The drawing shows a loudspeaker A and point C, where a listeer is positioned. A second loudspeaker B is located somewhere to the right of A. Both speakers vibrate in phase and are playing a 68.6-Hz tone. The speed of sound is 343 m/s. What is the closest to speaker A that speaker B can be located, so that the listener hears no sound?(adsbygoogle = window.adsbygoogle || []).push({});

Picture:

First, I connected C and B to make a triangle. I then bisect the triangle with a line straight down from C to make 2 triangles (with the extra point being D). I call AB line = x and CD line = y. Next, I set up the destructive interference where the point in which line AB has to be 1 wavelength and line CB is half wavelength. So, the difference between AC and CB has to be

CB - CA = (n*wavelength)/2

for smallest distance, n would have to be 1.

with this, I list all the known and unknowns which are

AD = AC*cos(60)

DB = AB-AD -> x-cos(60)

CD = AC*sin(60)

CB = sqrt(sin^2(60)+ (x-cos^2(60)))

So then:

wavelength/2 = sqrt(sin^2(60)+cos^2(60) + x^2 - 2*cos(60x))

I solve to get:

x^2 - (2*cos(60)x) + (1 + wavelength^2/4)

use quadratic equation to get

x = 2.845 and x = -1.845

The correct answer is 3.89 m.

So, I don't know if I'm missing something here. I hope you guys can help. Thanks.

**Physics Forums | Science Articles, Homework Help, Discussion**

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Homework Help: Sound problem: Destructive Interference

**Physics Forums | Science Articles, Homework Help, Discussion**