150 Problems d Sdutiom on Thermodynamics d Statiaticd Mechanicsvelocity of the macroscope motion of the gas be u(x, t) and the pressure
of the gas be p(x, t). Because the displacement of the piston is very small,
we can solve u(z, t) and p(x, t) approximately in the region 0 5 x 5 L
and consider u(0,t). The boundary conditions are p(0,t) = f(t)/A and
u(L, t) = 0. As f(t) is a sinusoidal function of t and the frequency is w, the
resulting u(z, t) and p(z, t) must be waves of frequency w and wave vector
k = w/c. In fact, u(z, t) and p(z, t) both satisfy the wave equation with
propagating velocity c. We can write
f(g) = Refoexp(iwt) ,
p = Re$(z) exp(iwt) ,
u = Reij(z) exp(iwt).Thus, to satisfy the boundary condition of p, we have$(x) = - fo cos(kz) + Xsin(ks) ,
Awhere X is to be determined.equationOn the other hand, the macroscope motion of fluid satisfies the Eulerau ap
Po- at = -- axwhere po is the average density, u is the velocity and p is the pressure. Then
G(x) = -(-posinkx+Xcoskz), ik where po = -. fo
WPO A
Using the boundary condition u(L) = 0, we haveX = po tan(kL).
Thusik %PO WL
c(x = 0) = -PO tan kL = -- tan -.
WP3 c PO C~(t) = Re(6(0)eiwt) = - (“tan C
CPO