Destructive interference of sound waves

[physicsfan01]

Homework Statement

Two loudspeakers are located 3.42m apart on an outdoor stage. A listener is 18.80m from one and 19.60m from the other. During the sound check, a signal generator drives the two speakers in phase with the same amplitude and frequency. The transmitted frequency is swept through the audible range (20 Hz to 20 kHz). What are the three lowest frequencies at which the listener will hear a minimum signal because of destructive interference.


2. The attempt at a solution

I know that (2n+1)/2 * wavelength = 19.6 - 18.8. So I try different values of n to find the wavelength then from this I try to calculate the frequencies. But I don't understand what the distance between the two speakers has to do with the problem.

 

[anathema]

I can offer some insight into the problem at hand. The distance between the two speakers is a crucial factor in determining the frequencies at which destructive interference will occur. This is because the distance between the two speakers affects the phase difference between the waves they produce.

When sound waves from two sources interfere, they can either add constructively (in phase) or cancel each other out destructively (out of phase). In order for destructive interference to occur, the waves from the two speakers must be out of phase by 180 degrees. This means that the distance between the two speakers must be equal to half of the wavelength of the sound wave.

In this case, the distance between the two speakers is 3.42m, which means that the wavelength of the sound wave must be 6.84m for destructive interference to occur. This can be calculated using the formula you mentioned, (2n+1)/2 * wavelength = 19.6 - 18.8.

Now, to find the three lowest frequencies at which destructive interference will occur, we can use the formula for wavelength, wavelength = speed of sound / frequency. The speed of sound is approximately 343 m/s, and we know that the wavelength is 6.84m. So, rearranging the formula, we get frequency = speed of sound / wavelength.

Substituting the values, we get the three lowest frequencies as 50 Hz, 25 Hz, and 16.67 Hz. These are the frequencies at which the sound waves from the two speakers will be out of phase by 180 degrees, resulting in destructive interference and a minimum signal at the listener's location.

I hope this helps to clarify the role of the distance between the two speakers in this problem. Let me know if you have any further questions or if you need clarification on any of the steps in this solution.

 

[Moetasim]

I can explain the concept of destructive interference of sound waves in this scenario. Destructive interference occurs when two waves of the same frequency and amplitude meet and result in a decrease in overall amplitude, or volume, of the sound. This happens because the peaks of one wave align with the troughs of the other wave, causing them to cancel each other out.

In this specific scenario, the distance between the two speakers is important because it determines the phase difference between the two waves. As the sound waves travel from the two speakers to the listener, they will have a certain phase difference based on the distance traveled. This phase difference can be calculated using the equation given in the problem ((2n+1)/2 * wavelength = 19.6 - 18.8), where n represents the number of wavelengths between the two speakers.

When the phase difference between the two waves is an odd multiple of half a wavelength (n=1,3,5...), destructive interference will occur at the listener's location. This means that the listener will hear a minimum signal at these specific frequencies, as the waves from the two speakers are canceling each other out. The three lowest frequencies at which this will occur can be calculated by solving for the wavelength and then using the formula v = f * wavelength, where v is the speed of sound (343 m/s).

In conclusion, the distance between the two speakers plays a crucial role in determining the frequencies at which destructive interference will occur and the listener will hear a minimum signal. By understanding this concept and using the given equation, the three lowest frequencies at which this will occur can be calculated.