Suppose we had two speakers, one placed some distance directly behind the other

Question

Suppose we had two speakers, one placed some distance directly behind the other and

both pointing along the same direction, facing us. Now, we connect the speakers to

the same source and turn them on producing a single continuous tone of 440 Hz. If the

speed of sound in air is about 343 m=s, then how far would we have to put the back

speaker behind the front one so that we never hear the tone when standing anywhere

directly in front of the speakers? (Note: you can assume that the speakers are small

enough that the front one does not aect the sound from the back one.) m

Explanation / Answer

first of all you have to find the wavelength of the sound produced.
since w=v/f where w = wavelength and f = frequency then you will have and v is velocity

w=343/440
= .779545 meters

now a wavelength is from 0 to peak (compression) to 0 to trough (rarefaction) and back to 0 as such:

/ ..one wavelength.. / / / / / / ..many wavelengths
../............................/ / / / / /

zero interference occurs when the compressions of the soundwaves coming from the first speaker co-incide with the rarefactions from the other speaker

so therefore the first speaker must emit half a wavelength before the other speakers wavelength co-incides with it.

i.e
.............S1.../ / / / /
................../ / / /

.........S2.... / / / / / /
................/ / / / /

if you line them up you will see the waves have cancelled each other out

For the purpose of the diagram you will have to imagine the two speakers in line with each other

The speed of sound in the air has no relevance on this question as it would not matter how fast the waves travelled but only that they travel at the same speed as each other.

The distance of half a wavelength in this case is

.779545/2

= 0.3897725 metres

so therefore the distance at which the second speaker must be placed behind the first speaker is

0.3897725 metres

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