I am familiar with the derivation of the speed of sound done in the style of, for example, this question:

Diatomic chain and speed of sound

I would like to derive that same result

$$ c^2=\frac{2ka^2}{m+M} $$ Obtained in that problem using $$ c= \lim_{q \to 0} \frac{\partial \omega}{\partial q} $$ but instead I want to derive the speed of sound using the Newton-Laplace formula $$ \frac{\partial P}{\partial \rho}=c^2 $$ What I tried so far:

For the density I put delta functions at each mass $$ \rho = \sum_n m_n \delta(x-u_n) $$ where the $ n $ th particle has mass $ m_n $ and equilibrium position $ u_n $. Note that this has correct units since the units of a delta function are the inverse of the units of the argument. For pressure I'm using something like $$ P= k( x_n-x_{n-1} ) $$ which has the right units so that $ P/\rho $ has units of speed squared but I need pressure to include some delta functions at each point without messing up the units. Perhaps a kronecker delta function would work?

I'm unsure of both my choice of $ P $ and of $ \rho $ and I don't see how to continue with the problem in any way that's likely to produce the $ a^2 $ factor I need in my answer.

This question is essentially equivalent to:

Speed of Sound in 1D using pressure and density

since diatomic vs monatomic is not important. However I don't find the answer posted there to be very useful.

Perhaps If I Fourier expand $ x_n $ as $$ x_n = \int_{- \pi}^\pi \frac{1}{2\pi} dq x_q e^{iqn-i\omega_q t} $$ and use that $$ \omega_q= \frac{k}{\mu}\pm \sqrt{ \frac{k^2}{\mu^2}-\frac{4k^2}{mM} sin^2(qa) } $$ To reiterate: how do I derive the speed of sound in a 1d chain using the formula $$ \frac{\partial P}{\partial \rho}=c^2 $$ Have I chosen my density and pressure correctly? Any assistance would be appreciated!

  • $\begingroup$ It might be worth remembering that the Newton-Laplace formula you provide is a macroscopic formula, and is only strictly valid for continua. I would recommend you define a macroscopic density and pressure (averaging over large numbers of elements), and then apply the formula to those quantities. $\endgroup$
    – Michael M
    Feb 11, 2021 at 18:11


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