IIT JEE , AIEEE Forum - Physics , Thermal Physics

computer001

EDITED:
sorry messed it up...pls refer my post some 3-4 posts below i think
dF=dP*A
dF=ma=-y*(P/V_o)Adv=-y*(P/V_o)(A)^2dx,P=P_o+mg/a

computer001

EDITED

eistien

gokul i didnt understand how u got the relation could you please explain!!!

Conjurer

Arey just see HC Verma example.

elastiboysai

I dont have HC Verma.
u shud realise dat
not all people have access to dat book.
If u have it or happen to know it, y dont u post the soln??

Conjurer

K man.. here it goes:

For adiabatic expansion: PV^gamma = C

Taking logs we get lnP + gammalnV = lnC
which gives dp/P + gamma dv/V = 0

dp = -gammadvP/V

NOw since frequency is small we can assume P = Po + mg/A and V = Vo

maA = - gamma Adx (Po + mg/A)/Vo

Now we can find omega.

PS: Didnt see HC Verma :) but dont rate if u dont like whatsoever.

computer001

im sorry i think i got wat u r sayon initially but i messed up the ans:
gamma=y
we can easily get dP=-y*(P/V_o)*dv
dF=dP*A
dF=ma=-y*(P/V_o)Adv=-y*(P/V_o)(A)^2dx,P=P_o+mg/a
a=-y*(P/V_o)(A)^2/m(dx)
so we have a =to -w^2x
w=root(y*(P/V_o)(A)^2/m)
= root{y/V_o*(P_o+mg/A)A^2/m}

computer001

to prove 1st part of my ans: v no tht PV^y=c..jus diff thts all

Ask & Discuss Questions with Community & Experts