Physics, 13.10.2020 05:01, hdjehd

Two speakers placed 0.94 m apart produce pure tones in sync with each other at a frequency of 1630 Hz. A microphone can be moved along a line parallel to the line joining the speakers and 9.4 m from it. An intensity maximum is measured a point P0 where the microphone is equidistant from the two speakers. As we move the microphone away from P0 to one side, we find intensity minima and maxima alternately. Take the speed of sound in air to be 344 m/s, and you can assume that the slits are close enough together that the equations that describe the interference pattern of light passing through two slits can be applied here. Required:
a. What is the distance, in meters, between Po and the first intensity minimum?
b. What is the distance, in meters, between Po and the first intensity maximum?
c. What is the distance, in meters, between Po and the second intensity minimum?
d. What is the distance, in meters, between Po and the second intensity maximum?

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answered: missymiss4273

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;

$sin(\theta) = \dfrac{Path \ difference}{d}$

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;

$\lambda = \dfrac{v}{f}$

Therefore;

$\lambda = \dfrac{344}{1,630} = \dfrac{172}{815} \approx 0.211$

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:

link

answered: joanasprinkman2262

a. approximately $1.1\; \rm m$ (first minimum.)

b. approximately $2.2\; \rm m$ (first maximum.)

c. approximately $3.4\; \rm m$ (second minimum.)

d. approximately $4.7\; \rm m$ (second maximum.)

Explanation:

Let represent the separation between the two speakers (the two "slits" based on the assumptions.)

Let $\theta$ represent the angle between:

the line joining the microphone and the center of the two speakers, andthe 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 P_0 would thus be $9.4\, \tan(\theta)$ meters.

Based on the assumptions and the equation from Young's double-slit experiment:

$\displaystyle \sin(\theta) = \frac{\text{path difference}}{d}$ .

Hence:

$\displaystyle \theta = \arcsin \left(\frac{\text{path difference}}{d}\right)$ .

The "path difference" in these two equations refers to the difference between the distances between the microphone and each of the two speakers. Let $\lambda$ denote the wavelength of this wave.

$\displaystyle \begin{array}{c|c} & \text{Path difference} \\ \cline{1-2}\text{First Minimum} & \lambda / 2 \\ \cline{1-2} \text{First Maximum} & \lambda \\\cline{1-2} \text{Second Minimum} & 3\,\lambda / 2 \\ \cline{1-2} \text{Second Maximum} & 2\, \lambda\end{array}$ .

Calculate the wavelength of this wave based on its frequency and its velocity:

$\displaystyle \lambda = \frac{v}{f} \approx 0.211\; \rm m$ .

Calculate $\theta$ for each of these path differences:

$\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}$ .