Calculating the wavelength of soundwaves

  • Thread starter leena19
  • Start date
  • Tags
    Wavelength
In summary, the microphone detects five sound maxima in rotating the speakers about the center of the microphone. The wavelength of the soundwave is 0.4m.
  • #1
leena19
186
0

Homework Statement



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

Homework Equations



v=f[tex]\lambda[/tex] I think

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.
 
Physics news on Phys.org
  • #2
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
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
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
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
leena19 said:
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?

There you go. Constructive interference. As it rotates there are 5 angles of the rotation between 0 and 90 that yield points of constructive interference.

Maybe look at the maximum difference in distance, the 2 meters, with the idea that between that and 0 you have 5 maxima?
 
  • #7
Sorry, I'm still having trouble understanding.

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

How do you get 2m as the maximum difference in distance and what does this have to do with the wavelength?Can you please explain
 
  • #8
leena19 said:
Sorry, I'm still having trouble understanding.

How do you get 2m as the maximum difference in distance and what does this have to do with the wavelength?Can you please explain

When the speakers rotate so that they are all in line with the microphone, 1 speaker is 2 m farther away.
 
  • #9
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
leena19 said:
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...

I think the problem can be interpreted 2 ways. They say in turning it they observe 5 maximums. The question is whether they count the initial one as part of the "in turning the contraption" regime, i.e. that there are 5 in addition to the first.

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
Thank you.
 

1. How do you calculate the wavelength of a soundwave?

The wavelength of a soundwave can be calculated by dividing the speed of sound (usually given in meters per second) by the frequency of the wave (in hertz). The formula is: wavelength = speed of sound / frequency.

2. What is the speed of sound?

The speed of sound is the rate at which sound waves travel through a medium, such as air or water. In dry air at room temperature, the speed of sound is approximately 343 meters per second.

3. Can the wavelength of a soundwave change?

Yes, the wavelength of a soundwave can change depending on the medium it is traveling through. In general, soundwaves travel faster through denser mediums, resulting in a shorter wavelength.

4. What is the relationship between wavelength and frequency?

Wavelength and frequency have an inverse relationship. This means that as the wavelength increases, the frequency decreases, and vice versa. This relationship can be seen in the formula for calculating wavelength: wavelength = speed of sound / frequency.

5. How is the wavelength of a soundwave measured?

The wavelength of a soundwave can be measured by using a ruler or measuring tape to measure the distance between two consecutive points on the wave, such as the peaks or troughs. This distance is equal to one full wavelength.

Similar threads

  • Introductory Physics Homework Help
Replies
8
Views
2K
  • Introductory Physics Homework Help
Replies
7
Views
1K
  • Introductory Physics Homework Help
Replies
1
Views
877
  • Introductory Physics Homework Help
Replies
1
Views
814
Replies
3
Views
2K
  • Introductory Physics Homework Help
Replies
6
Views
1K
  • Introductory Physics Homework Help
Replies
6
Views
2K
  • Introductory Physics Homework Help
Replies
1
Views
2K
  • Introductory Physics Homework Help
Replies
2
Views
934
  • Introductory Physics Homework Help
Replies
1
Views
2K
Back
Top