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Calculating the wavelength of soundwaves 
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#1
Jun1009, 01:05 PM

P: 186

1. The problem statement, all variables and given/known data
Two speakers connected to the same source of fixed frequency are placed 2m apart in a box.A sensitive microphone placed at a distance of 4m from their midpoint along the perpendicular bisector shows maximum response. The box is slowly rotated until the speakers are in line with the microphone.The distance between the midpoint of the speakers and the microphone remains unchanged.Exactly 5 maximum responses are observed in the microphone in doing this. The wavelength of the soundwave is, 1. 0.4m 2. 0.8m 3. 1m 4. 0.2m 5. 0.6m ans: 0.4m 2. Relevant equations v=f[tex]\lambda[/tex] I think 3. The attempt at a solution First of all, I don't quite understand the question, especially the part where they mention the change in response.The frequency is a constant,so I think [tex]\lambda[/tex] is also a constant but the number of loops maybe different. Then again do the 5 maximum responses mentioned here refer to beats heard? I'm really confused.Any help would be much appraciated. 


#2
Jun1009, 01:49 PM

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P: 5,341

As much as anything this looks like a matter of geometry.
Think of the speakers as on a 2 meter rod pivoted about the center. You initially know that there is a maximum at the initial configuration with the line of the speakers perpendicular to the displacement from the mic. Now you rotate the speakers on the rod and as you do you determine that it fades and then reaches a maximum a total of 5 times. These are not beats, because the frequency is constant. It rotates slowly so there is no doppler to worry about. It's sound maximums that you are counting. 


#3
Jun1109, 03:46 AM

P: 186

I'm sorry I still don't get it.
Do the maximum responses mentioned here refer to the antinodes? Taking the points at the speakers and the microphone as nodes I get, 5[tex]\lambda[/tex]/2=5m and so [tex]\lambda[/tex] as 2m at the final position of the speakers and a different wavelength of 4m at the initial position,which is obviously wrong. Can you help me work this out? please. 


#4
Jun1109, 08:23 AM

HW Helper
P: 5,341

Calculating the wavelength of soundwaves
The source is in phase. in the initial position then along the perpendicular bisector it takes the same time to reach the microphone  they add exactly in phase.
You rotate the contraption a bit and now one is further away, the other closer  the time difference over the different distances means they will be a little out of phase. As it rotates further, you get less and so on, until they begin to come back to arriving in phase again, in which case you detect another max. 


#5
Jun1209, 11:24 AM

P: 186

Ok,so this has to do with interference.But... what do I know about interference?
Not much,I'm afraid. All I know, is that interference is of two types; constructive interference, which occurs when two waves in phase move in the same direction , then the amplitude of the resultant wave is doubled and destructive interference ,when two waves are exactly out of phase so...the 5 maximums detected are the points where the waves are inphase,but, what is the relationship between this and the wavelength? 


#6
Jun1209, 12:01 PM

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P: 5,341

Maybe look at the maximum difference in distance, the 2 meters, with the idea that between that and 0 you have 5 maxima? 


#7
Jun1309, 07:54 AM

P: 186

Sorry, I'm still having trouble understanding.



#8
Jun1309, 09:02 AM

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#9
Jun1309, 10:52 AM

P: 186

Ok.I did a bit of reading today and from what I can understand,constructive interference occurs when the path difference=n[tex]\lambda[/tex],where n=0,1,2...
Since there are 5 maximum waves detected at the microphone,I'm thinking that the path difference=4[tex]\lambda[/tex].hope I'm right? At 90 degrees,as you mentioned there'a path difference of 2m, therefore 2m= 4[tex]\lambda[/tex].I get [tex]\lambda[/tex] as 0.5m,which is wrong so there cannot be a maximum at this position,ryt? so where do I go from here?Looking at the answer (0.4m)I know the path diff has to be 1.6m,but how do I get it? Once again I'm hoping you could help me.Please... 


#10
Jun1309, 11:23 AM

HW Helper
P: 5,341

One way suggests that there are 4λ, as you surmised. The fact that they don't offer .5m as a possible answer I think is constructive to the idea that there are 5λ and not 4λ along the way. 


#11
Jun1309, 12:26 PM

P: 186

Thank you.



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