IIT JEE , AIEEE Forum - Physics , Thermal Physics

magiclko

An ideal mono-atomic gas is enclosed in a fixed horizontal adiabatic cylinder of cross sectional area A. The cylinder is fitted with an adiabtaic piston of mass m (attached to one of a spring), which can move horizontally without friction inside the cylinder . in equilibrium, the spring is in natural length and pressure and volume are P and V respectively, If the atmosphereic pressure is also P, n the piston is slightly displaced from equlibrium and released, then find the frequency of small oscillations.

magiclko

no answer

gorakavipraveen

it is indeed a good question. In fact I am also looking for the solution eagerly. I GOT STUCKED UP BY LAST 2 STEPS I SUPPOSE. I want to get proceeded. EXPERTS PLEASE ANSWERE.

magiclko

how hav u done it???? i mean except the last two steps...can u gimme the rest solution!!

gorakavipraveen

LOOK FROM EQUATION, P1V1^r = P2V2^r thus there is a build up equation, P1L1^(5/3)= P2(L1+dL)^(5/3) as monoatomic, r=5/3.
So there is a pressure function P2 after compression of dL. And thus that pressure*A gives retarding force that opposes Kx. So on the whole the equation would turn as
Net force in terms of dL = Kx - [P1L1^(5/3)]/[(L+dL)^(5/3)] .
this is what i got upto. But I observed tht its not a SHM and thus cannot further proceed to find out the frequency.

I was actually in search of experts opinion as that might include a technique to calculate time period of a compleax non SHM like this.

gorakavipraveen

LOOK FROM EQUATION, P1V1^r = P2V2^r thus there is a build up equation, P1L1^(5/3)= P2(L1+dL)^(5/3) as monoatomic, r=5/3.
So there is a pressure function P2 after compression of dL. And thus that pressure*A gives retarding force that opposes Kx. So on the whole the equation would turn as
Net force in terms of dL = Kx - [{PL1^(5/3)]/[(L+dL)^(5/3)}-P]A .
this is what i got upto. But I observed tht its not a SHM and thus cannot further proceed to find out the frequency.

I was actually in search of experts opinion as that might include a technique to calculate time period of a compleax non SHM like this.

bhupesh

Dear
I am not explaining much and here i the solution

pv

_{1 =}p' (v₁+Ax)

( Ax is the change in volume due to the disturbance at the mean position )

p'=p/(1+ Ax/v₁ )

p'= p( 1-

Ax/v₁ ) (

<1 using binomial expansion )...................(1)

net force

{(p+kx/A) - p' } A = ma

substituting the value of p' from (1)

a = - (k+ pA²

/ v₁ ) x

hence

² =(k+ pA²

/ v₁ )

Now you can find the frequency

magiclko

yep sir, its the answer.... thanx

bhupesh

is fraction expanding it bonomially and neglecting thehigher terms

accidentaly I wrote

, 1 what i meant is

is a fraction

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