Manipulating a formula

  1. Jun 3, 2008 #1
    [SOLVED] Manipulating a formula

    1. The problem statement, all variables and given/known data
    "An instructor sets up an oscillating vertical mass-spring system (k=6.05 N/m). The maximum displacement is 81.7 cm and the maximum speed is 2.05 m/s. What is the mass of the oscillator?
    Answer: 0.961kg

    2. Relevant equations
    vmax=A*the square root of k/m
    (maximum speed of a mass-spring system)

    3. The attempt at a solution
    Well, I probably could solve this, if only I could manipulate the formula for m. I've tried m = the square root of V*A divide by k squared (sorry, I don't know how to use symbols) and other similar manipulations, but it just doesn't work for me. Note: I am a complete "n00b" at physics and am doing practice problems to improve.
     
  2. jcsd
  3. Jun 3, 2008 #2
    Two things i have to point out:

    a) Make sure all your metrics are in the right SI form (so km to meters etc etc...)

    b) In the first formula, you say that m is under the root, compared to vmax. There is no other root or square in the formula. When you would transform the equation to m =... , would it be possible to still have a root there? with something that is not squared underneath?
     
  4. Jun 3, 2008 #3
    a) Yes, when attempting I made sure that A was 0.817kg.

    b) I don't really understand what you are saying... I can try to explain the formula I expect I need to use better:

    Vmax = A√k/m

    That symbol being the square root, of course.
    The reason I mentioned "squaring" some things is because I assume the way to get things out of a square root (especially in this context) is to square them.
     
  5. Jun 3, 2008 #4
    First of all what do you mean with get the things out of the root? You have to square both sides so that the root drops out. Then bring m to one side using multiplication and division, you think you can do this with this info I gave you?
     
  6. Jun 3, 2008 #5
    Okay, while this does help and I feel I am on the verge of getting it right, i still didn't get it. I am left with equations such as m=vmax squared divide by A*K and variations of those variables being squared.

    Thanks for your patience, though, I realize I am very out of my league with physics.
     
  7. Jun 3, 2008 #6
    Hehe, this isnt that hard. You just made a small mistake in derivation of your formula for m out of the formula for vmax. Redo it, try and find your error. ;).
     
  8. Jun 3, 2008 #7
    Tried numerous times, getting things such as:

    m = vmax2/A(k)2
    m = SQUR(vmax2/A(k)2)
    m = vmax2/Ak
    m = vmax/A(k)2
    etc.

    But I'm still not getting it. :frown:
    I have a physics exam tomorrow and am practicing so that I can actually pass, and this question for whatever reason or other is kicking my ass.
     
  9. Jun 3, 2008 #8
    try writing down step by step on here. ill try and help you along.
     
  10. Jun 3, 2008 #9
    Ok

    a. vmax = A*SQUR(k/m)

    b. vmax2 = A * (k/m)2 (now out of the square root)

    c. vmax2/A = k/m2 (divided by A)

    d. vmax2/(Ak2) = m2 (divided by k, unsure if it should be squared or not)

    e. SQUR(vmax2/(AK2)) = m (not sure if this is necessary)
     
  11. Jun 3, 2008 #10
    ok. Your first step. From a to b. If you square both sides, keep in mind that a square cancels out a root, and variables that arent under a root get a square...
     
  12. Jun 3, 2008 #11
    Okay, so I see that I didn't square A... but when I do, I am left with the answer 0.4147...
     
  13. Jun 3, 2008 #12
    You didnt only forget to square A, but you also squared k/m when not necessary, because of the root. :)
     
  14. Jun 3, 2008 #13
    Also, to add...

    First fix your step from a-->b
    Also, from c-->d if you divide by k, how did m end up on the top and nothing changed on the left hand side.
     
  15. Jun 3, 2008 #14
    D'oh. Okay, well I've taken that into consideration and am now getting 0.922~.

    I thought there was something wrong with that, but I don't see what I should have done... i should multiply by m? :confused:
     
  16. Jun 3, 2008 #15
    Well, let me try something like this for an example, just some random variables that have nothing to do with anything and lets solve for p


    v*r/p = T/m

    1/p = T / (m*r*v)

    p = (m*r*v) / T <--- flipped the equation because the step before, it was 1/p

    EDIT: What you're really doing from step 2-->3 is...


    1/p *p = T / (m*r*v) *p (mult both sides by p)

    1 = T*p / (m*r*v) (mult both sides by (m*r*v))

    (m*r*v) = T*p (divide both sides by T)

    (m*r*v) / T = p
     
    Last edited: Jun 3, 2008
  17. Jun 3, 2008 #16
    Okay, I finally got it!

    A2*K/vmax2 = m = 0.9609 = 0.961kg

    Thank you Swatje, dirk_mec1, and lukas86, thank you very much!
    Off to more studying, but I will most likely be back soon :)

    Thanks again!
     
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