Question

Try this sweet physics question. I'm not bragging that I was able to solve this, but I am.

Answer 1

A certain mass of an ideal gas is enclosed in a cylinder of volume \(V_0\) fitted with a smooth heavy piston of mass \(m\) and cross-sectional area \(A\). The piston is displaced through a small distance downwards so as to compress the gas isothermally, and then it is let go. Take the atmospheric pressure as \(P\).

Find the time period of the resultant oscillations.

Answer 2

I've never seen a physics question like this before!!
Let me think about it

Answer 3

This sounds like a perpetual motion device lol

Answer 4

That is because it is in ideal conditions. But if we extract energy, it should stop its motion, so it isn't a perpetual motion in that sense (you cannot extract infinite energy).

Answer 5

start with the ideal gas law \(pV = nRT = const \) as isothermal

do \(d(pV) = 0 = pdV + V dp \implies dp = -\frac{p dV}{V}\)

from there, for ***small*** displacement x, the restoring "linear" force for shm is \(dp A\), so \( |dp A| = - m \ddot x\).

etc

Answer 6

I think that the system can be modeled with the subsequent 2-nd order ODE
\[\Large M\frac{{{d^2}\left( {PS} \right)}}{{d{t^2}}} = - \left( {P - {P_0} - Mg} \right)S\]
where \(S\) is the surface of the piston, \(M\) is its mass, and \(P_0\) is the atmospheric preesure

Answer 7

oops..pressure*

Answer 8

I get f''[x(t)] + xPA^2/mV = 0

where P = p + mg/A (p = atmospheric pressure)

Answer 9

looks similar to Michele's

Answer 10

and for the period i get T = 2πsqrt(mV/PA^2)

Answer 11

how does that look?

Answer 12

oops.. I have made a typo, here is the right equation

\[\Large \frac{{{d^2}z}}{{d{t^2}}} = \frac{{ - \left( {P - {P_0} - Mg} \right)S}}{M},\quad P = \frac{{nRT}}{{Sz}}\]

Answer 13

All I have to say is that being able to solve a question like this is my goal. Good job!

Answer 14

more precisely, we can write this:
\[\Large P = \frac{{nRT}}{{Sz}},\quad {P_0} + Mg = \frac{{nRT}}{{S{z_0}}}\]
and for little oscillations, namely \[\huge z \approx {z_0}\]
we have:
\[\huge \ddot z + \frac{{nRT}}{{Mz_0^2}}\left( {z - {z_0}} \right) = 0\]
therefore, the period of little oscillations is:
\[\huge T = 2\pi \left( {\frac{{{V_0}}}{S}} \right)\sqrt {\frac{M}{{nRT}}} \]

Answer 15

so @parthkohli what's the answer

Answer 16

he's disappeared and left us holding the baby!

Answer 17

Hey I'm back!

Answer 18

Basically you're not given \(n, T\) but I'm sure you can use the equation of state.

Answer 19

@alekos looks like you're right!

Answer 20

Awesome! thanks @parthkohli great question!

Answer 21

we can make this substitution:
\[\huge nRT = {P_0}{V_0}\]
so, we get:
\[\huge T = 2\pi \sqrt {\frac{{M{V_0}}}{{{P_0}{S^2}}}} \]

Answer 22

Substitute \(P = P_{atm} + mg/A\) for the actual final answer though.