Two identical speakers, labeled Speaker 1 and Speaker 2, are playing a tone with a frequency of 175 Hz in phase. The centers of the speakers are located 6.00 m apart. Five observers are standing on a line that is 6.00 m in front of the speakers, as shown in the figure. Each observer is separated by 1.00 m and Pam is directly in front of speaker 1. The speed of sound is 343 m/s for this problem.
At which observer will the sound be the loudest?
Pam, the observer directly in front of Speaker 1. Since the speakers are playing the same tone in phase, the sound waves will be in phase and will combine constructively at the point directly in front of Speaker 1. This will result in the loudest sound at this point.
To answer this question, we need to consider the concept of interference in sound waves. Interference occurs when two or more waves combine to form a resultant wave. There are two types of interference: constructive and destructive.
In constructive interference, the waves have the same frequency and are in phase with each other, meaning their crests and troughs align. This causes the amplitude of the resulting wave to increase.
In destructive interference, the waves also have the same frequency, but they are out of phase. This means that the crests of one wave align with the troughs of the other wave, resulting in a decreased or even zero amplitude.
In this scenario, we have two identical speakers playing a tone with a frequency of 175 Hz in phase. The speed of sound is given as 343 m/s, and the centers of the speakers are located 6.00 m apart. There are five observers standing on a line that is 6.00 m in front of the speakers, with each observer separated by 1.00 m.
To determine the interference pattern at each observer position, we need to calculate the path difference between the waves reaching that position.
Let's consider the observer positions from left to right, with Pam directly in front of Speaker 1.
Observer 1 (Pam):
Since Pam is directly in front of Speaker 1, the waves from both speakers have the same distance to travel. Therefore, there is no path difference, and the interference is constructive. The resulting wave will have a maximum or increased amplitude.
Observer 2:
To calculate the path difference, we need to consider the extra distance traveled by the wave from Speaker 2 compared to Speaker 1. From the given information, we know that the distance between the speakers is 6.00 m. Since Observer 2 is located 1.00 m to the right of Pam, the path difference can be calculated as 6.00 m - 1.00 m = 5.00 m.
The path difference of 5.00 m corresponds to a phase difference of 2π * (path difference / wavelength). Since we have the frequency of 175 Hz, we can calculate the wavelength using the formula: wavelength = speed of sound / frequency. Thus, wavelength = 343 m/s / 175 Hz = 1.96 m.
Substituting the values into the phase difference formula: phase difference = 2π * (5.00 m / 1.96 m) = 16π/5.
For constructive interference, the phase difference must be an integral multiple of 2π (i.e., 2π, 4π, 6π, etc.). In this case, the phase difference is not an integral multiple of 2π, so the interference at Observer 2 is destructive. The resulting wave will have a decreased or even zero amplitude.
Observer 3:
Observer 3's position is 1.00 m to the right of Observer 2. Therefore, the path difference is the same as Observer 2, which is 5.00 m.
Following the same steps as above, we can calculate the phase difference: phase difference = 2π * (5.00 m / 1.96 m) = 16π/5.
Again, the phase difference is not an integral multiple of 2π, so the interference at Observer 3 is also destructive.
Observer 4:
Observer 4's position is 1.00 m to the right of Observer 3. Therefore, the path difference is the same as Observer 2 and Observer 3, which is 5.00 m.
Calculating the phase difference: phase difference = 2π * (5.00 m / 1.96 m) = 16π/5.
Once again, the phase difference is not an integral multiple of 2π, so the interference at Observer 4 is destructive.
Observer 5:
Observer 5's position is 1.00 m to the right of Observer 4. Therefore, the path difference is the same as Observer 2, Observer 3, and Observer 4, which is 5.00 m.
Calculating the phase difference: phase difference = 2π * (5.00 m / 1.96 m) = 16π/5.
As before, the phase difference is not an integral multiple of 2π, so the interference at Observer 5 is also destructive.
To summarize, the interference pattern observed by the five observers is as follows:
- Observer 1 (Pam): Constructive interference
- Observer 2: Destructive interference
- Observer 3: Destructive interference
- Observer 4: Destructive interference
- Observer 5: Destructive interference