Respuesta :
Answer:
a. approximately [tex]1.1\; \rm m[/tex] (first minimum.)
b. approximately [tex]2.2\; \rm m[/tex] (first maximum.)
c. approximately [tex]3.4\; \rm m[/tex] (second minimum.)
d. approximately [tex]4.7\; \rm m[/tex] (second maximum.)
Explanation:
Let [tex]d[/tex] represent the separation between the two speakers (the two "slits" based on the assumptions.)
Let [tex]\theta[/tex] represent the angle between:
- the line joining the microphone and the center of the two speakers, and
- the line that goes through the center of the two speakers that is also normal to the line joining the two speakers.
The distance between the microphone and point [tex]P_0[/tex] would thus be [tex]9.4\, \tan(\theta)[/tex] meters.
Based on the assumptions and the equation from Young's double-slit experiment:
[tex]\displaystyle \sin(\theta) = \frac{\text{path difference}}{d}[/tex].
Hence:
[tex]\displaystyle \theta = \arcsin \left(\frac{\text{path difference}}{d}\right)[/tex].
The "path difference" in these two equations refers to the difference between the distances between the microphone and each of the two speakers. Let [tex]\lambda[/tex] denote the wavelength of this wave.
[tex]\displaystyle \begin{array}{c|c} & \text{Path difference} \\ \cline{1-2}\text{First Minimum} & \lambda / 2 \\ \cline{1-2} \text{First Maximum} & \lambda \\\cline{1-2} \text{Second Minimum} & 3\,\lambda / 2 \\ \cline{1-2} \text{Second Maximum} & 2\, \lambda\end{array}[/tex].
Calculate the wavelength of this wave based on its frequency and its velocity:
[tex]\displaystyle \lambda = \frac{v}{f} \approx 0.211\; \rm m[/tex].
Calculate [tex]\theta[/tex] for each of these path differences:
[tex]\displaystyle \begin{array}{c|c|c} & \text{Path difference} & \text{approximate of $\theta$} \\ \cline{1-3}\text{First Minimum} & \lambda / 2 & 0.112 \\ \cline{1-3} \text{First Maximum} & \lambda & 0.226\\\cline{1-3} \text{Second Minimum} & 3\,\lambda / 2 & 0.343\\ \cline{1-3} \text{Second Maximum} & 2\, \lambda & 0.466\end{array}[/tex].
In each of these case, the distance between the microphone and [tex]P_0[/tex] would be [tex]9.4\, \tan(\theta)[/tex]. Therefore:
- At the first minimum, the distance from [tex]P_0[/tex] is approximately [tex]1.1\; \rm m[/tex].
- At the first maximum, the distance from [tex]P_0[/tex] is approximately [tex]2.2\; \rm m[/tex].
- At the second minimum, the distance from [tex]P_0[/tex] is approximately [tex]3.4\; \rm m[/tex].
- At the second maximum, the distance from [tex]P_0[/tex] is approximately [tex]4.7\;\rm m[/tex].
Interference is the result when two or more waves combine
The distance between P₀ and
a. The first intensity minimum is approximately 1.06 m
b. The first intensity maximum is approximately 2.165 m
c. The second intensity minimum is approximately 3.36 m
d. The second intensity minimum is approximately 4.72 m
The reasons the above values are correct are given as follows:
The known parameters are;
The distance between the two speakers = 0.94 m
Frequency of the tone produced by the two speakers = 1,630 Hz (in sync)
The line along which the microphone moves is parallel to the line between the two speakers
The distance between the parallel lines above = 9.4 m
The speed of sound in air, v₀ = 344 m/s
The interference pattern of light passing between two slits is to be applied
a. Based on the arrangement, we have;
P₀ = 9.4 × tan(θ)
Where;
θ = The angle formed formed by the line from the microphone to the midpoint of the distance between the two speakers and the perpendicular bisector to the line joining the two speakers
Based on Young's double-slit experiment, we have;
[tex]sin(\theta) = \dfrac{Path \ difference}{d}[/tex]
Where;
d = The distance between the two speakers representing the slits
The path difference for a minimum is n × λ/2
Where n = 1, 3, 5,...,(set of odd numbers)
The path difference for a maximum intensity sound is n·λ
Where n = 1, 2. 3, ..., n (n is an integer)
The wavelength, is given as follows;
[tex]\lambda = \dfrac{v}{f}[/tex]
Therefore;
[tex]\lambda = \dfrac{344}{1,630} = \dfrac{172}{815} \approx 0.211[/tex]
The wavelength, λ ≈ 0.211
Therefore, the angle, θ, for the first minima, θ, ≈arcsine(0.211/(2×0.94))
First minima, λ/2, P₀ =9.4 × tan(arcsine(0.211/(2×0.94))) ≈ 1.06
First maxima, λ, P₀ =9.4 × tan(arcsine(0.211/(94))) ≈ 2.165
Second minima, 3·λ/2, P₀ = 9.4 × tan(arcsine(3*0.211/(2×0.94))) ≈ 3.36
Second maxima, 2·λ, P₀ = 9.4 × tan(arcsine(2*0.211/(0.94))) ≈ 4.72
Therefore;
a. The distance between P₀ and the first intensity minimum is ≈ 1.06 m
b. The distance between P₀ and the first intensity maximum is ≈ 2.165 m
c. The distance between P₀ and the second intensity minimum is ≈ 3.36 m
d. The distance between P₀ and the second intensity minimum is ≈ 4.72 m
Learn more about interference of sound waves here:
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