Soundwave Intensity & Sound Level

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The discussion focuses on calculating sound intensity and sound level from a given sound wave amplitude of 5 X 10^-5 N/m^2. Participants clarify that sound intensity can be derived from sound power and area, while sound pressure level is calculated using the formula L(dB) = 20*Log(P/Pref), resulting in approximately 8 dB. The correct sound intensity is found to be 6.3 X 10^-12 W/m^2, using the reference intensity of 10^-12 W/m^2. The original poster acknowledges a misunderstanding of the formulas used, highlighting the importance of accurate calculations for test preparation. The discussion emphasizes the relationship between sound pressure, intensity, and their respective calculations.
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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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