Question

Two stereo speakers mounted 4.52 m apart on a wall emit identical sound waves. You are standing at the opposite wall of the room at a point directly between the two speakers. You walk 2.11 m parallel to the wall to a location where you first notice that the sound intensity is much less. If the wall along which you are walking is 13.3 m from the wall with the speakers, what is the wavelength of the sound waves? (Answer: 1.4 m) can someone explain the process and formulas?

Answer #1

a little trigo is reqd

distance between speaker-1 and speaker-2 is 4.52

distance between the two wall is 13.3

while the person walking parallel to the speaker wall, he created a
difference distance (pd)

destructive weaker sound is at

pd = (n + 0.5)(lambda) n=0, 1, 2, 3, etc

pd = (0 + 0.5)(lambda)

there are two right triangles

triangle-1

side = 13.3

side = 4.52/2 + 2.11 = 4.37

d1 = sqrt(13.3^2 + 4.37^2) = 14

triangle-2

side = 13.3

side = 4.52/2 - 2.11 = 0.15

d2 = sqrt(13.3^2 + 0.15^2) = 13.3

pd = d1 - d2 = 0.7

(use abs value, it does not matter whether d1 is greater or smaller
than d2)

lambda = 0.7 / 0.5 = 1.4 m

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