I am trying to find the equation that explains why we can say the pressure exerted by a wave (squared) is proportional to the power. I see this claim made a lot but I can't tell where it comes from.
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2 Answers
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First, let's consider a pressure transducer, such as a microphone.
When we use it to record sound, in practice we are logging a voltage. For a linear system, each increase in pressure results in an equal increase in voltage. For electrical power, $P = V^2/R$.
A convention is to ignore the resistance/impedance (i.e., $R$, in Ohms) by setting $R = 1$. This leads to reporting the measured value as proportional to the actual acoustic power (see Norton and Karczub 2003, Fundamentals of Noise and Vibration Analysis for Engineers):
...the spectra of continuous signals are referred to as power spectral densities because they have units related to power and that the spectra of transient signals are referred to as energy spectral densities because they have units relating to energy. This is an important point - one which warrants further discussion. A power spectral density has units of (volts)^2 per hertz or V^2 s. Thus the area under a power spectral density curve has units of (volts)^2 which is proportional to power (i.e. electrical power is ∝ V^2).
Now we proceed by analogy to actually answer your question.
Matter resists acoustic flow, a physical quantity we call acoustic impedance, $Z$. Impedance varies depending on material properties (such as mass, elasticity, and geometric configuration.) So we can write the above equation for a purely acoustic scenario as:
$$P = \frac{p^2}{Z}$$
Or, accepting that $Z$ is variable (and often empirically determined), it may be simpler to say $P \propto p^2$.
answered Nov 15, 2022 at 2:31
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Introduction
Citing Heinrich Kuttruff from Acoustics - An Introduction (of course you can find this information in other textbooks)
Any oscillatory motion contains energy, namely, kinetic energy stored in the mass element, and potential energy stored in the spring representing the restoring force. The same holds for the motion of particles in sound waves.
Energy density
The aforementioned citation can lead to a derivation of the energy in a travelling wave. Following the derivation in the textbook (section 3.4 - Intensity and energy density of sound waves in fluids), we reach the formula
$$ w = \rho_{0} \frac{\tilde{v}^{2}}{2} + \frac{\tilde{p}^{2}}{2 \rho_{0} c^{2}} \tag{1} \label{energy-density} $$
where $w$ denotes the energy density, $\rho_{0}$ the density of the medium, the $\tilde{\cdot}$ symbol denotes that the value is an RMS value, $c$ is the wave propagation speed, $v$ is the particle velocity and $p$ the pressure.
Plane waves
We consider to be known that for plane waves the particle velocity can be calculated by the pressure like
$$ v = \frac{p}{\rho_{0} c} \tag{2} \label{plane-wave-particle-velocity} $$
Plugging in equation \eqref{plane-wave-particle-velocity} into equation \eqref{energy-density} to replace $v$ we get
$$ w = \rho_{0} \frac{\left( \frac{\tilde{p}}{\rho_{0} c} \right)^{2}}{2} + \frac{\tilde{p}^{2}}{2 \rho_{0} c^{2}} \implies w = \rho_{0} \frac{\frac{\tilde{p}^{2}}{\rho_{0}^{2} c^{2}}}{2} + \frac{\tilde{p}^{2}}{2 \rho_{0} c^{2}} \implies w = \rho_{0} \frac{\tilde{p}^{2}}{2 \rho_{0}^{2} c^{2}} + \frac{\tilde{p}^{2}}{2 \rho_{0} c^{2}} \implies w = \frac{\tilde{p}^{2}}{2 \rho_{0} c^{2}} + \frac{\tilde{p}^{2}}{2 \rho_{0} c^{2}} \implies w = 2 \frac{\tilde{p}^{2}}{2 \rho_{0} c^{2}} \implies w = \frac{\tilde{p}^{2}}{\rho_{0} c^{2}} \tag{3} \label{plane-wave-energy-density} $$
From equation \eqref{plane-wave-energy-density} we can make at least two notes. One is that for a simple harmonic signals, the kinetic energy (given by the part associated with the particle velocity) and the potential energy (given by the part associated with the pressure) are equal and that energy density is proportional to the square of the pressure.
Power
Since power is directly proportional to the energy of a function (it is the energy per unit time by definition) we conclude from equation \eqref{plane-wave-energy-density} that the power is proportional to the square of the pressure amplitude (although RMS values are more appropriate here).
Non-plane waves
We have used plane waves to reach this derivation and for good reason. They are the simplest waves we can formulate and resemble the simple harmonic oscillation in the acoustical domain. Furthermore, they also serve as the basis for the decomposition of complex wavefields which gives them great value.
Nevertheless, there are other formulations of acoustical waves that are very useful. A great example is the spherical waves originating from point sources. The point here is that the energy density in most (if not all) linear mechanical/acoustical waves is given by the sum of the kinetic and potential energy in the motion.
Without going into the derivations here I'll state the equations for the pressure and particle velocity for a spherical wave to make the distinction between those and plane waves.
$$ p \left( r , t \right) = \frac{j \omega \rho Q}{4 \pi r} e^{j \left(\omega t - k r \right)} \tag{4} \label{spherical-pressure} $$
$$ u \left( r , t \right) = \frac{j k Q}{4 \pi r} \left( 1 + \frac{1}{j k r} \right) e^{j \left(\omega t - k r \right)} = \frac{p \left( r, t \right)}{\rho c} \left( 1 + \frac{1}{j k r} \right) \tag{5} \label{spherical-particle-velocity} $$
where $j$ is the imaginary unit for which $j^{2} = -1$ holds true, $\omega = 2 \pi f$ with $f$ the temporal frequency is the radial frequency, $\rho$ the medium density, $Q$ is the "strength" of the source (actually is the volume velocity of the radially vibrating sphere), $k = \frac{\omega}{c} = \frac{2 \pi}{\lambda}$ is the wavenumber and $c$ is the speed of sound (magnitude of group velocity) and $\lambda$ is the wavelength for which $\lambda = \frac{c}{f}$ holds. Finally, $r$ denotes the distance from the source since the formulation is in spherical coordinates.
As you can see from equations \eqref{spherical-pressure} and \eqref{spherical-particle-velocity} the two quantities have components that are in quadrature for all distances from the source apart from $\infty$.
This introduces the notion of active and reactive intensity. The active intensity is associated with the radiated energy which propagates away from the source. The reactive intensity is associated with the energy that is "stored" in the soundfield close to the source and oscillates without propagating away from the latter.
Since in plane waves there is only active intensity, all the power calculated is propagating. In most cases, the active component of the total intensity is of practical interest and this is why this is usually the one calculated.
Since in "large" distances from the source even the spherical waves will approximate/resemble plane waves, the active intensity (and thus the power associated with it) can be calculated at the far-field by integrating the intensity on a sphere around the source (the same of course applies to the power since they are directly related).
As you can see, the relations that hold for plane waves are very useful in calculating the power of other kinds of waves too. This is yet another reason that plane-waves are useful.
Point to take from here is that the active intensity (power propagating away from the source) can be calculated with the use of equation \eqref{plane-wave-energy-density} for other sources too.
