Solve Sound Wave Problem: Frequency of Sound in Air (336 m/s)

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Homework Help Overview

The problem involves a sound source and a detector with a cardboard barrier affecting the intensity of sound waves. The objective is to determine the frequency of the sound emitted based on the observed changes in intensity as the barrier is moved. The velocity of sound in air is given as 336 m/s.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to relate the path difference of sound waves to the observed intensity changes, using equations involving wavelength and integer values for maxima and minima. Some participants clarify the relationship between consecutive maxima and minima, suggesting a specific interpretation of the path difference.

Discussion Status

Participants are exploring different interpretations of the path difference and its relation to the wavelength. Some guidance has been provided regarding the relationship between maxima and minima, but no consensus has been reached on the values of n and m used in the calculations.

Contextual Notes

The original poster references a previous thread for context and mentions the inability to attach a figure, which may limit the clarity of the problem setup. There is also a note of uncertainty regarding the selection of integer values for n and m in the calculations.

Saitama
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Homework Statement


A source of sound S and a detector D are placed at some distance from one another. A big cradboard is placed near the detector and perpendicular to line SD as shown in figure. It is gradually moved away and it is found that the intensity changes from a maximum to a minimum as the board is moved through a distance of 20 cm. Find the frequency of the sound emitted. Velocity of sound in air is 336 m/s.

Homework Equations


The Attempt at a Solution


I have posted a similar problem in the past.
https://www.physicsforums.com/showthread.php?t=684072&highlight=sound+detector+wall

(The figure is present in the linked thread. I can't seem to attach it to the current thread :confused: )

I tried the current problem the following way:

Let the cardboard be initially at a distance ##x## from S and let D be at a distance ##y## from S. The path difference between the two waves reaching D is ##2(x-y)##. For a maximum, the path difference must be ##n\lambda## i.e ##2(x-y)=n\lambda\,\, (*)## where n is a non-negative integer. When the cardboard is moved by 20 cm, the path difference is ##2(x-y)-40##. For a minimum, this must be a equal to ##(m+1/2)\lambda## where m is a non-negative integer. Using (*), I get:
$$n\lambda-40=\left(m+\frac{1}{2}\right)\lambda$$
Solving for ##\lambda##
$$\lambda=\frac{40}{n-m-1/2}$$
If I assume ##n=1## and ##m=0##, I get the given answer i.e 420 Hz but if I assume some other values, I get different answers. :confused:

Any help is appreciated. Thanks!
 
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The key thing is that the path difference between a consecutive maxima and minima is λ/2 .

Now when the detector is moved 20 cm away ,the path difference introduced is 40 cm .

i.e 40 = λ/2 or λ = 80cm .

f=c/λ = 420 Hz .
 
Last edited:
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See how you can attach an old picture


attachment.php?attachmentid=57630&d=1365435100.png


Or you can attach it again, by changing the name.

Tania is right, the keyword is "consecutive" n and m differ by one.

ehild
 
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Thanks a lot Tanya and ehild! :smile:

I had a similar dilemma with the other problems I am currently going through but now they are all solved, thanks! :)
 

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