Solving Billy and Speakers: 16m, 0.8m, 3000 Hz

  • Thread starter axgalloway
  • Start date
So, from that point, you have a right triangle with one leg 16m, the other leg 0.05715m. Not sure how that triangle is configured in relation to the wall, but you should be able to find the length of the hypotenuse.In summary, the problem involves two loud speakers located 16m from a wall, playing a 3000 Hz sound in phase. The speed of sound is 343 m/s and the wavelength is calculated to be 0.1143 meters. To find the distance that Bill needs to walk along the wall to hear the m=1 maximum, a right triangle can be formed with one leg being 16m and the other being 0.05715m
  • #1
axgalloway
6
0

Homework Statement



Two lound speakers are each located 16m from a wall and are 0.8m apart. They play a 3000 Hz sound, both speakers are in phase. Bill is located at the wall at the m=0 maximum of sound interference. If he walks along the wall, how far does he have to walk to hear m=1 maximum? (The speed of sound is 343 m/s.)


Homework Equations



2pi * (x2-x1) / lambda = 2mpi
vsound = lambda*f


The Attempt at a Solution



343 m/s = lambda*3000Hz
lambda= .1143 meters

Beyond that, I do not know how to solve it.
Thank you.
 
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  • #2
Sounds like you have a little geometry problem. Don't you just need to find the point away from the mid-line whose distance from each speaker is different by half a wavelength?
 
  • #3


I would first clarify the problem by asking for additional information such as the direction of the sound waves and the orientation of the speakers. This will help determine the specific equation to use in solving the problem.

Assuming that the sound waves are traveling in a straight line and the speakers are oriented towards the wall, we can use the equation for constructive interference to solve for the distance that Bill needs to walk to reach the m=1 maximum.

Constructive interference occurs when the path difference between the two sound waves is equal to an integer multiple of the wavelength (λ). In this case, we can use the following equation:

2π * (x2-x1) / λ = 2π * m

Where:
x2 = distance from the second speaker to the wall (16m)
x1 = distance from the first speaker to the wall (16m+0.8m = 16.8m)
λ = wavelength (0.1143m)
m = order of maximum interference (1)

By substituting the given values, we can solve for x2:

2π * (x2-16.8m) / 0.1143m = 2π * 1
x2-16.8m = 0.1143m
x2 = 16.9143m

Therefore, Bill needs to walk 0.9143m along the wall to reach the m=1 maximum of sound interference. This can also be confirmed by using the equation for the speed of sound to calculate the time it takes for the sound wave to travel the additional distance of 0.9143m:

Time = Distance / Velocity
Time = 0.9143m / 343m/s = 0.0027s

Since the frequency of the sound is 3000Hz, the time for one cycle is 1/3000s = 0.000333s. Therefore, in 0.0027s, the sound wave will complete 8 cycles, which corresponds to the m=1 maximum of interference.
 

1. How do you solve for the sound frequency given the distance and size of the speakers?

To solve for the sound frequency, we can use the formula f = c/λ, where c is the speed of sound (approximately 343 m/s) and λ is the wavelength. In this case, we can calculate the wavelength by dividing the distance between the speakers (16m) by the number of wavelengths (0.8m). This gives us a wavelength of 20m. Plugging this into the formula, we get a frequency of 3000 Hz.

2. How does the size of the speakers affect the sound frequency?

The size of the speakers does not directly affect the sound frequency. However, it can indirectly impact the frequency by affecting the wavelength. Smaller speakers may produce shorter wavelengths, resulting in higher frequencies, while larger speakers may produce longer wavelengths and lower frequencies.

3. Can the distance between the speakers be adjusted to change the sound frequency?

Yes, the distance between the speakers can be adjusted to change the sound frequency. As mentioned earlier, the wavelength is directly proportional to the distance between the speakers. Therefore, increasing the distance will result in a longer wavelength and a lower frequency, while decreasing the distance will result in a shorter wavelength and a higher frequency.

4. How accurate is the calculated sound frequency?

The calculated sound frequency is fairly accurate, but it may not be exact due to various factors such as air temperature, humidity, and the materials surrounding the speakers. These factors can affect the speed of sound and the wavelength, resulting in a slightly different frequency than the calculated one.

5. What other factors should be considered when solving for sound frequency?

In addition to the distance and size of the speakers, other factors that should be considered when solving for sound frequency include the speed of sound, the wavelength, and the medium through which the sound is traveling. These factors can vary depending on the environment and can impact the accuracy of the calculated frequency.

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