Two loudspeakers, finding amplitude.

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SUMMARY

To achieve an amplitude of 1.60 times that of each loudspeaker alone, the distance between two in-phase loudspeakers emitting 1000 Hz sound waves must be calculated using vector addition of amplitudes. The correct formula is (1.6a)^2 = a^2 + a^2 + 2*a*a*cos(phi), where phi represents the phase difference determined by the distance x and the wavelength. The wavelength is calculated using the speed of sound (343 m/s) and the frequency (1000 Hz), resulting in a wavelength of 0.343 m. The final distance x is determined to be approximately 0.0506 m.

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Homework Statement


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 1.60 times that of each speaker alone?

Homework Equations


None given, but I assume:
v=f[tex]\lambda[/tex]
A = 2asin(kx)

The Attempt at a Solution



1.6a = 2asin(kx)

Found k=(2pi)/wavelength = (2pi/0.343)
using v = 343 m/s

arcsin(0.8)=18.318x

x = 0.0506m

But this answer is wrong. Help?
 
Last edited by a moderator:
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You have to add the amplitudes as vectors.
(1.6*a)^2 = a^2 + a^2 + 2*a*a*cos(phi) where phi is the phase difference = 2*pi*x/wavelength
 
Thanks!
 

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