(adsbygoogle = window.adsbygoogle || []).push({}); [Solved] Two Speakers - Sound Maximum and Sound Minimum Problem

Hi, I am having difficulty solving the following problem:

1. The problem statement, all variables and given/known data

Two loudspeakers5.0 mapart are playing the same frequency. If you stand13.0 min front of the plane of the speakers, centered between them, you hear a sound of maximum intensity. As you walk parallel to the plane of the speakers, staying 13.0 m in front of them, you first hear a minimum of sound intensity when you are directly in front of one of the speakers.

What is the frequency of the sound? Assume a sound speed of340 m/s.

2. Relevant equations

Sound Maximum:

L1 - L2 = n[tex]\lambda[/tex]

Sound Minimum:

L1' - L2 = (n+[tex]\frac{1}{2}[/tex])[tex]\lambda[/tex]

Frequency:

f = [tex]\frac{v}{\lambda}[/tex]

3. The attempt at a solution

Sound Maximum:

L1 - L2 = n[tex]\lambda[/tex]

L2 = 13.0 m

L1 = [tex]\sqrt{13.0^{2}+2.50^{2}}[/tex] = 13.23820229

L[tex]_{1}[/tex] - L[tex]_{2}[/tex] = n[tex]\lambda[/tex]

13.23820229 - 13 = n[tex]\lambda[/tex]

n[tex]\lambda[/tex] = 0.23820229

Sound Minimum

L1' - L2 =(n + [tex]\frac{1}{2}[/tex])[tex]\lambda[/tex]

L2 = 13.0 m

L1' = [tex]\sqrt{13.0^{2}+5.0^{2}}[/tex] = 13.92838828

Sub in n[tex]\lambda[/tex]= 0.23820229:

L1' - L2 = (n + [tex]\frac{1}{2}[/tex])[tex]\lambda[/tex]

13.92838828 - 13 = n[tex]\lambda[/tex] + [tex]\lambda[/tex]/2

[tex]\lambda[/tex]/2 = 0.92838828 - 0.23820229

[tex]\lambda[/tex] = 1.380371974

Sub in [tex]\lambda[/tex] = 1.380371974:

f = [tex]\frac{v}{\lambda}[/tex]

f = [tex]\frac{340}{1.380371974}[/tex]

f = 246.3104195 Hz

I'm not sure if my approach is wrong or if I'm interpreting the question incorrectly. Any help would be greatly appreciated!

Thanks.

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# Two Speakers - Sound Maximum and Sound Minimum Problem

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