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Frequency of vibration

  1. Apr 25, 2010 #1

    DR1

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    1. The problem statement, all variables and given/known data

    a mass attached to the lower end of a vertical spring causes the spring to extend by 25mm to its equilibrium position. The mass is then displaced a further 20mm and released. a trace of the vibration and time mesurements are taken. From these mesurements it can be seen that the displacement from the equilibrium position is 19.2mm when the time is 0.05s

    A) calculate the expected frequency of vibration

    B) calculate the maximum acceleration of the mass

    C) calculate the maximum velocity of the mass

    2. Relevant equations

    i belive the frequency formule to be f=1/T Hz, or the displacement formule x=Asin(wt+o) both of which im sure are part of the solution but i can not see how to use them with the info given

    3. The attempt at a solution

    i am having an problem knowing were to start with this as i belive the key is finding out the weight of the Mass any help or pointers would be greatfully recived

    many thanks
     
  2. jcsd
  3. Apr 25, 2010 #2

    kuruman

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    You need to plug in the formula that you have for x(t) to find ω. If it is frequency f that you are looking for, them f = 2π/ω. Begin by identifying the symbols in the equation

    x(t) = A sin(ωt+δ)

    What is A, what is δ? Use the fact that at t = 0, the mass is at 20 mm from the equilibrium position and is not moving (instantaneously).
     
  4. Apr 25, 2010 #3

    DR1

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    ok so A is amplitude which is the maximum displacement = 0.02m

    so putting all the info in gets 0.020=0.020sin(w0+angle)
    therefore 0.02/0.02=sinw which is 0.017 this does not look right to me i am obviously missing something please help
     
  5. Apr 25, 2010 #4

    kuruman

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    No. Try again. What do you get if you solve 0.02 = 0.02 sin(ω*0+δ)? Any number multiplied by zero is zero, so there is no ω in the expression.
     
  6. Apr 25, 2010 #5

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    ok so i am still mot following im afraid i am using the figures as
    displacement (x) = 0.020
    time (t)=0
    Amplitude (A)=0.020
    initial Angle (O)= 0
    are these the correct figures to use or am i best using the figures once released ie

    x=0.0192
    t=0.05
    A=0.02
    O=0

    i am trying to reaarange the formule for w but failing miserably as it is giving me silly answers

    im sorry if i am missing the obvious
     
  7. Apr 25, 2010 #6

    kuruman

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    0.02 sin(ω*0+δ) = 0.02
    sin(0+δ) = 0.02/0.02
    sin(δ) = 1
    δ = ?
     
  8. Apr 25, 2010 #7

    DR1

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    δ=90
    if i were to then put this with the figures gained once released
    0.0192=0.02sin(w*0.05+90) how do i get w on its own? do i need to mutiply out the barckets
     
  9. Apr 25, 2010 #8

    kuruman

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    ω*0.05+90=arcsin(0.0192/0.02)
     
  10. Apr 25, 2010 #9

    DR1

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    ok so reaaranging for w i am getting

    (arcsin(0.0192/0.02)/0.05)-90=1384.796 this sounds to big to me were am i going wrong
     
  11. Apr 25, 2010 #10

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    alternatively if i do ( Arcsin(0.0192/0.02))/90.05 = 0.82rads-1 this sounds better? could you confirm either way
     
  12. Apr 25, 2010 #11

    kuruman

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    You put in the numbers incorrectly. Also, depending on the settings of your calculator, the arcsine may return the angle in degrees or radians. It's up to you to figure this out - I don't have your calculator in front of me.

    Smaller but not necessarily better. How did that 90.05 get in the denominator? It seems you need to improve your algebra skills.
     
  13. Apr 25, 2010 #12

    DR1

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    I am really struggling with this, my maths skills have never been particularly great as I am sure I am demonstrating!

    ω*0.05+90=arcsin(0.0192/0.02)

    You stated the above, to get w on its own I cannot work out whether I group up 0.05+90 and then put it in the denominator, or if I move both individually, and therefore put 0.05 in the denominator and minus 90 from the overall result?

    I no longer have the book for my calculator so will try and find a manual online re the rads/degrees issue

    Thanks for your assistance, it is appreciated, I know you must be getting frustrated with me!
     
  14. Apr 25, 2010 #13

    kuruman

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    If you are calculating the arcsine in radians, convert the 90 to π/2. Then move 90 (or π/2) over to other side and change its sign. Now divide everything on the right side and on the left side by 0.05. What do you get?
     
  15. Apr 25, 2010 #14

    DR1

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    Am i right in say that my calculator is working out the arcsine in degrees as the result I get is 73.74?
     
  16. Apr 25, 2010 #15

    kuruman

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    Yes you are right in that regard. Now proceed with the calculation of ω as I indicated.
     
  17. Apr 25, 2010 #16

    DR1

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    (73.74-90)/0.05 = -325.2

    alternatively if I do the rads option (having now found a guide for my calculator!) i get

    (1.287002218-(π/2))/0.05 = 16.38

    am i right in concluding that the first answer is in degrees and the second one is in rads so its the second one that I want?
     
  18. Apr 25, 2010 #17

    kuruman

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    Correct, it is rads that you want, but your numbers are wrong. What do you get when you subtract π/2 from 1.287?
     
  19. Apr 25, 2010 #18

    DR1

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    -0.2838?

    then dividing by 0.05 gives -5.676 as my answer?
     
  20. Apr 25, 2010 #19

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    and then dividing that by 2π gives me the answer in hertz?
     
  21. Apr 25, 2010 #20

    kuruman

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    Right. However, if you solve for the frequency, you get a negative number and that's not cool. Note that after one period T has gone by, the angle returns to the exact same value of -0.2838. This means that, for any time t, we can write
    ω(t - T) = -0.2838.
    Take out the parentheses and you get
    ωt - ωT = -0.2838.
    Now ωT = 2π. Put that in the equation, move it over to the other side and change its sign. Then solve for ω. You should get a positive value.

    I know this sounds complicated. There is a simpler way to write x(t), but you should stick with this description until is it is finished. I don't want you to be unnecessarily confused.

    Good luck. I have to sign off for now.

    **** Edit ****
    This posting refers to DR1's posting #18.
     
    Last edited: Apr 25, 2010
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