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Energy Density vs Total Energy

  • Thread starter cosine
  • Start date
4
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Here is a detail that makes me doubt:

1. Homework Statement
I am given the equation for a standing wave with a dependence in 't' and 'z' only... I am told that the wave is propagating in a confined space (2*2*2) m
Q1. Calculate the energy density for n=1,2,3. I didn't have any pb with this.
Q2. Calculate the total energy for n=1,2,3. (!!!)


2. Homework Equations



3. The Attempt at a Solution
Ok, I can't find anything in my lecture notes about total energy for waves. So I went back to the definition of energy density (h) which the amount of energy per unit of volume.
Therefore if I multiply the volume by h I should get the energy (right?)
The volume here is 8 m^3 BUT since the wave in question is only in 1 direction (z) do I still have to multiply by V or only by L=2m instead (for this particular case)

Thanks for your answer(s)
1. Homework Statement



2. Homework Equations



3. The Attempt at a Solution
 

Answers and Replies

dextercioby
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Well, since the integrated function has no "x" and "y" dependence, then the integration wrt them should be trivial, right ?
 
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Well, since the integrated function has no "x" and "y" dependence, then the integration wrt them should be trivial, right ?
trivial... maybe not or perhaps I would not be posting...

The wave equation has no dependence in 'x' and 'y' indeed,however I am not integrated that function to get the total energy.
I got the energy density (Question 1) which is h = rho * (A*Omega/2)^2
A being the amplitude of the wave. The formula is correct according to my book (which does not mention total energy of sound waves btw)

If I check the units of 'h' I get J/m^3 therefore I should multiply by a volume to get the total energy (in J) trivial right? I guess that why I posted, I was surprised to get an energy density in J/m^3 calculated from a wave that's propagating in 1 direction only...
 

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