Sound frequency from distance b/w first minimum and maximum

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The problem involves two speakers separated by 1.5 m, with a listener moving toward point P from point O, 12 m away. The distance between the first minimum and the next maximum is 0.4 m, and the speed of sound is given as 344 m/s. The key equations involve the frequency (ƒ = v/λ) and the phase shifts corresponding to minima and maxima. The approximation of sin(Θ) as tan(Θ) is suggested to simplify calculations, particularly for small angles, but its application in this context remains uncertain. Clarification on the use of this approximation and guidance on determining the first minimum's location are sought.
kennyadams1317
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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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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 \sin \theta \approx \tan \theta, this is a valid approximation for small \theta.

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

\sin \theta = \frac{\mathrm{opposite}}{\mathrm{hypotenuse}}

\cos \theta = \frac{\mathrm{adjacent}}{\mathrm{hypotenuse}}

\tan \theta = \frac{\mathrm{opposite}}{\mathrm{adjacent}}

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 c = \sqrt{a^2 + b^2}, 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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The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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