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

AI Thread Summary
The discussion revolves around calculating the frequency of sound emitted by a source based on the observed intensity changes when a cardboard is moved away from a detector. The path difference for maximum and minimum intensity is analyzed, leading to the relationship between wavelength and the distance moved. The calculation shows that when the cardboard is moved 20 cm, the corresponding wavelength is determined to be 80 cm, resulting in a frequency of 420 Hz using the speed of sound in air (336 m/s). Participants clarify that the key to solving the problem is recognizing the relationship between consecutive maxima and minima. The thread concludes with the original poster expressing gratitude for the assistance received.
Saitama
Messages
4,244
Reaction score
93

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!
 
Physics news on Phys.org
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:
  • Like
Likes 1 person
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
 
  • Like
Likes 1 person
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! :)
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Back
Top