Distances between wave sources

May 2016
114
1
Ireland
Screenshot (6).png

hello
i am stuck on this question
I am not getting any of the answers here,
does anyone know what to do?
thanks
 

romsek

Math Team
Sep 2015
2,961
1,674
USA
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.
 

romsek

Math Team
Sep 2015
2,961
1,674
USA
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.