Thermodynamics and simple harmonic motion

In summary, the conversation discusses a horizontal piston-cylinder system containing ideal gas surrounded by air at a certain temperature and pressure. It is shown that if the piston is slightly displaced, it will execute simple harmonic motion with a specific angular frequency. This is derived using equations such as PV=nRT and F=ma, and assuming that the gas temperature remains constant. The final expression for force on the piston can be simplified by neglecting higher order terms due to the displacement being small.
  • #1
vladimir69
130
0

Homework Statement


A horizontal piston-cylinder system containing n mole of ideal gas is surrounded by air at temperature [itex]T_{0}[/itex] and pressure [itex]P_{0}[/itex]. If the piston is displaced slightly from equilibrium, show that it executes simple harmonic motion with angular frequency [itex]\omega=\frac{AP_{0}}{\sqrt{MnRT_{0}}}[/itex], where A and M are the piston area and mass, respectively. Assume the gas temperature remains constant.


Homework Equations


PV=nRT
F=Ma
P=F/A
x=amount by which piston moves
L=length of cylinder
F = force on piston

The Attempt at a Solution


[tex]V_{0}=AL[/tex]
[tex]V=A(L-x)[/tex]
[tex]F_{net}=A(P_{0}-P)[/tex]
[tex]L=\frac{nRT_{0}}{P_{0}A}[/tex]
[tex]P=\frac{ALP_{0}}{V}[/tex]
Popping all this into the mix gives
[tex]F_{net}=\frac{-P_{0}^2A^2x}{nRT_{0}-P_{0}Ax}[/tex]
[tex]M\frac{d^2x}{dt^2}=\frac{-P_{0}^2A^2x}{nRT_{0}-P_{0}Ax}[/tex]
If I get rid of the [itex]P_{0}Ax}[/itex] term then we arrive at the correct answer but not sure why one should throw away this term

Thanks
 
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  • #2
The denominator is equal to [tex]
nRT_0 (1-x/L)
[/tex]

It is assumed that the piston is only slightly displaced from equilibrium, so that x/L<<1. That means that its second or higher power can be neglected with respect to it.
We often apply this method in Physics when working with small quantities: use Taylor expansion and keep linear terms.
Your expression of force can be written in terms of (x/L),

[tex]F=AP_0(1-\frac{1}{1-x/L})
[/tex]

Expanding the the fraction with respect to x/L :

[tex]F=AP_0(1-(1+x/L+(x/L)^2+(x/L)^3+...))

[/tex]

Omitting all terms but linear, we get:

[tex]F=-AP_0(x/L)

[/tex]

ehild
 
  • #3
ok
thanks for that
 

Related to Thermodynamics and simple harmonic motion

1. What is the definition of thermodynamics?

Thermodynamics is the branch of physics that deals with the relationships and conversions between different forms of energy, such as heat, work, and internal energy, within a system.

2. How does the first law of thermodynamics relate to simple harmonic motion?

The first law of thermodynamics states that energy cannot be created or destroyed, only transferred or converted. In simple harmonic motion, the energy of a system oscillates between kinetic and potential energy, but the total energy remains constant.

3. What is the difference between adiabatic and isothermal processes in thermodynamics?

Adiabatic processes occur without the transfer of heat, while isothermal processes occur at a constant temperature. In simple harmonic motion, the motion can be considered adiabatic because no heat is transferred between the system and its surroundings.

4. How does the second law of thermodynamics relate to simple harmonic motion?

The second law of thermodynamics states that the total entropy of a closed system will always increase over time. In simple harmonic motion, the friction and damping forces acting on the system result in a decrease in energy and an increase in entropy.

5. Can thermodynamics be applied to non-ideal systems?

Yes, thermodynamics can be applied to non-ideal systems. In fact, the study of non-ideal systems and their behavior is an important area of research in thermodynamics. Simple harmonic motion can also occur in non-ideal systems where there may be additional forces or factors affecting the motion.

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