...I get it now, but that does not bode well for my test results =|. Thanks again for your help.
 

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