Thermodynamics and harmonic motion

  • #1
pedrovisk
7
0
Thread moved from the technical forums to the schoolwork forums
TL;DR Summary: Problem said that the ball moves in a harmonic motion and asked to prove it. The process is adiabatic

Problem said that the ball moves in a harmonic motion and asked to prove it. The process is adiabatic.

I did the development, but at certain point I'm having a problem. The right answer is that the P(Vx) and Vx are P/V (the initial pressure made by the gas and the initial volume), but my equations leads to P(Vx) and Vx. Where is my mistake?

Does the final volume and the final pressure are very close to the original volume and pression that is safe to assume they're almost equal?

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  • #2
pedrovisk said:
The right answer is that the P(Vx) and Vx are P/V (the initial pressure made by the gas and the initial volume), but my equations leads to P(Vx) and Vx.
I don't understand what this means.
 
  • #3
vela said:
I don't understand what this means.
Elaborate, please.

But got the answer in another forum, so no need for reply.
 
  • #4
pedrovisk said:
Elaborate, please.
For example, I would interpret "##P(V_x)## and ##V_x## are ##P/V##" to mean ##P(V_x)=V_x=P/V##, which doesn't make sense.
 
  • #5
vela said:
For example, I would interpret "##P(V_x)## and ##V_x## are ##P/V##" to mean ##P(V_x)=V_x=P/V##, which doesn't make sense.
The exercise is based on a real experiment named "Ruchhardt experiment". This experiment aims to calculate \gamma (Cp/Cv). The ball in the bottle follows SHM. So it is possible calculate \gamma finding the period of this SHM.

The equation of the force that makes the ball go back to equilibrium is ##dF = -\gamma .A^2.(P/V).dx##. P and V are the pressure and volume when the acceleration of the ball is 0.

As you can see, I got the equation "right". Problem is that instead o P/V (which are constants), I got ##P(V_x)/V_x##.

The question, that is the part you did not understand, is why the equation has ##P/V## instead of ##P(V_x)/V_x)##. It happens that the variation of ##P(V_x)## and ##(V_x)## are so small that it can be considered as equal to the initial pressure and volume, just like I said in the end of the original post.
 
  • #6
I can't see the figure. Just a big black screen.
 
  • #7
Chestermiller said:
I can't see the figure. Just a big black screen.
The figure has been deleted, which is why we ban the use of external image servers.

Thread locked.
 

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