Why Does Destructive Interference Occur at Different Frequencies?

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SUMMARY

Destructive interference occurs when the path difference between sound waves from two speakers results in a phase difference of half a wavelength. In this discussion, the speakers are 2.36m apart, and a listener is positioned 2.8m from one speaker and 3.4m from the other, leading to a path difference of 0.6m. The correct wavelength for destructive interference is calculated as 0.12m, resulting in a frequency of 2858.33Hz using the sound velocity of 343 m/s. The participant's initial calculation was incorrect due to a misunderstanding of the relationship between path difference and wavelength.

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  • Understanding of sound wave properties
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  • Basic mathematics for calculating path differences
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  • Learn about the relationship between wavelength, frequency, and velocity of sound
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Destructive Interference: Please Help!

Two loudspeakers are 2.36m apart. A person stands 2.8m from one speaker and 3.4m from the other. What is the lowest frequency at which destructive interference will occur at this point?

My answer is wrong, but this is what I did:
3.4m-2.8m=0.6m, so the person is .6m farther from one speaker than the other.
In order to sense destructive Interference, the person, must be one-half wavelength closer to or farther from one speaker than the other, the wavelength must be twice the distance difference. Therefore, Wavelength=0.12m
Using the equation Wavelength =velocity/frequency, I solved for frequency. (velocity of sound is 343 m/s).
My answer: 2858.33Hz

PLease help! Thanks!
 
Physics news on Phys.org
What's 2 times 0.6? :wink:
 
Haha, thanks! Those careless mistakes make such a difference!
 

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