Imagine you are in an open field where two loudspeakers are set up and connected to the same amplifier so that they emit sound waves in phase at 688 Hz. Take the speed of sound in air to be 344 m/s. If you are 3.00 m from speaker A directly to your right and 3.50 m from speaker B directly to your left. What is the shortest distance d you need to walk forward to be at a point where you cannot hear the speakers?

In order to have destructive interference, the path difference from the sources to the listener must be an odd multiple of half wavelengths, as follows:

d = (2n+1) * λ/2

In orfer to know which is the wavelength, we can use the relationship between propagation speed (in this case speed of sound), frequency and wavelength:

v= λ*f ⇒ λ = v/f = 344 m/s / 688 1/sec = 0.5 m

So, the path difference must be, at least, λ/2:

d = b-a = λ/2, where b is the distance to the speaker B, and a, the distance to the speaker A.

Applying Pithagorean Theorem, as the perpendicular distance d (which is our unknown) is the same for the triangles defined by the horizontal distance to the listener, and the straight line from the new position to the sources, we can write:

d² = a²- (3.0)²

d² = b²- (3.5)²

As the left sides are equal, so do right sides:

a² - (3.0)² = b² - (3.5)²

⇒ b² - a² = (3.5)² - (3.0)² = 3.25

We can replace (b²- a²) as follows:

b² - a² = (b+a)(b-a) = 3.25 (2)

We know that b-a = λ/2 = 0.25

Replacing in (2), we have:

b+a = 3.25 / 0.25 = 13 m

b-a = 0.25 m

Adding both sides:

2*b = 13.25 m ⇒ b= 13.25 /2 = 6.63 m

⇒ d² = (6.63)² - (3.5)² = 31.6 m²

⇒ d=√31.6 m² = 5.62 m