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

• Shakas
In summary, in this conversation, two loudspeakers are placed 2.5 m apart and a person stands 3.0 m from one speaker and 3.8 m from the other. The lowest frequency at which destructive interference will occur at this point is calculated to be 214.375 Hz. The next two frequencies that also result in destructive interference are calculated to be 643.125 Hz and 1071.875 Hz, which are triple and 5x the original frequency, respectively. This is because for destructive interference to occur, the listener must be at a point where there is one half-wavelength path difference between the two sources. Therefore, doubling the frequency would result in constructive interference, and to achieve destructive interference
Shakas

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.

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:
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!

1. What is the purpose of this homework assignment?

The purpose of this homework assignment is to practice finding the lowest and next two frequencies at a distance of 3 meters and 3.8 meters.

2. Why is it important to know the lowest and next two frequencies at a specific distance?

Knowing the lowest and next two frequencies at a specific distance can help determine the characteristics of sound waves and how they interact with their environment. This information can be useful in various fields such as acoustics, engineering, and communication.

3. How do you calculate the lowest and next two frequencies at a distance of 3 meters and 3.8 meters?

The lowest frequency can be calculated by dividing the speed of sound (343 m/s) by the wavelength (3 m or 3.8 m). The next two frequencies can be found by multiplying the lowest frequency by 2 and 3, respectively.

4. What factors can affect the lowest and next two frequencies of sound waves at a specific distance?

The speed of sound, temperature, and medium through which the sound travels can all affect the lowest and next two frequencies of sound waves at a specific distance. In addition, the source of the sound and any obstacles or barriers in the environment can also impact these frequencies.

5. How can knowing the lowest and next two frequencies at different distances be applied in real life?

Knowing the lowest and next two frequencies at different distances can be applied in various real-life situations such as designing sound systems, predicting sound propagation in different environments, and optimizing communication systems. This information can also be useful in fields such as music, sonar, and ultrasound technology.

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