Physics: Suppose that the separation between two speakers A and B is 6.60 m and the speakers are vibrating in-phase. They are playing identical 126-Hz tones and the speed of sound is 343 m/s. An observer is seated at a position directly facing speaker B in such a way that his line of sight extending to B is perpendicular to the imaginary line between A and B. What is the largest possible distance between speaker B and the observer, such that he observes destructive interference

This answer is the same for all elements, a single atom of any element is still that element....

Suppose that the separation between two speakers

Mw3spartan17

11.02.2022

Complete question

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The largest possible distance is $e = 15.33 \ m$

Explanation:

From the question we are told that

The separation between Speaker at position A and B is AB = 6.60 m

The frequency of the tune which the speaker are playing is $f = 126 \ Hz$

The speed of sound is $v = 343 \ m/s$

Generally the wavelength of the tune playing is mathematically represented as

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

=> $\lambda = \frac{343}{ 126}$

=> $\lambda = 2.72 \ m$

Let the observer be at position D

Generally the distance A and is mathematically evaluated using Pythagoras theorem as

$AC = \sqrt{AB ^2 + BC^2}$

Let BC = e

$AC = \sqrt{6.60 ^2 + e^2}$

Generally the path difference between the first and the second speaker from the observer point of view is mathematically represented as

P = AC - BC

=> $P = \sqrt{6.60 ^2 + e^2} - e$

Generally the condition for destructive interference is mathematically represented as

$P = (2n - 1 )\frac{\lambda}{2}$

Here n is the order of the fringe which is one

=> $\sqrt{6.60 ^2 + e^2} - e = (2 * 1 - 1 )\frac{2.72}{2}$

=> $\sqrt{6.60 ^2 + e^2} - e = 1.36$

=> 6.60 ^2 + e^2 =( 1.36 +e)^2

=> 6.60 ^2 + e^2 =1.8496 + 2.72e +e^2

=> $e = 15.33 \ m$

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