Why are high frequencies reflected more than low frequencies off an 'acoustically hard' surface such as concrete? I basically understand that the amount of reflection is determined by the impedance ratio between the two mediums (i.e air/concrete). But does that mean that high frequency sound waves have a lower acoustic impedance than low frequencies (hence why high frequencies are reflected and more than low frequencies?).
[Physics] Frequency dependance of sound wave reflection
absorptionacousticsfrequencyreflectionwaves
Best Answer
First off, much of this answer is based on the amazingly helpful University of New South Wales website on acoustic impedance. This is a topic I'm very interested in and this is a great opportunity to reduce the complex topic of impedance down for my own learning purposes as well as to provide an answer.
You ask,
Actually, specific acoustic impedance, $z$, is a property of the medium, not of the waves themselves.
It has both a real part, resistance $r(\omega)$ and an imaginary part, reactance $x(\omega)$:
$$z(\omega) = r(\omega) + i \space x(\omega)$$
Where $\omega$ is the angular frequency, $2 \pi f$.
Reactance has a positive and a negative part: compliance (capacative) reactance $x_C(\omega)$, and inertive (inductive) reactance $x_L(\omega)$:
$$x(\omega) = x_L(\omega) - x_C(\omega)$$
Without going into exhaustive detail, I want to point out the relevant fact that inertiance, associated with the density of the medium, is proportional to frequency:
$$x_L(\omega) \propto \omega$$
While compliance, associated with the elasticity of the medium, is inversely proportional to frequency:
$$x_C(\omega) \propto \frac{1}{\omega}$$
In other words,
In the example of your concrete wall - a dense, inelastic media relative to air - we might expect more of an impedance missmatch (between the two media) for higher-frequency energy than for the lower frequency energy. This would result in the higher frequency energy being reflected more.