What is the frequency (in Hz) of the tone being emitted?

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 7.15 m from one of the speakers and 7.17 m from the other. What is the frequency (in Hz) of the tone being emitted? (Assume air temperature to be 20°C.)

Answer 1

Hi Emma,

The key in this problem is that they tell us where the location is that there is no sound... in other words, the waves of sound being emitted are destructively interfering. They are canceling each other out and they are 180 degrees out of phase. We need to start by figuring out the wavelength of the wave from the distances they provide us, and then we can convert this wavelength into a frequency.

Let's start by looking at the point where there is no sound. The first wave travels 7.15 meters and the other travels 7.17 meters. One wave only traveled 7.17-7.15 = .02 meters more than the other. Since this is the point where they destructively interfere, this distance has to be half of the wavelength. If it were anything but half the wavelength, the waves would not totally cancel each other out.

L₁-L₂=ΔL=.02 meters

ΔL = λ/2

Therefore, λ=.04 meters

We also know that the speed of light relates frequency and wavelength by v=f*λ

so.....

330 m/s = .04 meters * f

Frequency = 8250 Hz.