Question

An ideal gas enclosed in a cylindrical container supports a freely moving piston of mass `M`. The piston and the cylinder have equal cross-sectional area `A`. When the piston is in equilibrium, the volume of the gas is `V_(0)` and its pressure is `P_(0)`. The piston is slightly displaced from the equilibrium position and released. Assuming that the system is completely isolated from its surrounding, the piston executes a simple harmonic motion with frequency
A. `(1)/(2pi)sqrt((MV_(0))/(AgammaP_(0)))`
B. `(1)/(2pi)(AgammaP_(0))/(V_(0)M)`
C. `(1)/(2pi)(V_(0)MP_(0))/(A^(2)gamma)`
D. `(1)/(2pi)sqrt((A^(2)gammaP_(0))/(MV_(0)))`

Answer 1

Correct Answer - D

FBD off piston at equilibrium

FBD of piston when piston is pushed down a distance x

`(P_(0)+dP)A-(P_("atm")A+Mg)=M(d^(2)x)/(dt^(2))` . . (ii)
As the system is completely isolated from its surrounding therefore the change is adiabatic.
For an adiabatic process.
`PV^(gamma)=`constant `thereforeV^(gmma)dP+PgammaV^(gamma-1)dV=0`
or `dP=-(gammaPdV)/(V)thereforedP=-(gammaP_(0)(Ax))/(V_(0))` . .. (iii)
Using (i) and (iii) in (ii), we get
`M(d^(2)x)/(dt^(2))=-(gammaP_(0)A^(2))/(V_(0))x` or `(d^(2)x)/(dt^(2))=-(gammaP_(0)A^(2))/(MV_(0))x`
Comparing it with standard equation of SHM,
`(d^(2)x)/(dt^(2))=-omega^(2)x`
we get, `omega^(2)=(gamma)/(2pi)=(1)/(2pi)sqrt((gammaP_(0)A^(2))/(MV_(0)))`
Frequency, `upsilon=(omega)/(2pi)=(1)/(2pi)sqrt((gammaP_(0)A^(2))/(MV_(0)))`