Quick Sound Question

  • Thread starter Shakas
  • Start date
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!
 
Last edited:

dynamicsolo

Homework Helper
1,649
4
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!
For the lowest frequency, the listener is standing at a point where there is one half-wavelength path difference between the two sources. If you double the frequency, what happens to the wavelength of the emitted sound? What would the path difference be now in terms of the new wavelength? Would that give you destructive or constructive interference?

[EDIT: It occurs to me that this is similar to the situation for the question of the wavelengths that lead to standing waves in a half-open pipe of a specified length...]
 
Last edited:
Oh, I think I got it. If I were to double the frequency, then the wavelength would be halved. But since the guy is already standing at half the wave length, he would hear constructive interference. So to get destructive interference he has to be at 3/2 the wave length. Then you have to multiply the wavelength by 2/3. Same deal with the next frequency.

Thanks for the help!
 

Want to reply to this thread?

"Quick Sound Question" You must log in or register to reply here.

Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving

Top Threads

Top