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An acoustically rigid boundary has zero perturbation normal velocity. For a reflected acoustic plane wave, pressure amplitudes add in phase and the reflection coefficient is $+1$. It is the infinite-surface acoustic impedance limit. A pressure-release boundary instead fixes the pressure perturbation to zero and reflects with coefficient $-1$.

Use the prescribed harmonic convention $e^{i\omega t}$. For a propagating acoustic plane wave, let $k=(\omega /c_0)\cos \theta$ and $\ell =(\omega /c_0)\sin \theta >0$. The incident and reflected pressure amplitudes in the upper half-space have vertical factors $e^{i\ell y}$ and $e^{-i\ell y}$ respectively:The linear homentropic acoustic equations imply $i\omega {\rho }_0{v}_y=-{\partial }_y{p}^{\prime }$. At the surface, the normal velocity is therefore $\ell (R-A)/({\rho }_0\omega )=-V$, while the pressure amplitude is $P=A+R$. The surface acoustic impedance condition $P=ZV$ givesHere $Z_f$ is the normal acoustic impedance; the angle in this question is measured from the horizontal, not the normal. For a passive acoustic impedance, the mean power absorbed per unit area is $\frac{1}{2}\mathrm{Re}(P{V}^{\ast })=\frac{1}{2}\mathrm{Re}Z|V{|}^2\ge 0$. The four limiting cases have distinct meanings:

Away from a transmitted grazing or critical angle, ${\rho }_1{c}_1\gg {\rho }_2{c}_2$ gives a small normal acoustic impedance ratio $\delta$. ThusThe reflected energy is nearly all the incident energy, while the pressure phase depends on the plate response. Large $|\beta |$ gives $R\simeq +1$, resembling a rigid boundary; $\beta \simeq 0$ gives $R\simeq -1$, resembling a pressure-release boundary. Bulk impedance contrast alone does not ensure $\delta \ll 1$ arbitrarily close to a transmitted critical angle, since the factor $1/\cos {\theta }_2$ can be large.

For fixed frequency and finite plate parameters, $\beta$ contains the incident normal wavevector and therefore tends to zero as $\theta \to \pi /2$. If $c_1\ne c_2$, the outgoing $l_2$ tends to a nonzero real or imaginary value, whereas $l_1\to 0$, so $\delta \to 0$ too. ConsequentlyThe lower-side boundary pressure $1+R$ tends to zero, regardless of how large the fixed $m,B$ are. The angle must become closer to grazing as those parameters grow; the assertion is not uniform in an unbounded plate mass or rigidity.

If $c_1=c_2$, however, $l_2=l_1$ and $\delta ={\rho }_2/{\rho }_1$ remains constant. The exact grazing limit is $(\delta -1)/(\delta +1)$, rather than exactly $-1$ for finite contrast. Under the stated large-contrast approximation it is still arbitrarily close to the pressure-release value, with boundary pressure $2\delta /(1+\delta )$. Thus the printed pressure-release description is an approximation in that equal-speed case, not an exact zero-pressure identity. If a transmitted wave becomes evanescent, the outgoing square-root convention above continues to give the unequal-speed grazing limit.