= Solution
Substitution into the <wave equation> gives $\partial_x^2\psi+\partial_y^2\psi=-(k_x^2+k_y^2)\psi$ and $\partial_t^2\psi=-\omega^2\psi$. Therefore $\omega^2=c^2(k_x^2+k_y^2)$. The fixed edges at zero are automatically satisfied by the sine factors; the opposite edges require
$$
k_x=\frac{p\pi}{a},\quad k_y=\frac{q\pi}{b},\quad p,q=1,2,\ldots,\qquad
\boxed{f_{pq}=\frac c2\sqrt{\frac{p^2}{a^2}+\frac{q^2}{b^2}}.}
$$
Neither integer may vanish, because that would make the entire displacement zero.
The element's <mass> is $\rho\,dx\,dy$, and its <speed> is $-\omega A\sin(k_xx)\sin(k_yy)\sin(\omega t)$. Squaring and multiplying by half its <mass> gives the requested <energy> <density>. The two sine-squared spatial integrals are $a/2$ and $b/2$, so
$$
\boxed{K(t)=\frac{\rho ab\omega^2A^2}{8}\sin^2(\omega t),\qquad
\overline K=\frac{\rho ab\omega^2A^2}{16}.}
$$
There is no <dissipation>. The total <mechanical energy> equals the maximum <kinetic energy>, giving
$$
\boxed{U(t)=\frac{\rho ab\omega^2A^2}{8}\cos^2(\omega t).}
$$
Equivalently, integrate the small-slope tension <energy> $\rho c^2|\nabla\psi|^2/2$ and use the <dispersion relation>.
For $b=4a/5$, the two lowest <modes> are $(p,q)=(1,1)$ and $(2,1)$:
$$
\boxed{f_{11}=\frac{\sqrt{41}\,c}{8a},\qquad
f_{21}=\frac{\sqrt{89}\,c}{8a}.}
$$
The fundamental has one <antinode> at $(a/2,b/2)$ and no interior nodal line. The next <mode> has a <node> along $x=a/2$ and opposite-sign <antinodes> at $(a/4,b/2)$ and $(3a/4,b/2)$. Every boundary is a <node>.
\Image[/past-exam-of-the-physics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2013/paper-1-ia2013-membrane-modes.png]
{title=Original displacement maps for the two lowest modes of a rectangular membrane, showing nodal lines and signed antinodes}
The square fundamental has $f=c\sqrt2/(2\beta a)$. Equating it to $f_{11}$ gives
$$
\boxed{\beta=\sqrt{\frac{32}{41}}=0.88345.}
$$
At the same peak displacement <amplitude> and <frequency>, time-averaged <kinetic energy> is proportional to membrane area. Therefore
$$
\boxed{\frac{\overline K_{\rm square}}{\overline K_{\rm rectangle}}
=\frac{\beta^2a^2}{a(0.8a)}=\frac{40}{41}.}
$$
The reciprocal ratio is $41/40$ if the rectangle is placed first.
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