Question

The problem says the speakers are in phase. However, I only get the correct answer using the out of phase equation. How is that possible??

Larry, Moe and Curly stand in a line with a spacing of 1.00 m. Larry is 3.00 m in front of a pair of stereo speakers 0.800 m apart. The speakers produce a single-frequency tone, vibrating in phase with each other. What are the two lowest frequencies that allow Larry to hear a loud tone while Moe and Curly hear very little?

How do I know what eq to use?

In phase Equation 1: D1-D2=N(V/F)

Out of Phase equation 2: D1-D2=N(V/F)

The problem says the speakers are in phase. However, I only get the correct answer using the out of phase equation. How is that possible??

Answer 1

If Larry is standing in the middle then Moe and Curly are each standing 1m away from him. You have to find the distance between each speaker and either Moe or Curly. Draw a line from each speaker to Moe and call them X1 and X2 (the distances will be the same for either Moe or Curly). To find X1 (from the closer speaker):

X1=sqrt(L^2+y^2)
y = 1-.4= 0.6m because Larry is standing in the middle of the 2 speakers you can find the vertical distance between Moe/Curly and the speakers

X2 is from the speaker that's farther away:
X2=sqrt(L^2+z^2)
z=1+.4= 1.4m

since Curly and Moe can't hear the sound, it's destructive interference and the difference in the path length will have to equal 1/2(lambda). If it were constructive the path distance would equal lambda

X2 - X1 = 1/2(lambda)
2(X2 - X1) = lambda

frequency= v/lambda the v of sound in air is 343 m/s. This answer will be in Hz, so divide by 1000 to get kHz.

1/2(lambda) is the first lowest frequency, the next is 3/2(lambda).
X2 - X1 = 3/2(lambda) solve for lambda and plug it into the frequency equation again, divide by 1000 and you should have your answer