A person stands directly in front of two speakers

Question

A person stands directly in front of two speakers that are emitting the same pure tone. The person then moves to one side until no sound is heard. At that point, the person is 6.65 m from one of the speakers and 6.67 m from the other. What is the frequency (in Hz) of the tone being emitted? (Assume air temperature to be 20°C.)

Answer 1

Initially the sounds were in phase as the path difference was 0. When he moves to the first MINIMUM the path difference is λ/2 =6.67 m - 6.65 m = 0.02 m. Assuming v - 343 m/s, f = v/f = 343 /.02 = 1.7 x 10⁴ Hz