# Distances between wave sources

#### romsek

Math Team
Hrm... I'm showing that placing the speakers an odd number of wavelengths apart will cause destructive interference at the point 3/4 of the way between the speakers.

$\lambda = \dfrac{v}{\nu} = \dfrac{343}{880} \approx 0.389773~ m$

oh... it doesn't have to be a single wavelength. Look for odd multiples of the wavelength among your list of answers.

• markosheehan and topsquark

#### romsek

Math Team
Hrm... I'm showing that placing the speakers an odd number of wavelengths apart will cause destructive interference at the point 3/4 of the way between the speakers.
Why this is.

If we place the speakers a single wavelength apart. Then at $\dfrac 3 4 \lambda$ we have the phase of the wave from the far speaker being $\dfrac 3 4 \cdot 2 \pi = \dfrac{3\pi}{2}~rad$
Likewise the phase of the near speaker is $\dfrac 1 4 \cdot 2 \pi = \dfrac \pi 2~rad$

The phase difference between these two is $\dfrac{3\pi}{2}-\dfrac{\pi}{2}=\pi$ which is what we want for destructive interference. Any odd multiple of $\pi$ will work as well.

What happens when we use a multiple of the wavelength as the spacing is that a spacing of $n \lambda$ yields of phase difference of $n\pi$ at the distance $\dfrac 3 4$ from one speaker to the other.

Thus we need to space the speakers by odd multiples of the wavelength to get odd multiples of $\pi$ phase difference and thus destructive interference. Even multiples result in full constructive interference.

• markosheehan and topsquark