Sound levels and bernoulli's problems

[owl]

Hi there! Any help would be greatly appreciated. Thanks!

Homework Statement

If a sound level of 0 dB at 1000 Hz corresponds to a maximum gauge pressure (sound amplitude) of 10^-9 atm, what is the maximum gauge pressure in a 50dB sound at the same frequency? (answer: 3x10^-7 atm)

Homework Equations

L (in dB) = 10 log (I/I₀)

The Attempt at a Solution

If it is 0 dB that must mean that I₀, ie the reference intensity is 10^-9 atm right?

So ie 0 = 10 log (10^-9/10^-9)
So to get an answer at 50dB... it would be 50 = 10 log ( ??/10^-9)

but then I'm getting 50/10 = log ( ??/10^-9)
10^5 = ?? / 10^-9 (going 10^x both sides to get rid of the log)
which leaves me with ?? = 10^14 which can't be (and isn't) right...?

And the other one...

Homework Statement

What is the pressure drop due to the Bernoulli effect as water goes into a 3 cm diameter nozzle from a 9cm diameter fire hose while carrying a flow of 5 x 10^-4 m³s^-1? (answer: 247 Pa)

Homework Equations

So the bernoulli equation is P₁ + 1/2 pv₁² = P₂ + 1/2 pv₂²
Leaving out the gravitational parts as horizontal flow is assumed...
(the little p being density = 1000 kgm^3 for water and the big P being pressure)

The Attempt at a Solution

Well rearranging the bernoulli equation to get P₂ - P₁ I get
change in P = 1/2 p ( [Q / A₁ ]² - [ Q / A₂ ]² ] ) = 15.44 Pa?

The answer shows up as 247 Pa which I noticed to be approximately 15.44 squared... but I can't seem to come up with anything other than this!

Any help would be really appreciated. Thanks!

 

[Jhero]

Hi there! I'm happy to help with your questions.

For the first problem, you are on the right track with using the formula L (in dB) = 10 log (I/I0). However, I think you may have made a mistake when you set up the equation for 50 dB. It should be:

50 = 10 log (P/10^-9)

where P is the maximum gauge pressure for 50 dB. From there, you can solve for P by rearranging the equation and using logarithm rules. Your final answer should be P = 3 x 10^-7 atm.

For the second problem, your approach is correct. However, you made a small error in your calculation. The correct formula for pressure drop due to Bernoulli effect is:

change in P = 1/2 p ( [Q / A1 ]^2 - [ Q / A2 ]^2 )

Notice that the square should be on the entire fraction, not just the numerator. Using this formula, you should get the correct answer of 247 Pa.

I hope this helps! Let me know if you have any further questions. Good luck with your studies!