Is this correct? I'm looking for other sources of the result.

If pressure at a point is sinusoidal

$$ p=p_0 sin(\omega t) $$

then particle displacement is

$$ \boldsymbol{s} = {1 \over {\omega^2 \rho}} \boldsymbol\nabla p $$

where $\rho$ is the fluid's density.

I derived it by taking the 2nd time integral of particle acceleration ${\boldsymbol{a}} = {-1\over\rho} \boldsymbol\nabla p$.

  • $\begingroup$ It looks good, as long as you are not accounting for any viscous effects, which would change the conservation of momentum equation you are using. $\endgroup$
    – Michael M
    Jan 24 at 13:52


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