Complete question
The diagram for this question is shown on the first uploaded image
Answer:
The largest possible distance is [tex]e = 15.33 \ m[/tex]
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 [tex]f = 126 \ Hz[/tex]
The speed of sound is [tex]v = 343 \ m/s[/tex]
Generally the wavelength of the tune playing is mathematically represented as
[tex]\lambda = \frac{v}{f}[/tex]
=> [tex]\lambda = \frac{343}{ 126}[/tex]
=> [tex]\lambda = 2.72 \ m[/tex]
Let the observer be at position D
Generally the distance A and is mathematically evaluated using Pythagoras theorem as
[tex]AC = \sqrt{AB ^2 + BC^2}[/tex]
Let BC = e
So
[tex]AC = \sqrt{6.60 ^2 + e^2}[/tex]
Generally the path difference between the first and the second speaker from the observer point of view is mathematically represented as
[tex]P = AC - BC[/tex]
=> [tex]P = \sqrt{6.60 ^2 + e^2} - e[/tex]
Generally the condition for destructive interference is mathematically represented as
[tex]P = (2n - 1 )\frac{\lambda}{2}[/tex]
Here n is the order of the fringe which is one
=> [tex]\sqrt{6.60 ^2 + e^2} - e = (2 * 1 - 1 )\frac{2.72}{2}[/tex]
=> [tex]\sqrt{6.60 ^2 + e^2} - e = 1.36[/tex]
=> [tex]6.60 ^2 + e^2 =( 1.36 +e)^2[/tex]
=> [tex]6.60 ^2 + e^2 =1.8496 + 2.72e +e^2[/tex]
=> [tex]e = 15.33 \ m[/tex]
