= Solution
For spherical symmetry, $r\widetilde p$ satisfies the one-dimensional <wave equation>, so an outgoing solution has the form $A(t-r/c)/r$. To relate $A$ to the source, use a <velocity potential> $\Phi$ with $u_r=\partial_r\Phi$ and $\widetilde p=-\rho\partial_t\Phi$. The outgoing monopole potential
$$
\Phi=-\frac{q(t-r/c)}{4\pi\rho r}
$$
has $4\pi r^2\rho u_r\to q(t)$ as $r\to0$, fixing its mass-flux normalization. It follows that
$$
\boxed{\widetilde p(r,t)=\frac{q'(t-r/c)}{4\pi r}.}
$$
For an element with displacement $f_0e^{i\omega t}$, its volume outflow is $i\omega f_0h^2e^{i\omega t}$ and mass outflow is $\rho$ times this. Thus $q'=-\rho\omega^2f_0h^2e^{i\omega t}$. A rigid wall imposes zero normal velocity away from the source. The <method of images> uses a source of the same sign on the other side; a source on the plane and its image coincide, doubling the full-space monopole field. Equivalently its mass flux spreads over a hemisphere rather than a sphere. Hence
$$
\boxed{\widetilde p=-\frac{2\rho\omega^2f_0h^2}{4\pi R}e^{i\omega(t-R/c)}.}
$$
The small patch relative to wavelength justifies the point source, and the small displacement justifies linear acoustics.
Superposition over the aperture gives the <Rayleigh integral for baffled acoustic radiation>
$$
\widetilde p(t,x,y,z)=-\frac{\rho\omega^2}{2\pi}
\iint\frac{f(y',z')}{R}e^{i\omega(t-R/c)}\,dy'\,dz',
\quad R=\sqrt{x^2+(y-y')^2+(z-z')^2}.
$$
For $r\gg L$, $R=r-(yy'+zz')/r+O(L^2/r)$ and $R^{-1}=r^{-1}[1+O(L/r)]$. The additional condition $r\gg\omega L^2/c$ makes the omitted phase small, so with the printed <Fourier transform> convention
$$
\boxed{\widetilde p\simeq-\frac{\rho\omega^2}{2\pi r}e^{i\omega(t-r/c)}\widehat f(m,n),
\quad m=-\frac{\omega y}{rc},\quad n=-\frac{\omega z}{rc}.}
$$
The minus signs in $m,n$ arise because the outgoing phase contributes $+i\omega(yy'+zz')/(rc)$.
For the rectangle the two integrals factor, giving
$$
\boxed{\widehat f(m,n)=\frac{4\sin(am)\sin(bn)}{mn}
=4ab\,\operatorname{sinc}(am)\operatorname{sinc}(bn),}
$$
where $\operatorname{sinc}u=\sin u/u$ and its zero-argument limit is one. If $k=\omega/c$, $kb\ll1$ implies $|bn|\leq kb\ll1$, so the $z$-direction factor is nearly one throughout the radiation hemisphere. With $ka$ of order one the $y$-direction factor varies appreciably: the long dimension gives stronger directivity in that angular direction, while the short dimension radiates broadly. The forward amplitude is proportional to the area $4ab$. Nulls have $ka\,y/r$ equal to nonzero integer multiples of $\pi$ when those values are physically accessible; order-one $ka$ does not automatically imply that the first null lies in the hemisphere.
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