= Solution
The fluid velocity at the piston equals its velocity,
$$
u_p(t)=\frac d{dt}\bigl[-\epsilon t(1-t)\bigr]
=-\epsilon(1-2t).
$$
Using $c-c_0=(\gamma-1)u/2$ in part (ii) gives
$$
\frac{\delta p}{p_0}
=\gamma\frac u{c_0}
+\frac{\gamma(\gamma+1)}4\frac{u^2}{c_0^2}
+O(\epsilon^3).
$$
Over $0\leq t\leq1$,
$$
\langle u_p\rangle=0,
\qquad
\langle u_p^2\rangle
=\epsilon^2\int_0^1(1-2t)^2\,dt
=\frac{\epsilon^2}{3}.
$$
Thus
$$
\boxed{
\left\langle\frac{\delta p}{p_0}\right\rangle
=\frac{\gamma(\gamma+1)}{12}
\frac{\epsilon^2}{c_0^2}
+O(\epsilon^3)}.
$$
The linear pressure oscillation has zero mean, while the quadratic compressibility produces a positive mean pressure, the finite-amplitude acoustic analogue of radiation pressure.
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