Acoustic beats interference vs. path difference interference I'm trying to understand the differences between two kinds of waves interference. The first one is a time-interference, only due to frequency difference of the two waves with the same amplitude, which gives origin to acoustic beats. The other one is a spatial interference, originated by the different path followed by two waves with the same frequency. Usually these two are clear to me but sometimes it is not clear which one is acting.
In this example (exercise from Serway-Jewett) I think it is quite difficult to choose which of the two kind of interference is happening. I'm not looking for the resolution, I just report it as an example.

A loudspeaker at the front of a room and an identical loudspeaker at
  the rear of the room are being driven by the same oscillator at $456 Hz$. A student walks at a uniform rate of $1.50 m/s$ along the length of
  the room. She hears a single tone, repeatedly becoming louder and
  softer. Model these variations as beats between the
  Doppler-shifted sounds the student receives. Calculate the number of
  beats the student hears each second.

It is explicitly said that it is a phenomenon of beats, originated by the Doppler shift of frequency, and I'm ok with this. 
But could the same situation be explained ignoring the Doppler effect, but considering that the two waves, with the same frequency interfere with each other constructively or distructively in different positions and the student hears these variations because she is changing her position indeed?
Is there a clear distinction between the two interference situations in similar cases? I actually do not see any reason to choose one interference model or the other in the problem in the example.
Am I missing something? 
 A: Yes, the two approaches are equivalent. As you've noted, there's only one physical effect going on here, interference, and the standard beat frequency / path length interference formulas are just special cases of the same thing. 
There are some restrictions. Both beat frequency and path length interference only make physical sense when the sources are coherent and the frequencies are close to each other. In the path length case, we often also assume that the frequencies are exactly equal, for convenience. (You can accommodate a frequency difference by adding a time-varying phase difference.)
Switching between the two approaches can be useful for problem solving. For example, suppose you have a big row of evenly spaced speakers, each one tuned $1 \, \text{Hz}$ higher than the previous one, and you want to run to avoid hearing any beats.
This problem looks messy, but you can solve it quickly if you instead view it from a path length interference perspective. There, you have a diffraction grating with time-varying phase difference between the slits. The solution is to run at the same speed a diffraction maximum is moving.
A: The exercise seems not too difficult. With $f_0 = 456 \text{Hz}$, $c$ the speed of sound an $v$ the speed of the observer, you just have to find the beat frequency by adding the wave equations for 2 different sound waves with frequencies $f_1=f_0(1-\frac{v}{c})$ and $f_2=f_0(1+\frac{v}{c})$. So this is just an exercise in applying the Doppler effect, and you do not need to worry about standing waves.
But if I understand your question, you are not asking the answer of the exercise, but wheter if the observer would be moving very slow (so you can neglect the Doppler effect), it would still be possible for this person to hear variations in the intensity, dependent on the position. The answer is yes, and you even won't need two speakers.


*

*Consider the case of an 1D wave originating from a single speaker in a room (see figure). If the length of the room $l$ happens to be a multiple of the wavelength $\lambda$ of the wave (via $f = c\cdot \lambda$), a standing wave may occur. The intensity differences this person will hear will be separated by a distance $\frac{\lambda}{4}$. For 456 Hz, the lowest and highest intensities will be about 16 cm separated from each other (I used 300 m/s for the speed of sound). Furthermore, the speaker surface has to be preferably at an antinode of the standing wave, since if it is at another place, the wave will be less intense.





*I do not think that the position of the speaker may influence this (but I'm not sure). You also do not have to strike a guitar string where the antinode of the standing wave will be.

*Adding a second speaker that emits a sound with the same frequency does not change the situation for the beats. The intensity of the peaks may double, as you deliver more energy to the system. But it will not affect the distance between the nodes.
So if the length of your room is equal to a multiple of the wavelength, this will be the case. In open air, or in a room dat does not fit this condition, an observer would not hear these 'spatial beats'.
EDIT
On the path difference situation, as this was not clear in my initial answer.
If, and only if the speakers emit sound in a coherent way, interference will occur. The situation will be very similar to the well-known double slit experiment for light (you have two coherent point sources, and it does not matter whether they emit light or sound). Without walls around the set-up, you can easily calculate minima and maxima for the resulting wave. If you take the room into account, it will influence the pattern by reflection at the walls, but I think the mathematics will become quite involved. 
This path difference effect is exploited in so-called 'line arrays' at concerts: interference between different speakers in such setups enables audio engineers to direct the sound waves to the audience. The wave from a combined source will be more directional than the intensity of a single source. So instead of wasting part of the power to send sound waves e.g. up into the air, or, at a festival, towards another concert, you deliver most of the power at the audience of the concert. 
