I have a loudspeaker. Let's say that I feed a signal to it, say, a pure sine wave:
$$f(t) = a \sin(\omega t).$$
How does the sound pressure of the resulting sound relate to the amplitude $a$? Is it proportional to the amplitude $a$? Does it depend on the frequency, too, or only on the amplitude? (I've read that sound pressure level is the log of the square of the sound pressure, but I'm trying to work out the relationship to the amplitude of the signal.)
Best Answer
The relationship between SPL, pressure, velocity, displacement and driving voltage are standard in the audio/loudspeaker literature.
First, the sound pressure level (SPL) in decibels (dB) $$ \text{SPL} = 20 \log_{10} \frac{p}{p_\mathrm{ref}} $$ where $p_\mathrm{ref} = 2 \times 10^{-5} N/m^2$ is a reference pressure, and $p$ is the absolute pressure produced from the loudspeaker cone motion.
Now, the question is about feeding a sinusoidal signal to the loudspeaker. Normally, we think of this signal as a voltage source (an audio amplifier acts as a voltage source): $$ V(t) = V_0 \sin \omega t $$ The connection between the driving voltage and the cone displacement $x$ depends on the properties of the loudspeaker motor (the magnet, voice-coil and suspension), but these two are linearly related. If you double the driving voltage, you double the cone amplitude. So. let's say that we apply a voltage that drives the cone with amplitude $x_0$
$$ x = x_0 \sin \omega t $$ Note that we do not use $a$ as in the OP because $a$ is normally reserved for the cone acceleration. So, we rephrase the OP's question as "how can we relate the cone amplitude $x_0$ to the sound pressure level". The answer to this question is simple (and not explained correctly in the other answer). The velocity and acceleration of the cone are just derivatives of the cone displacement $$ \begin{eqnarray} v &=& \frac{dx}{dt} = \omega x_0 \cos \omega t \\ a &=& \frac{dv}{dt} = -\omega^2 x_0 \sin \omega t \\ \end{eqnarray} $$ We need the acceleration here because the pressure from the cone is written as $$ p = \frac{\rho S_D}{2 \pi r} \, a $$ Here, $\rho$ is the density of air, $S_D$ is the surface area of the cone, and $r$ is the distance from the cone. This means that the pressure is proportional to the cone acceleration. So, in terms of the cone displacement, we have $$ p = -\frac{\rho S_D}{2 \pi r} \,\omega^2 x $$ I believe that this answers the OP's question. It means that the pressure produced by the cone is proportional to the displacement times the square of the frequency. Finally, we can connect the pressure to the amplifier driving voltage $V_0$: $$ p = \frac{\rho S_D}{2 \pi r} \frac{V_0 B\ell}{R_E M_\mathrm{ms}} \sin \omega t $$ This expression is true for a purely resistive loudspeaker motor with DC resistance $R_E$, force factor $B\ell$ and moving mass $M_\mathrm{ms}$ operating well above the resonant frequency.