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# Wave interference question Watch

1. Hello there

I am having difficulties with this. The answer is 1/4landa but I am not too sure why.

Help would be really appreciated
2. In this question, you're looking at the phase of the sound. Initially, you're told that the sound arrives back at a maximum, which means that the sound is in phase. However, when it's shifted the sound comes back at a minimum, which means that it is in antiphase (Or, if you'd like, as far out of phase as possible). There are two things you should consider:

- How much does the distance the sound travels need to be reduced by in order to make the sound out of phase?
- The sound is travelling across a certain distance, but it's travelling across that distance both there and back. Bearing this in mind, what will changing this distance by x do to the overall distance travelled?
3. The distance between a maximum (antinode) and a minimum (node) is where the path difference between the interfering waves is half a wavelength. For this set up the reflected wave has to make a two way journey mic to reflector back to mic so 2x=1/2 lambda.
4. (Original post by Joinedup)
The distance between a maximum (antinode) and a minimum (node) is where the path difference between the interfering waves is half a wavelength. For this set up the reflected wave has to make a two way journey mic to reflector back to mic so 2x=1/2 lambda.
sorry but why is it 2x=1/2 lambda

5. Look at the above diagram.

Although sound is NOT a transverse wave we shall imagine we are dealing with some form of electromagnetic radiation instead of sound. The concept is still the same.

Note that one full wavelength is marked as lambda. Half a wavelength is a minima. Between half a wavelength and 0 we have a maxima.

Distance between ADJACENT maxima and minima is therefore lambda/4.

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6. (Original post by upthegunners)
sorry but why is it 2x=1/2 lambda
the standing wave is caused by a wave travelling left to right interfering with it's reflection travelling right to left.
change in path length of the reflected wave to a point (the mic) which caused a change from a max to the next min equals lambda/2, half a wavelength, the difference between constructive and destructive interference.

Change in path length of the reflected wave by moving the reflector a distance of X is 2X (two way journey of the reflected wave).

We can express this as an equation relating X to lambda.
2x=lambda/2

we can now divide both sides of the eqn 2x=lambda/2 by 2...

X=lambda/4

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