
#1
Jan3010, 05:40 PM

P: 2

Hi, I am having difficulty solving the following problem:
1. The problem statement, all variables and given/known data Two loudspeakers 5.0 m apart are playing the same frequency. If you stand 13.0 m in 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 of 340 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. 



#2
Jan3110, 07:51 AM

HW Helper
P: 4,442

In the central position the two speakers are at equal distance. So the path difference is zero. In between the first and the second position, there is neither a maximum nor a minimum. So at the second position ( l1'  l2) = λ/2.




#3
Jan3110, 08:57 AM

P: 2

Ooh.. no wonder. Thank you very much!



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