Some one Pressure Amplitude

[Jtappan]

Some one please help! Pressure Amplitude

Homework Statement

(a) What is the pressure amplitude of a sound wave with an intensity level of 124.5 dB in air?

_____ N/m2
(b) What force does this exert on an eardrum of area 0.520 10-4 m2?

_____ mN



3. The Attempt at a Solution


if P is amplitude of pressure, then sound intensity (I) is proportional to P^2
I/Io = (P/Po)^2
dB = sound intensity level = 10 log[I/Io]
124.5 = 10 log[I/Io] = 10 log[P/Po]^2 = 20 log(P/Po)
log(P/Po) = 124.5/20 = 6.225
P = Po [10]^6.225
Po = reference pressure amplitude for human ear = 2*10^-5 N/m^2
P = 2*10^-5*[10]^6.225
P = 33.576 N/m^2
----------------
Force on eardrum = P * area = 33.576*0.52*10^4 N
F = 1.746 milli N




This is what I did. But it is not giving me the correct answer...can anyone help me?

 

[Shooting Star]

This is pressure tactics, posting it three times.

33.5 N/m^2 seems to be correct. Are you getting both answers wrong?

 

[ogb p]

Hi there, thank you for your question. Based on your calculations, it seems like you have the right approach to solving this problem. However, there may be a small mistake in your final calculation.

When solving for the pressure amplitude (P), you used the reference pressure amplitude for the human ear (Po) as 2*10^-5 N/m^2. However, this value is actually the reference pressure amplitude for sound in air at sea level, not specifically for the human ear. The reference pressure amplitude for the human ear is typically given as 20 micro Pascals (μPa) or 2*10^-5 Pa.

So, when solving for P, you should use Po = 2*10^-5 Pa instead of 2*10^-5 N/m^2. This will give you a pressure amplitude of 33.576 μPa or 33.576*10^-6 N/m^2. Using this value in your final calculation for the force on the eardrum, you should get a result of 1.746 μN or 1.746*10^-3 mN.

I hope this helps! Let me know if you have any further questions.