Two loudspeakers in a plane are 2.0 m apart and in phase with each other. Both emit 700 Hz sound waves into a room where the speed of sound is 341 m/s. A listener stands 5.0 m in front of the loudspeakers and 2.0 m to one side of the center line. Is the interference at this point completely constructive, completely destructive, or in between

interference is between destructive  and constructive

Explanation:

The interference of two sound waves periodicity in phase by the speakers is

          Δr =

in this case they indicate that the frequency is f = 700 Hz, the wave speed is

            v =λ f

            λ = v / f

            λ = 341/700

            λ = 0.487 m

Let's use the Pythagorean theorem to find the distance that each wave travels

        r₁ =

let's measure the distance from speaker 1

          r₁ =

          r₁ = 5,099 m

the distance from the second speaker

          r₂ = \sqrt{x^2 + y^2}

          r₂ =

          r₂= 5.831 m

the difference in the way is

          Δr = r₂ -r₁

          Δr = 5,831 - 5,099

          Δr = 0.732 m

          = Δr /λ

         \frac{ \phi }{2\pi } = 0.732 / 0.487

         \frac{ \phi }{2\pi } = 1.50

this is the phase difference this phase difference is approximately

            Ф= =1.57,

so the interference is between destructive ( Ф = π) and constructive (Ф=2π)

hypothesising or postulating

your answer is, orbit around the nucleus