Distance needed to walk in order to hear local maximum

In summary, the person needs to walk 1.00 m away from the first speaker to hear the first local maximum in sound intensity.
  • #1
yoloswag69
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0

Homework Statement


A person stands in an open space listening to the sound from two speakers. The speakers generate sound with a frequency of 489.5 Hz, the speed of sound in air is 343 m/s. The speakers are 2.00 m apart and the person walks away from one of the speakers along a line that is perpendicular to a line drawn between the two speakers as shown in the diagram. The person starts 1.00 m from the first speaker. What distance does the person need to walk along the straight line in order to hear the first local maximum in sound intensity due to the interference of the two waves?

Homework Equations


y = Asin(kx - wt)
v = f*lambda
c^2 = a^2 + b^2 (Maybe??)

The Attempt at a Solution


Since there is an interference between the two waves, i thought you can calculate the addition of the two wave equations. Since both waves have the same equation, it will come out to y = 2Asin(kx-wt), where k = 2pi*343m/s/489.5Hz and k = 2pi*489.5Hz. Since y = 0 at the first wavelength, which is 343m/s/489.5Hz, you can substitute that into find t, but realized that i was going nowhere because i don't know the amplitude... Am i going in the right direction by adding the two equations?

Also, how does wave superposition work if one wave comes in at an angle to the other? Are there different equations for the addition of them? Or do the waves just not superimpose?
Screen Shot 2016-06-05 at 1.05.41 PM.png
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Last edited:
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  • #2
Please, define your variables. There also seems to be a conflict.
yoloswag69 said:
k = 2pi*343m/s/489.5Hz and k = 2pi*489.5Hz. Since y = 0
 
  • #3
Okay so i think i may have been on the wrong track. I think you need to use Path difference to calculate the distance travelled. So that means Absolute Value of (1+d) - (sqrt(4 + (1+d)^2) = n*lambda. But now i need to find another equation in d or n to eliminate either one. Does anyone else think this is the right track or have any other clues?
 
  • #4
Tom.G said:
Please, define your variables. There also seems to be a conflict.
k = 2pi/lambda = 8.96681985
w = 2pi*f = 979pi.
But i don't think this is the right way anymore
 
  • #5
Tom.G said:
Please, define your variables. There also seems to be a conflict.
whats the conflict?
 
  • #6
yoloswag69 said:
whats the conflict?
"k" is defined as being two different values.
yoloswag69 said:
k = 2pi*343m/s/489.5Hz and k = 2pi*489.5Hz.
yoloswag69 said:
I think you need to use Path difference to calculate the distance travelled.
Yes, that is the most straight forward way.
 
  • #7
Tom.G said:
"k" is defined as being two different values.
.
Sorry the second k is meant to be w. If you use path difference, how do you know what the value of n is?
 
  • #8
Ignoring n for the moment, what is the relative phase needed between two waves for their sum to be maximum?
 
  • #9
Tom.G said:
Ignoring n for the moment, what is the relative phase needed between two waves for their sum to be maximum?
there should be no relative phase, if their sum needs to be maximum, since through constructive interference the maximum sum is when they are in phase, i think
 
  • #10
Right again.
So what should the ratio of their distances be for them to arrive in phase?
 
  • #11
Tom.G said:
Right again.
So what should the ratio of their distances be for them to arrive in phase?
d1/d2 = lambda? I am not sure
 
  • #12
Distances being different by one wavelength would certainly do it. What about two wavelengths, or 10 wavelengths? Would that work? What about 10.5 wavelengths?
 
  • #13
Tom.G said:
Distances being different by one wavelength would certainly do it. What about two wavelengths, or 10 wavelengths? Would that work?
i think it would, but they would be later local maximums, and the question is asking for first one
 
  • #14
yoloswag69 said:
i think it would, but they would be later local maximums, and the question is asking for first one
10.5 wavelengths wouldn't work though, since it wouldn't indicate a meeting between two crests, so it wouldn't be a maximum?
 
  • #15
yoloswag69 said:
i think it would, but they would be later local maximums, and the question is asking for first one
What you need to do is to first work at the initial position and set up the equation ## k (d_2 - d_1) = n \lambda## (here you know all the parameters except "n") and solve for that "n". I called it it n but it may not be an integer here (if the initial position is not a maximum, it won't be integer). For the sake of the argument, let's say it comes out to be 2.34 (I just made that up to explain my point). Then you find the position along the line indicated where you have ##k(d_2 - d_1) = N \lambda ## where N is now the nearest integer above the "n" (in my example, N would then be equal to 3). Now ##d_2## and ##d_1## are unknowns (but are related, of course) and you can find the answer.
 
