# Two loudspeakers, finding amplitude.

1. Nov 27, 2007

### hydrocodone

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

2. Relevant equations
None given, but I assume:
v=f$$\lambda$$
A = 2asin(kx)

3. 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: Nov 27, 2007
2. Nov 27, 2007

### rl.bhat

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

3. Nov 27, 2007

Thanks!