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Pressure variation with sound waves equation

  • #1
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Homework Statement


P(average) for a speaker is 10 W. Gamma is 1.4 (ratio of specific heats), molar mass is 28.8 g/mol, air temperature is 50F, and pressure is 1atm. Find Pmax at 100
I have this equation that gives Intensity = (Pmax^2)/(2*Rho*v) where rho is density, and v is speed of sound through this air

Homework Equations


All but one value has already been found:
Solved:
Rho = .001239 g/cm^3 = 1.239 kg/m^3
v = 338.28 m/s
I = 7.9577x10^(-5) W/m^2
(all reasonable values)

The Attempt at a Solution


I'm just trying to solve for Pmax, very simple, but the units dont work out. So the equation becomes
Pmax = sqrt( 2 * I * Rho * v )
and the units for pressure become sqrt( J * kg / (m^4 * s^2) ) which is (square Joules) * (square kg) /( m^2 * s)
which doesn't make any sense to me, and the answer becomes 0.258 but intuitively this one doesnt seem reasonable, shouldn't the maximum pressure only be VERY slightly different than the regular pressure? Shouldn't the units be some kind of Force per area?
 

Answers and Replies

  • #2
TSny
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Pmax = sqrt( 2 * I * Rho * v )
and the units for pressure become sqrt( J * kg / (m^4 * s^2) ) which is (square Joules) * (square kg) /( m^2 * s)
which doesn't make any sense to me,
Simplify further by expressing Joules in terms of kg, m ,s.
and and the answer becomes 0.258 but intuitively this one doesnt seem reasonable, shouldn't the maximum pressure only be VERY slightly different than the regular pressure?
What do you consider to be the "regular pressure" in SI units?
 
  • #3
62
9
I would consider Newtons per square meter to be SI units. But what would be the .258 for Pmax that i calculated for? I know it cant be kPa, Pa, atm, or Torr/mmHg because Pmax should be very close to 101.3kPa , 1.0atm, or 760Torr, but 0.25 isnt a number close to any of those.
Isnt it in its simpliest form with sqrt(J kg)/ (m^2 * s^2) ?
Is a Newton the same thing as a J * kg maybe?
 
  • #4
TSny
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I would consider Newtons per square meter to be SI units. But what would be the .258 for Pmax that i calculated for? I know it cant be kPa, Pa, atm, or Torr/mmHg because Pmax should be very close to 101.3kPa , 1.0atm, or 760Torr, but 0.25 isnt a number close to any of those.
Does Pmax denote the maximum absolute pressure, or the maximum amount that the pressure deviates from the unperturbed pressure?
Isnt it in its simpliest form with sqrt(J kg)/ (m^2 * s^2) ?
Should the seconds be squared here? I'm assuming that only the J and the kg are inside the square root.

There are different ways you can show that your units boil down to the units for pressure. If you want to express Joules in terms of Newtons, then that would be OK. You would then still need to know the relation between the Newton and the units of kg, m, and s.
Is a Newton the same thing as a J * kg maybe?
No.
 
  • #5
62
9
Opps i meant sqrt(J kg)/ (m^2 * s ), and yes, with only the square root on on J and kg. Im pretty sure P_max denotes the maximum absolute pressure. Do you happen to know if this equation is true? It was given by the professor so I assume it is, but there is no such thing as "root kg" or "root joules" lol. Do you happen to know what units this .258 is? If so, I could put all the units in my notes and figure out from there
 
  • #6
TSny
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Opps i meant sqrt(J kg)/ (m^2 * s ), and yes, with only the square root on on J and kg.
OK

Take any formula you know for energy, such as the formula for kinetic energy of a particle. Do a unit analysis to discover how a Joule can be expressed in terms of kg, m , and s. Then you can use this to see how sqrt(J kg)/ (m^2 * s ) boils down to some combination of kg, m, and s.

Do the same for pressure ##P = F/A## to see how pressure boils down to some combination of kg, m, and s. You should find that it all checks out.


Im pretty sure P_max denotes the maximum absolute pressure. Do you happen to know if this equation is true?
Pmax is the maximum variation in the pressure from the normal atmospheric pressure. It will be very small compared to the normal atmospheric pressure.
 
  • #7
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9
Okay, so then the answer of .258 has the potential to make sense. Doing the unit analysis for pressure gives F/A = N / m^2 = kg / (m * s^2) but after I break my equation down in those units, i get kg/s.
I'm one for learning through struggle, and youve been a great help, but is the .258 from using W/m^2, kg/m^3, and m/s in Pascals? I dont have time to spend an entire day trying to figure out this one very small fraction of a homework problem lol. I believe it would be the only unit to make sense because a .25 fluctuation in kPa would be very large for a speaker 100m away. Pascals 98% must be the correct answer
 
  • #8
TSny
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Yes, Pascal is the correct unit for the pressure.
 

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