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Two in-phase loudspeakers emit identical 1000 Hz sound waves along the x-axis. What distance should one speaker be placed behind the other for the sound to have an amplitude 1.5 times that of each speaker alone?

Short Answer

Expert verified

The distance would be 7.99 cm.

Step by step solution

01

The concept of interference in amplitude 

The interference of amplitude represents the coincidental paths of the moving waves.

02

The explanation of the numerical areas 

The interference of the phase is sourced with amplitude where A=1.5A0will provide A and the differences in the term are ∆∅. The equation will be localid="1649082665213" A=2A0cos∆∅2

The phase difference is ∆∅=xλ·2π

After combining the values the outcome will be localid="1649082764169" A=2A0cosππλ

Based on the amplitude A=1.5A0the substitution will be,A2A0=1.52=cosπxλ

Thus, πxλ=cos-1(34)

The outcome of the distance will be based on λ=vfwhich is,

role="math" localid="1649083161043" x=λπcos-1(34)=vÏ€´Úcos-1(34)x=3431000Ï€cos-1(34)=7.99cm

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