Resonance

This is a question I usually ended up confused at.
A uniform tube 60 cm long, stands vertically with lower end dipping into water. When its length above water is 14.8 cm and successively again when it is 48 cm, the tube resonates to a vibrating tuning fork of frequency 512Hz. Determine the lowest frequency to which this tube can resonate when it is taken out of water approximated to the nearest integer.

(You can check your answer at the judge.)


-Shivanker

23 comments:

  1. AnonymousApril 25, 2011 at 9:26 PM

    can't believe that the answer ain't
    283

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  2. ShivankerApril 25, 2011 at 9:30 PM

    it ain't.. :)

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  3. shaanApril 25, 2011 at 11:25 PM

    This comment has been removed by a blog administrator.

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  4. ShivankerApril 25, 2011 at 11:37 PM

    give me ur email id shaan.. i can only email it to you..

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  5. shaanApril 25, 2011 at 11:57 PM

    Thanks buddy,

    wonderful question..

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  6. mayankApril 26, 2011 at 9:58 AM

    This comment has been removed by a blog administrator.

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  7. ShivankerApril 26, 2011 at 10:25 AM

    This comment has been removed by the author.

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  8. mayankApril 26, 2011 at 10:26 AM

    sorry, I dont have any gmail id.

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  9. mayankApril 26, 2011 at 10:27 AM

    Ok, no tension then. We will discuss the question here only today evening so, that everyone would have seen the question.

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  10. Sambhav GuptaApril 26, 2011 at 12:54 PM

    the question ain't difficult, but very easy to confuse...(confused me too)

    people MUST try this one

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  11. mayankApril 26, 2011 at 1:04 PM

    This comment has been removed by a blog administrator.

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  12. Sambhav GuptaApril 26, 2011 at 1:24 PM

    Determine the lowest frequency to which this tube can resonate when it "is taken out of water" approximated to the nearest integer.

    it is no longer a closed end one

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  13. mayankApril 26, 2011 at 2:23 PM

    jab medium alag hai to it is closed end-if it is just above the water and touching it. If it is at sufficient height from water then it is open pipe.

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  14. Sambhav GuptaApril 26, 2011 at 2:47 PM

    i cudn't tell if you were telling or asking, but yes, i agree to what you said

    actually whenever there is a change of medium, mechanical waves are partly reflected and partly absorbed, but there can't be any 'displacement' at the interface of two media, so it has to be a node, and we treat such a pipe as a closed end one

    when it is completely out of water, both ends have max displacement and both act as antinodes, and the pipe is an opened end one

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  15. mayankApril 26, 2011 at 4:41 PM

    so, now as it is 'just above' i.e.a change in medium so, shouldn't we consider it as a closed pipe.

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  16. Sambhav GuptaApril 26, 2011 at 4:47 PM

    'just above' ?

    i didn't get you?
    it has been completely taken out of water, only medium in picture is air

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  17. mayankApril 26, 2011 at 4:55 PM

    ok fine then

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  18. shivam gargshyaApril 28, 2011 at 1:11 PM

    This comment has been removed by the author.

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  19. AnonymousApril 28, 2011 at 1:13 PM

    okay here as i did £=4*14.8 speed=4*14.8*512 now new frequency = 4*14.8*512/2*60 which gives 252

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  20. ShivankerApril 28, 2011 at 1:44 PM

    here's the solution..

    lambda/4 = 0.148 + e(end correction)
    3lambda/4 = 0.48 +e

    hence lambda=0.664, e=0.018

    =>> minimum frequency for open tube resonance = c/lambda = c/2(effective length of tube) = 0.664*512 / 2*(0.60 + 2*0.018) = 267 Hz

    Cheers!

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  21. AnonymousApril 28, 2011 at 2:29 PM

    question was really beautiful pehle socha tha ki do bar length kyun de rakhi hai but then think shayad n nikalne ke liye de ho and last post maine hi anominous ke naam se ki thi

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  22. shivamApril 28, 2011 at 2:32 PM

    abe yaar phir anomnious its shivam

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  23. Pulkit MendirattaApril 28, 2011 at 9:26 PM

    Nice work shivanker !

    ReplyDelete

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