Two speakers are driven by a common oscillator at 870 Hz and face each other at a distance of 1.20 m. Locate the points along a line joining the two speakers where relative minima of pressure amplitude would be expected. (Use v = 343 m/s. Choose one speaker as the origin and give your answers in order of increasing distance from this speaker.)
To locate the points along a line joining the two speakers where relative minima of pressure amplitude would be expected, we need to use the concept of interference of waves.
First, let's understand the phenomenon of interference. When two waves meet, they superpose, meaning their amplitudes add up or cancel each other out, depending on their phase relationship. In the case of sound waves, when two waves meet in phase (crest aligns with crest or trough aligns with trough), they reinforce each other, resulting in constructive interference and increased amplitude. On the other hand, when two waves meet out of phase (crest aligns with trough), they cancel each other out, resulting in destructive interference and decreased amplitude.
In our setup, we have two speakers driven by a common oscillator at 870 Hz. Since the speakers are driven by the same source, they produce sound waves with the same frequency and wavelength.
To determine the points along the line where relative minima of pressure amplitude would be expected, we need to consider the path difference between the waves produced by the two speakers. This path difference will determine the phase relationship between the waves and whether they interfere constructively or destructively.
Let's consider the two extreme cases:
1. Constructive Interference: In this case, the path difference between the two waves is an integer multiple of the wavelength. That is, the path difference can be written as Δx = nλ, where n is an integer and λ is the wavelength.
2. Destructive Interference: In this case, the path difference between the two waves is an odd integer multiple of half the wavelength. That is, the path difference can be written as Δx = (n + 1/2)λ, where n is an integer and λ is the wavelength.
Given that the distance between the two speakers is 1.20 m and the sound velocity is 343 m/s, we can calculate the wavelength using the formula:
λ = v/f
where λ is the wavelength, v is the sound velocity, and f is the frequency.
Substituting the given values, we have:
λ = 343/870
λ ≈ 0.394 m
Now, we can determine the points along the line where relative minima of pressure amplitude would be expected. We start from one of the speakers and increment the distance by λ/2 for each minima.
Let's assume the speaker at the origin as Speaker A. From Speaker A, the first minimum would be located at a distance of λ/2, the second minimum would be at a distance of λ, the third minimum at a distance of 3λ/2, and so on.
Therefore, the points along the line, in order of increasing distance from Speaker A, where relative minima of pressure amplitude would be expected are:
- λ/2 = 0.197 m (First minimum from Speaker A)
- λ = 0.394 m (Second minimum from Speaker A)
- 3λ/2 = 0.591 m (Third minimum from Speaker A)
- 2λ = 0.788 m (Fourth minimum from Speaker A)
- 5λ/2 = 0.985 m (Fifth minimum from Speaker A)
- ...
Note: The pattern of minima repeats after each λ, so you can continue the pattern indefinitely by adding subsequent multiples of λ/2 to the distances mentioned above.
To locate the points where relative minima of pressure amplitude would be expected between two speakers driven by a common oscillator, we can use the concept of constructive and destructive interference.
The distance between the two speakers is 1.20 m, and the wavelength can be calculated using the formula:
λ = v / f
where:
λ is the wavelength
v is the speed of sound (343 m/s)
f is the frequency (870 Hz)
Plugging in the values, we get:
λ = 343 / 870
λ ≈ 0.394 m
In order to locate the points of destructive interference (relative minima), we need to find the positions where the path difference between the two speakers is equal to an odd number of half wavelengths (λ/2).
Let's assume the distance from the first speaker (the origin) to the point of interest is x. Then, the distance from the second speaker to the same point is 1.20 - x.
The path difference is equal to the difference in distances traveled by the sound waves from each speaker, which can be calculated as:
Δx = (1.20 - x) - x
Δx = 1.20 - 2x
To find the relative minima, we set Δx equal to an odd number of half wavelengths:
1.20 - 2x = (2n + 1) * (λ/2)
where n is an integer representing the order of the destructive interference.
Now, let's solve for x using the equation:
1.20 - 2x = (2n + 1) * (0.394/2)
Rearranging the equation, we find:
x = (1.20 - (2n + 1) * (0.394/2))/2
Using this equation, we can plug in different values for n to find the positions of the relative minima.
For example, when n = 0:
x = (1.20 - (2(0) + 1) * (0.394/2))/2
x ≈ 0.906 m
When n = 1:
x = (1.20 - (2(1) + 1) * (0.394/2))/2
x ≈ 0.453 m
Similarly, you can calculate x for higher values of n to find the positions of additional relative minima.
So, the points along the line joining the two speakers where the relative minima of pressure amplitude would be expected, in order of increasing distance from the first speaker (origin), are approximately 0.453 m and 0.906 m.