Interference - Q done, is it right?

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SUMMARY

The discussion centers on the calculation of frequency using the wavelength and distance in an interference problem. The user initially calculated the wavelength (λ) as 0.698m and derived the frequency (f) as 487 Hz using the formula v=fλ, where v is the speed of sound at 340 m/s. Upon review, it was confirmed that the correct approach involves recognizing that multiple frequencies can be derived from the equation (2n-1)λ/2=d, leading to a more comprehensive understanding of the problem.

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Hi all, I have a question here that I've done but it hasn't got an answer provided. I'm sure it's easy stuff but haven't looked at this material in a while so
just want to double check.

(The things in red are what I added to the original diagram)

intereferenceexampleQ.jpg


RE-DONE

d = √17

d + λ/2 = 2√5

λ/2 = 2√5 -d

λ = 4√5 -2(√17)

λ = 0.698m

Using v=fλ

∴f = 340/0.698 = 487 Hz

Is this correct?

Thanks in advance
 
Last edited:
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d=\sqrt{1^2+4^2}=\sqrt{17}
 
szynkasz said:
d=\sqrt{1^2+4^2}=\sqrt{17}

Woops, muppet at work. No idea how that came about, thanks.

Looking over it I've done a couple of very stupid errors such as line 4.

First post RE-DONE
 
Last edited:
Correct, but you get more frequencies:

(2n-1)\frac{\lambda}{2}=d\Rightarrow f=(2n-1)\frac{v}{2d},\,n=1,2,3,...
 
szynkasz said:
Correct, but you get more frequencies:

(2n-1)\frac{\lambda}{2}=d\Rightarrow f=(2n-1)\frac{v}{2d},\,n=1,2,3,...

Thanks a lot, big help
 

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