Sound frequency from distance b/w first minimum and maximum

In summary, the problem involves two speakers separated by 1.5 m and a listener moving towards a point where they encounter a series of alternating minima and maxima. The distance between the first minimum and the next maximum is 0.4 m. Using the speed of sound in air and the approximation of sin(Θ) = tan(Θ), the frequency of the speakers can be determined. The trigonometric approximation can be used to simplify calculations by substituting the hypotenuse with the adjacent, which in this problem is a known value of 12 m.
  • #1
kennyadams1317
2
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Homework Statement


Two speakers are arranged as shown. For this problem, assume that point O is 12 m along the center line and the speakers are separated by a distance of 1.5 m. As the listener moves toward point P from point O, a series of alternating minima and maxima is encountered. The distance between the first minimum and the next maximum is 0.4 m. Using 344 m/s as the speed of sound in air, determine the frequency of the speakers. (Use the approximation of sin(Θ)=tan(Θ).

Speakers_zpsksnibpl8.jpg


Homework Equations


ƒ = v/λ
r1 = distance from top speaker to observer
r2 = distance from bottom speaker to observer

The Attempt at a Solution


Assumptions:
Δr = λ/2 for first minimum (corresponding to phase shift of π)
Δr = λ for first maximum after first minimum (corresponding to phase shift of 2π)

I guess the trig/algebra is tripping me up here. The question specifically calls out that sin can be approximated as tan. I'm not sure how to use this, what I assume to be, useful information. I think I need to find the location of the first minimum before I can solve, but I'm not sure. I've tried to determine this using pythagorean ratios, but I keep coming up with lousy answers. Would anyone be willing to point me in the right direction on how to get through the first step of this? If the problem can be solved without finding the location of the first minimum, let me know that as well please.
 
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  • #2
kennyadams1317 said:

Homework Statement


Two speakers are arranged as shown. For this problem, assume that point O is 12 m along the center line and the speakers are separated by a distance of 1.5 m. As the listener moves toward point P from point O, a series of alternating minima and maxima is encountered. The distance between the first minimum and the next maximum is 0.4 m. Using 344 m/s as the speed of sound in air, determine the frequency of the speakers. (Use the approximation of sin(Θ)=tan(Θ).

Speakers_zpsksnibpl8.jpg


Homework Equations


ƒ = v/λ
r1 = distance from top speaker to observer
r2 = distance from bottom speaker to observer

The Attempt at a Solution


Assumptions:
Δr = λ/2 for first minimum (corresponding to phase shift of π)
Δr = λ for first maximum after first minimum (corresponding to phase shift of 2π)

I guess the trig/algebra is tripping me up here. The question specifically calls out that sin can be approximated as tan. I'm not sure how to use this, what I assume to be, useful information. I think I need to find the location of the first minimum before I can solve, but I'm not sure. I've tried to determine this using pythagorean ratios, but I keep coming up with lousy answers. Would anyone be willing to point me in the right direction on how to get through the first step of this? If the problem can be solved without finding the location of the first minimum, let me know that as well please.

In terms of the [itex] \sin \theta \approx \tan \theta [/itex], this is a valid approximation for small [itex] \theta [/itex].

To realize how it might help you in this problem, first review the following:

[itex] \sin \theta = \frac{\mathrm{opposite}}{\mathrm{hypotenuse}} [/itex]

[itex] \cos \theta = \frac{\mathrm{adjacent}}{\mathrm{hypotenuse}} [/itex]

[itex] \tan \theta = \frac{\mathrm{opposite}}{\mathrm{adjacent}} [/itex]

The approximation helps [might help] you because if you want to calculate the sine of an angle, you might have to calculate the hypotenuse all the time using the ugly [itex] c = \sqrt{a^2 + b^2} [/itex], but if instead you use the tangent approximation, you can substitute the hypotenuse with the adjacent. You already know what the adjacent is: it's 12 m for all your calculations. :wink: (one part of the figure indicates it's 8 m, but I think that's a mistake in the figure.)

[Edit: although thinking about it a little more, I'm not completely sure if that approximation would help with this particular problem without making one or two other additional approximations. Hmm.]
 
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1. What is the relationship between sound frequency and distance between the first minimum and maximum?

The sound frequency and the distance between the first minimum and maximum are inversely proportional. This means that as the distance between the first minimum and maximum increases, the sound frequency decreases and vice versa.

2. How is sound frequency affected by the distance between the first minimum and maximum?

The sound frequency is affected by the distance between the first minimum and maximum because the distance determines the wavelength of the sound wave. As the distance increases, the wavelength increases and therefore, the sound frequency decreases.

3. Can the sound frequency be calculated from the distance between the first minimum and maximum?

Yes, the sound frequency can be calculated from the distance between the first minimum and maximum using the formula: frequency = speed of sound / wavelength. The speed of sound is a constant value and the wavelength can be determined from the distance between the first minimum and maximum.

4. Does the medium through which sound travels affect the distance between the first minimum and maximum?

Yes, the medium through which sound travels can affect the distance between the first minimum and maximum. The speed of sound is different in different mediums, so the distance between the first minimum and maximum may vary depending on the medium.

5. How does the distance between the first minimum and maximum relate to the quality of sound?

The distance between the first minimum and maximum does not directly affect the quality of sound. However, it can indirectly impact the quality by affecting the sound frequency. The lower the sound frequency, the deeper the sound and vice versa. So, a larger distance between the first minimum and maximum may result in a deeper or lower quality sound.

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