Quick Sound Question Homework: Lowest & Next Two Freqs at 3m & 3.8m

  • Thread starter Thread starter Shakas
  • Start date Start date
  • Tags Tags
    Sound
AI Thread Summary
The discussion revolves around calculating the lowest frequency and the next two frequencies for destructive interference between two loudspeakers positioned 2.5 m apart, with a listener standing at specific distances from each speaker. The lowest frequency identified is 214.375 Hz, while the subsequent frequencies are 643.125 Hz and 1071.875 Hz, derived from multiplying the fundamental frequency by three and five, respectively. A key point of confusion is the relationship between overtone frequencies and their multiples, with the original poster questioning why the next frequencies are not simply double or triple the fundamental frequency. Clarification is provided that doubling the frequency results in halving the wavelength, thus altering the path difference and leading to constructive interference instead of destructive. Ultimately, understanding the relationship between frequency, wavelength, and path difference is crucial for solving such interference problems.
Shakas
Messages
5
Reaction score
0

Homework Statement



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?

Homework Equations



v=lambda*f

v = velocity (m/s)
lambda = wavelength (m)
f = frequency (Hz)

The Attempt at a Solution



Alright, so part A understand. The difference between the distances would be half the wavelength, double it, plug it into 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:
Physics news on Phys.org
Shakas said:
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!
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top