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

[Saitama]

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 )

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.

Any help is appreciated. Thanks!

 

[Tanya Sharma]

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 .

 

[ehild]

See how you can attach an old picture

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

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

ehild

 

[Saitama]

Thanks a lot Tanya and ehild!

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

 

[HalfLostAndSearching]

Your approach is correct, but you are missing a key piece of information: the distance between the source and detector. Without this information, it is impossible to determine the exact frequency of the sound emitted. The frequency will depend on the wavelength, which in turn depends on the distance between the source and detector.

To solve this problem, you will need to use the formula for the speed of sound in air (336 m/s) to find the wavelength of the sound. Then, using the path difference equation you have already derived, you can solve for the frequency. However, keep in mind that there are multiple possible values for the wavelength and frequency, so you will need to use some trial and error to find the correct one that fits the given information.

Alternatively, if you know the distance between the source and detector, you can use the formula v = fλ to solve for the frequency directly.