= Solution
Between collisions the <momentum> is $p_0=\sqrt{2mE}$ or $-p_0$. In <phase space> the cycle runs right along the upper horizontal segment from $q=0$ to $q=b$, jumps vertically to $-p_0$ at the right wall, runs left along the lower segment, and jumps back at the left wall. The jumps have $dq=0$. Therefore
$$
\boxed{I=\oint p\,dq=2b\sqrt{2mE}.}
$$
<Differentiation> gives $dI/dE=2bm/\sqrt{2mE}=2b/v$, exactly the out-and-back period. Under slow wall motion the action $I$ is the <adiabatic invariant>; a convention using $I/(2\pi)$ is equivalent. Hence
$$
\boxed{E(b)=E(b_0)\left(\frac{b_0}{b}\right)^2.}
$$
Compression raises the energy and expansion lowers it, provided the wall speed is small compared with the particle speed and the change per cycle is small.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2006/ii/paper-1-bouncing-action.png]
{title=One elastic-bounce cycle in position-momentum <phase space>}
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