Question

To loudspeakers face each other, vibrate in phase, and produce identical 520

Answer 1

f = 520 Hz, u = 343 m/s, so wavelength ? = u/f

f1 = 2.06 Hz

find v

distance between speakers = d

distance between one speaker and the listener = x
path difference = (d - x) - x = d - 2x

assume at x he hears loud, so d - 2x = k? = ku/f        (1)

after time t = 1/f1, he hears next loud, so

d - 2(x + vt) = (k - 1)u/f      (2)

(1) - (2):

2vt = u/f

2v/f1 = u/f

so v = u*f1/(2f) =343*2.06/2*520 = 0.6794 m/s

Answer 2

I think we can assume a standing wave pattern here
each quiet point would be 1/2 wave length apart
using 343 m/s
1/2 lambda = 1/2 (343/520) = 0.3298 m
now you are moving at 2.06*0.3298 m/s = 0.6794 m/s

Answer 3

You've calculated the wavelength of the original waves correctly: lambda = 343 m/sec / (520/sec) ~ 65.96 cm. And you know all the sound trapped between the speakers has that wavelength.
Someone walking through that sound field will hear the loudness changing due to the standing waves set up (neglecting the fact that standing in a sound field will change the field itself). However, the ear can't tell the difference between pressure in a sound wave which is *below* ambient, and pressure *above* ambient (another way of saying that is that your ears don't detect phase, just amplitude and frequency). So if you were to walk along a whole wavelength of the sound, starting at a null (place of no sound), you'd hear
i) starting at a null (node): no sound;
ii) walk 1/4 wavelength: lots of sound; you've reached an antinode;
iii) walk 1/4 more wavelength: no sound; you're at another null;
iv) walk 1/4 more wavelength: lots of sound; you've reached another antinode; note that if you drew the "static" wave, this antinode would be 180 deg. out of phase with that other antinode (in step ii), but your ears don't care; the wave is changing back and forth 520 times a second; what you draw is a snapshot in time;
v) walk 1/4 more wavelength: back to a null.