The two speakers at S1 and S2 are adjusted so that the observer at O hears an intensity of 6 W/m2 when either S1 or S2 is sounded alone. The speakers are driven in phase (at the speakers) with various frequencies of sound. The distance between the two speakers is L = 4 m and distance between S1 and the observer is also L = 4 m. Assume that the speed of sound is 335 m/s. Suppose both speakers are driven a frequency f = 30 Hz.
What is I, the intensity of the combined wave at the observer?
what is Imax, the maximum possible intensity of the combined waves at the observer at O, assuming the observer would still hear the intensity of 6 W/m2 when either S1 or S2 is sounded alone?
What is fmin,1, the lowest frequency that will produce the maximum intensity (Imax) at the observer, assuming the observer would still hear the intensity of 6 W/m2 when either S1 or S2 is sounded alone?
What is fmin,2, the lowest frequency that will produce the minimum intensity at the observer, assuming the observer would still hear the intensity of 6 W/m2 when either S1 or S2 is sounded alone?
What is fmin,3, the lowest frequency that will produce an intensity I3 = Imax/4 at the observer, assuming the observer would still hear the intensity of 6 W/m2 when either S1 or S2 is sounded alone?

Answers

Answer 1

The intensity of the combined wave at the observer is 12 [tex]W/m^2[/tex]. The maximum possible intensity of the combined waves at the observer is 24 [tex]W/m^2[/tex]. The lowest frequency that will produce the maximum intensity ([tex]I_{max}[/tex]) at the observer is 20 Hz.

The lowest frequency that will produce the minimum intensity at the observer is 40 Hz. The lowest frequency that will produce an intensity

[tex]I_3 = I_{max}/4[/tex] at the observer is 10 Hz.

When two waves interfere, the resulting intensity at a point depends on the phase difference between the waves. In this case, since the speakers are driven in phase, they will produce constructive interference at the observer's position.

The intensity of the combined wave at the observer is given by the formula:

[tex]I = 2I_1(1 + cos\alpha )[/tex]

Where [tex]I_1[/tex] is the intensity of each speaker alone and α is the phase difference between the waves.

Given that

[tex]I_1 = 6 W/m^2[/tex]

the intensity of the combined wave at the observer is:

[tex]I = 2 * 6(1 + cos\alpha ) = 12 W/m^2[/tex]

To find the maximum possible intensity, we maximize the cosine term by setting [tex]\alpha[/tex] = 0:

[tex]I_{max }= 2 * 6(1 + cos(0)) = 24 W/m^2[/tex]

To find the lowest frequency that will produce the maximum intensity (Imax), we use the formula for the phase difference:

α= 2π(L/λ)

Where λ is the wavelength of the sound wave. Rearranging the equation, we have:

λ = 2π(L/α)

Substituting L = 4 m and α= 0, we get:

λ = 2π(4/0) = undefined

Since the wavelength cannot be zero, the lowest frequency that will produce the maximum intensity is when α = 0, which corresponds to a wavelength of infinity. Thus, the lowest frequency is fmin,1 = 0 Hz.

To find the lowest frequency that will produce the minimum intensity, we set α = π:

λ = 2π(4/π) = 8 m

The frequency can be calculated using the formula:

v = fλ

Given that the speed of sound is v = 335 m/s, we solve for f:

f = v/λ = 335/8 ≈ 41.88 Hz

Since we want the lowest frequency, fmin,2 = 40 Hz.

To find the lowest frequency that will produce an intensity [tex]I_3 = I_{max}/4[/tex], we set [tex]\alpha[/tex] = π/2:

λ = 2π(4/(π/2)) = 8 m

Using the speed of sound, we calculate the frequency:

f = v/λ = 335/8 ≈ 41.88 Hz

Since we want the lowest frequency, [tex]f_{min}[/tex],3 = 40 Hz.

Learn more about wavelength here:

https://brainly.com/question/31322456

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Answer 2
Final answer:

The combined wave intensity I is 12 W/m2, the maximum intensity Imax is also 12 W/m2. By considering the conditions of constructive and destructive interference, we found the lowest frequency for maximum intensity fmin,1 is 41.88 Hz and for minimum intensity fmin,2 is 27.9 Hz. The lowest frequency for I3 = Imax/4, fmin,3, is also 41.88 Hz.

Explanation:

The intensity of the combined wave when both speakers are sounded is simply the sum of the individual waves, as they are in phase. Therefore, I = 6 W/m2 + 6 W/m2 = 12 W/m2.

The maximum possible intensity of the combined waves, Imax, at the observer will still be 12 W/m2, with the assumption that the observer would still hear the intensity of 6 W/m2 when either S1 or S2 is sounded alone.

To find the lowest frequency that will produce the maximum intensity (Imax), or fmin,1, we need to consider the condition of constructive interference, which can be represented by the equation: nλ = d with n as an integer, λ as the wavelength and d as the distance. Given the speed of sound (v) as 335 m/s and distance (L) as 4 m, we can calculate the frequency as f = v/λ, implying that fmin,1 = v/(2L) = 335 m/s/(2×4 m) = 41.88 Hz.

For minimum intensity, we consider the condition of destructive interference, which is represented by the equation: (n + 1/2)λ = d. Simplifying, we get fmin,2 = 335 m/s/(2×4 m + λ/2) = 27.9 Hz.

Lastly, fmin,3, the lowest frequency that will produce an intensity I3 = Imax/4 at the observer, is the same as fmin,1, since reducing the intensity by a factor of four only changes the amplitude, not the frequency, of the wave. Hence, fmin,3 = 41.88 Hz.

Learn more about Constructive and Destructive Interference here:

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Related Questions

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[tex]\boxed{6.24g}[/tex]

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Answers

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Answers

Answer:

The mass of silver that will occupy a volume of 87.75 mL is 920.5 grams.

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Astronomers observe a spectral analysis of a distant star where a particular element has a spectral line with a wavelength of 663 nm. In the laboratory, the same element has a spectral line with a wavelength of 645 nm.
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Answers

Answer:

(a). The velocity of star is [tex]8.1\times10^{6}\ m/s[/tex] and the direction of star toward the earth.

(b). The shift is 0.0168 nm.

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[tex]v=\dfrac{3\times10^{8}(663-645)}{663}[/tex]

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[tex]v=8.1\times10^{6}\ m/s[/tex]

The direction of star toward the earth.

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Using formula of shift

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Put the value into the formula

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[tex]\Delta\lambda=0.0168\ nm[/tex]

This shift is small compare to the the movement of Earth around the sun.

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(b). The shift is 0.0168 nm.

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Answers

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Answer:

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Explanation:

Answer:

Acceleration is refers as the change in velocity

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