Chapter 17: Q. 59 (page 486)

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

The concept of interference in amplitude

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

The explanation of the numerical areas

The interference of the phase is sourced with amplitude where $A = 1.5 A_{0}$ will provide A and the differences in the term are $∆ \emptyset$ . The equation will be localid="1649082665213" $A = 2 A_{0} \cos (\frac{∆ \emptyset}{2})$

The phase difference is $∆ \emptyset = \frac{x}{λ} \cdot 2 π$

After combining the values the outcome will be localid="1649082764169" $A = 2 A_{0} \cos (\frac{ππ}{λ})$

Based on the amplitude $A = 1.5 A_{0}$ the substitution will be, $\frac{A}{2 A_{0}} = \frac{1.5}{2} = \cos (\frac{π x}{λ})$

Thus, $(\frac{π x}{λ}) = \cos^{- 1} (\frac{3}{4})$

The outcome of the distance will be based on $λ = \frac{v}{f}$ which is,

role="math" localid="1649083161043" $x = \frac{λ}{π} \cos^{- 1} (\frac{3}{4}) = \frac{v}{πf} \cos^{- 1} (\frac{3}{4}) x = \frac{343}{1000 π} \cos^{- 1} (\frac{3}{4}) = 7.99 c m$