  • #16
nrqed said:
What you need to do is to first work at the initial position and set up the equation ## k (d_2 - d_1) = n \lambda## (here you know all the parameters except "n") and solve for that "n". I called it it n but it may not be an integer here (if the initial position is not a maximum, it won't be integer). For the sake of the argument, let's say it comes out to be 2.34 (I just made that up to explain my point). Then you find the position along the line indicated where you have ##k(d_2 - d_1) = N \lambda ## where N is now the nearest integer above the "n" (in my example, N would then be equal to 3). Now ##d_2## and ##d_1## are unknowns (but are related, of course) and you can find the answer.
Okay i understand now, so you have to find the first "n" after the intial position which points to a maximum. Also just clarifying, in the equation ## k (d_2 - d_1) = n \lambda## k = 2*pi\lambda right? and d2 and d1 are sqrt(5) and 1 respectively? When you substitute these values into the first equation, you get n, then round up to the nearest integer above n and substitute it back but this time solve for d1 and d2?
 
  • #17
yoloswag69 said:
Okay i understand now, so you have to find the first "n" after the intial position which points to a maximum. Also just clarifying, in the equation ## k (d_2 - d_1) = n \lambda## k = 2*pi\lambda right? and d2 and d1 are sqrt(5) and 1 respectively?
Yes, everything you wrote is correct. EDIT: as long as you meant ##k = 2 \pi/ \lambda##
 
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  • #18
Replies in reverse order.

Right, 10.5 would not work.
True. I was trying to point out the general rule. I think you have already found the wavelength to be less than the speaker spacing. That means the difference in distances from observer to speakers will be greater than one.

I see nrqed joined.
yoloswag69 said:
k(d2−d1)=nλk(d2−d1)=nλ k (d_2 - d_1) = n \lambda k = 2*pi\lambda
Is that 2*pi factor really appropriate?
 
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  • #19
nrqed said:
Yes, everything you wrote is correct. EDIT: as long as you meant ##k = 2 \pi/ \lambda##
If i substitute the rounded value of N back into the equation, i would get d2 - d1 = Nlambda/k. Are d2 and d1 given by sqrt(4 + (1+d)^2) and (1 + d)? If so the equation gives a value of 0.9999 for d.
Tom.G said:
Replies in reverse order.

Right, 10.5 would not work.
True. I was trying to point out the general rule. I think you have already found the wavelength to be less than the speaker spacing. That means the difference in distances from observer to speakers will be greater than one.

I see nrqed joined.

Is that 2*pi factor really appropriate?
I think so, isn't the value of k given by 2pi/lambda?
 
  • #20
yoloswag69 said:
k given by 2pi/lambda

d2 - d1 = Nlambda/k.

The problem I see is that d1, d2, λ already have units of linear distance. What units do you get by multiplying linear distance by 2*pi?
 
  • #21
Tom.G said:
The problem I see is that d1, d2, λ already have units of linear distance. What units do you get by multiplying linear distance by 2*pi?
ohhh i see, you get rad*m i think
 
  • #22
Tom.G said:
The problem I see is that d1, d2, λ already have units of linear distance. What units do you get by multiplying linear distance by 2*pi?
the units don't change, ##2 \pi ## is a pure number, it is dimensionless.
 
  • #23
@yoloswag69
I suggest you check your answer.
In Post #19:
yoloswag69 said:
If so the equation gives a value of 0.9999 for d.
That would put the person 1+d or about 2 meters from one speaker, and how far from the other speaker?
Are the fractional parts of the distances, measured in Wavelengths, the same, thus allowing the waves to reinforce each other?

Lets get terms settled:
Are 'λ' and 'Lambda' the same thing and equal to the Wavelength? What is the Wavelength?

