Distance along beam calculation
My attempt:

Consider the following diagram showing the beam with length L, sensors A and B on each side of the beam and an arbitrary position x representing the location of the discontinuity. First consider the case of a signal source emanating from x equally toward A and B, with zero loss in the beam. The overall distance of the beam itself is such that the delay from A to B does not cause the phase of a signal at it's sampling rate to exceed $2\pi$ (otherwise alias solutions will exist). Smaller phases at this frequency would be allowable, but this will allow for the best precision. 

Thus similar to the transfer function of any cable the transfer function from x to A over the normalized radian frequency of $0$ to $2\pi$ has a magnitude of $1$ and a linear phase that extends from $0$ to $-\frac{2\pi}{x/L}$. Similarly the transfer function from x to B would have a magnitude of $1$ and a phase that extends from $0$ to $-\frac{2\pi}{(L-x)/L}$

These are given as:

$$H_{xA}(\omega) = e^{-j\omega L/x}$$

$$H_{xB}(\omega) = e^{-j\omega L/(L-x)}$$

And the ratio of these two is given as:

$$H(\omega) = H_{xA}(\omega)/H_{xB}(\omega) = e^{-j\omega L (L-x)/x}$$

With $\omega$ as the normalized radian frequency in units of radians/sample ($\omega$ extends from $0$ to $2\pi$ with $2\pi$ representing the sampling rate assuming the beam is sized such that the phase delay of the sampling rate was $2\pi$ (or this can be modified accordingly for shorter delays). 



Consider a single tone given as $e^{j \omega_n}$ that is originating at $x$. The signal received at A compared to the signal received at B is given by $H(\omega_n)$ to be $e^{-j\omega_n L (L-x)/x}$.

This would appear at A as $e^{-j\omega_n L/x}$ and at B as $e^{-j\omega_n L/(L-x)}$.

Specifically if we followed the guidelines above on length versus sampling rate, the phase of this comparison signal is unique to $\omega_n$ and $x$ for a given $L$ as $\omega_n L/(L-x)$.



Given that and to the extent we can model the discontinuity as a source that emanates equally in all directions, I would recommend the following approach:

Measure A as $a(t)$ and compute the DFT to be $A(\omega)$

Measure B as $b(t)$ and compute the DFT to be $B(\omega)$

Compute the comparison signal $R(\omega) = A(\omega)/ B(\omega)$

Compare the phase of $R\omega$ to the phase of $H\omega$ and solve for x for each bin:

$$arg(R(\omega)) = \omega L/(L-x)$$

$$x(\omega) = L\bigg(  1 - \frac{\omega} {arg(R(\omega))} \bigg)$$

Where $\omega = 2\pi k/N$ for DFT bins $k = 0$ to $N-1$.

This approach can be extended for lengths with phase shifts less than $2\pi$ at $f_s$ as well as actual transmission loss factors (since lossless was assumed here). 

Note that two or more discontinuities that react to the same frequency will not be resolvable: their relative phase contributions will add and so will therefore appear to be coming from another location (such as images in a reflection).  

We can calculate this for your specific example:

$$f_s = 9000 \text{ Hz}$$
$$L = 74 \text{ m}$$
$$v = 12000 \text{ m/s}$$

The wavelength in meters of the sampling frequency is $v/f_s$ = $12000/9000 = 1.33 \text{ m}$

This means a the maximum length of the beam is 1.33 meters for a unique solution over the full sampling rate given this sampling rate given. However all frequencies up to point where their delay would be $2\pi$ would be resolvable and this frequency would be:

$$f_{max} = \frac{v}{L} = \frac{12000}{74} \approx 162.2 \text{ Hz}$$

Given this, there is no reason to sample higher than twice this frequency but it doesn't hurt to do so. We would expect to see repeating images of everything from DC to 162.2 Hz at the higher frequencies.