(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

Two loudspeakers are 2.5 m apart. A person stands 3.0 m from one speaker and 3.8 m from the other. Assume the air temperature is 20°C.

(a) What is the lowest frequency at which destructive interference will occur at this point?

(b) Calculate the next two frequencies that also result in destructive interference at this point.

second lowest frequency?

third lowest frequency?

2. Relevant equations

v=lambda*f

v = velocity (m/s)

lambda = wavelength (m)

f = frequency (Hz)

3. The attempt at a solution

Alright, so part A understand. The difference between the distances would be half the wavelength, double it, plug it in to the equation, find out that

Part A: f = 214.375 Hz.

Answers to part B: 643.125 Hz (triple original answer) and 1071.875 (5x the original answer)

Now what I don't get is why the next two frequencies are triple/5x my original answer instead of double/triple. I thought that an overtone would be where the next frequency would be, which I thought was double/triple/etc. the fundamental frequency.

Any help is greatly appreciated!

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# Quick Sound Question

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