In the OP you have 'k = 2pi*343m/s/489.5Hz', which is equivalent to 2*pi*Wavelength.
In Post #16 you have 'k = 2*pi\lambda'; which I assume is a typo of '\' vs '/'.
In either case, the 2*pi is unnecessary and only gets in the way. (the same is true for 'k', for that matter. but you do need 'λ')
 
Last edited:
  • #24
yoloswag69 said:
Okay i understand now, so you have to find the first "n" after the intial position which points to a maximum. Also just clarifying, in the equation ## k (d_2 - d_1) = n \lambda## k = 2*pi\lambda right? and d2 and d1 are sqrt(5) and 1 respectively? When you substitute these values into the first equation, you get n, then round up to the nearest integer above n and substitute it back but this time solve for d1 and d2?
I made a mistake, I apologize! My bad. I should have written ##(d_2 - d_1) = n \lambda## without any k there (I was actually thinking of ## k (d_2 - d_1) = 2 \pi n##). My apologies!
 
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  • #25
Tom.G said:
@yoloswag69
I suggest you check your answer.
In Post #19:

That would put the person 1+d or about 2 meters from one speaker, and how far from the other speaker?
Are the fractional parts of the distances, measured in Wavelengths, the same, thus allowing the waves to reinforce each other?

Lets get terms settled:
Are 'λ' and 'Lambda' the same thing and equal to the Wavelength? What is the Wavelength?

In the OP you have 'k = 2pi*343m/s/489.5Hz', which is equivalent to 2*pi*Wavelength.
In Post #16 you have 'k = 2*pi\lambda'; which I assume is a typo of '\' vs '/'.
In either case, the 2*pi is unnecessary and only gets in the way. (the same is true for 'k', for that matter. but you do need 'λ')
nqred said the k is to be left out of the formula now
 
  • #26
nrqed said:
I made a mistake, I apologize! My bad. I should have written ##(d_2 - d_1) = n \lambda## without any k there (I was actually thinking of ## k (d_2 - d_1) = 2 \pi n##). My apologies!
oh right that makes more sense. But when i work it out without the k now i get the distance = -0.27, when the answer is 1.50. I've looked over it and can't see what's wrong. So this is the method i used. (d2 - d1) = n*lambda. d2,d1 and lambda are known so solve for n. Then round n up to the nearest integer. Substitute this value as N back into the equation so it now becomes (d2 - d1) = N*lambda. d2 becomes sqrt(4 + x^2), d1 becomes x, where x = 1 + distance moved. When i solve this equation i get the value for x to be 0.726397, that means the distance moved would be -0.2736, which is obviously incorrect. Not sure what's wrong.
 
  • #27
yoloswag69 said:
Then round n up to the nearest integer.
As you walk away from the speakers, does d2-d1 increase or decrease?
 
  • #28
Since Post #3 isn't working:
yoloswag69 said:
(1+d) - (sqrt(4 + (1+d)^2) = n*lambda
I think we need to go back to post #23:
Tom.G said:
Are the fractional parts of the distances, measured in Wavelengths, the same, thus allowing the waves to reinforce each other?
Rephrasing Post #23 yields (D2 mod λ) = (D1 mod λ) [where λ = Wavelength, D1 = 1+x, x = distance to walk]
Then write D2 in terms of D1 to give a single equation to solve.
The rational is that integer wavelengths are not required in both paths, only that the phases match at some intersection.
 

1. How is the distance needed to walk in order to hear local maximum calculated?

The distance needed to walk in order to hear local maximum is calculated by measuring the distance between the source of the sound and the listener, and taking into account any obstacles that may affect the sound transmission.

2. What factors affect the distance needed to walk in order to hear local maximum?

The distance needed to walk in order to hear local maximum can be affected by various factors such as atmospheric conditions, background noise, sound frequency, and the direction and angle of the sound source.

3. How does the terrain affect the distance needed to walk in order to hear local maximum?

The terrain can greatly affect the distance needed to walk in order to hear local maximum. For example, sound may travel farther and clearer over flat terrain compared to hilly or mountainous terrain.

4. Is there a specific formula to determine the distance needed to walk in order to hear local maximum?

There is no specific formula to determine the distance needed to walk in order to hear local maximum as it can vary depending on the factors mentioned above. However, there are general guidelines and equations that can be used to estimate the distance.

5. How can the distance needed to walk in order to hear local maximum be minimized?

To minimize the distance needed to walk in order to hear local maximum, it is important to consider the factors mentioned above and try to eliminate or reduce any obstacles or sources of interference. Additionally, using amplification or directional speakers can also help to reduce the distance needed to walk.

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