150 Problems d Sdutiom on Thermodynamics d Statiaticd Mechanics
velocity 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) ,
A
where X is to be determined.
equation
On the other hand, the macroscope motion of fluid satisfies the Euler
au ap
Po- at = -- ax
where 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 have
X = po tan(kL).
Thus
ik %PO WL
c(x = 0) = -PO tan kL = -- tan -.
WP3 c PO C
~(t) = Re(6(0)eiwt) = - (“tan C
CPO