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Oscillations in a spider web

  1. Sep 11, 2014 #1
    1. The problem statement, all variables and given/known data

    A small fly of mass 0.29g is caught in a spider's web. The web oscillates predominately with a frequency of 4.4Hz .

    a) What is the value of the effective spring stiffness constant k for the web?

    b) At what frequency would you expect the web to oscillate if an insect of mass 0.50g were trapped?

    2. Relevant equations

    f = (1/2Pi)(k/m)^.5

    3. The attempt at a solution

    a) rearranging the equation

    k =m(2Pif)^2 = (.00029kg)(2Pi*4.4Hz)^2 = .22 N/m

    mastering physics says I am right on this one. So I figured for part b I just rearrange the equation for f and plug in .0005 for m and .22 for k

    f=(1/2Pi)(k/m)^.5 = (1/2Pi)(.22N/m/.0005kg)^.5 = 33Hz

    but mastering physics says i am wrong. Where is my mistake?
     
  2. jcsd
  3. Sep 11, 2014 #2

    BvU

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    The expression is correct. An estimate (sqrt(2000/5) = 20) yields 20/π, around 3.
    As expected: twice as heavy, 1.4 times lower freq.
     
  4. Sep 11, 2014 #3

    BiGyElLoWhAt

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    Hmm... I'm not sure, looks like right to me. using w = root(k/m) and w = 2pif i get the same thing. Did you try 32.9hz?
     
  5. Sep 11, 2014 #4
    it said 32.9 was wrong also. and it says in the question it wants 2 sig figs =[
     
  6. Sep 11, 2014 #5

    BiGyElLoWhAt

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    Have you tried adding another sig fig? use .22 for k and .0005 for m and take it out 2 places.
     
  7. Sep 11, 2014 #6
    without rounding the answer is 32.94930172. Im not sure if thats what you mean
     
  8. Sep 11, 2014 #7

    BiGyElLoWhAt

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    Yea that's about what I got, but it wants 2 sig figs and 33 isn't working, so maybe they mis stated and want 2 after the decimal? I'm honestly not sure, toothpaste. Your algebra looks good, we got the same answer, and assuming it did in fact say that .22N/m was the correct k value (I didn't double check that), I'd probably just punch it in the face and call it a day, honestly.

    Btw, I've been meaning to say something for a while, but I think your name is effing hilarious XD
     
  9. Sep 11, 2014 #8
    yeah its telling me .22 N/m is correct on the first part. =[ so frustrating.

    and thank you! :D i love this username!

    anyways thanks for your help i guess ill email my professor, maybe there is a problem with mastering physics.
     
  10. Sep 11, 2014 #9

    BvU

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    Listen up, you two: if a.29 g fly swings with 4.5 Hz a HEAVIER fly swings at a LOWER freq. So anything above 4.4 Hz is plain wrong.
     
  11. Sep 11, 2014 #10
    By increasing the mass the frequency must decrease. You should be aware that 30 something does not make sense if the original frequency was 4.4 Hz.

    Check your units and calculations.
     
  12. Sep 11, 2014 #11

    AlephZero

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    (1/2Pi)(.22N/m/.0005kg)^.5 is correct.

    You must be doing something wrong with your calculator. You are getting the decimal point in the wrong place.
     
  13. Sep 11, 2014 #12

    BiGyElLoWhAt

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    Yea I see what you guys are saying, I kind of feel like a noob for not seeing that. That's my bad everyone =[

    I just think it's weird we both got the same (wrong) thing.
     
  14. Sep 12, 2014 #13
    I forgot parenthesis on my calculator. I entered 20.976/2Pi instead of 20.976/(2Pi)... sorry guys. I got 3.3 this time and mastering physics accepted it
     
  15. Sep 12, 2014 #14

    BvU

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    My point is that at all times you should make a reasonable guess about the outcome:
    1. The square root can't be far from 20,
    2. The freq has to be lower than 4.4 by a factor of around 1.5
     
  16. Sep 13, 2014 #15
    Yea i should have noticed that. Sometimes the intuition aspect of it doesnt come very easily to me
     
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