# Longitudinal Wave Equation from Transverse One

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1. Apr 26, 2015

### bananabandana

1. The problem statement, all variables and given/known data
Part ii)
2. Relevant equations

3. The attempt at a solution
So I try to conserve volume as it suggests in the hint. I take the initial volume of the region to be given by:
$$h \times \delta x \times l = (\delta x + \eta) (h+\Psi) l$$
Where l is just some fixed, constant length which can immediately be canceled. Expanding:
$$\Psi \delta x = - \eta (\psi + h)$$
But $h>> \psi \implies (\psi+h) \approx h$
$$\Psi \delta x = - \eta h$$
For small values of $\eta$ (which is implied by the fact that $\psi$ is small? ) we can make the statement:
$$\eta \approx \frac{\partial \eta}{ \partial x} \delta x$$
So that:
$$\Psi \approx - h\frac{\partial \eta}{\partial x}$$

Well, I got to the result, but I'm just not sure that this approach is correct - for instance should I not put $\delta \Psi$ instead of $\Psi$ - but then I have second differentials and I get the wrong answer... - also not entirely sure how to justify the assumption that if $h >> \psi$ then $\eta$ must be small... or is that okay because we are just approximating?

Thanks!

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2. Apr 26, 2015

### Delta²

I think you shouldnt use $\delta \psi$ cause the change in the transverse direction of the volume is simply $\psi(x)$ however you should use $\delta \eta$ cause the change in the longitudinal direction of the volume is $\eta(x+\delta x)-\eta(x)=\delta \eta$.

Last edited: Apr 26, 2015