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Homework Help: Simple harmonic motion spring question

  1. Apr 26, 2009 #1
    A spring with spring constant 14.5 n/m hangs from the ceiling.
    A 430g ball is attached to the spring and allowed to come to rest.
    It is then pulled down 7.20cm and released.

    What is the time constant if the ball's amplitude has decreased to 3.70 after 52.0 oscillations?

    3. The attempt at a solution

    w = squareroot (14.5 n/m / 0.43kg)

    vmax = 2pi * 3.7 / T
    vmax = w*a
    w*a = 2pi*3.7/ T
    solve for T

    i know it's damping

    not sure what it means by the time constant, nor what formula to use to solve for it

    help would be appreciated, and an explanation of how you got to the awnser
  2. jcsd
  3. Apr 26, 2009 #2


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    Hi vorcil! :smile:

    An exponential process A(t) = A0e-λt has decay constant λ and time constant (or mean lifetime) 1/λ :wink:
  4. Apr 26, 2009 #3

    so 3.7 cm = 7.2cm * e^ (lavender t)

    how do i get lavender??????
    and what am i suppose to be solving for, is it t?

    or because lavender = 1/t do i go e^ (lavender * 1/lavender)
    and solve for lavender, then solve for t??????????
  5. Apr 26, 2009 #4


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    erm :redface: … it's spelt (and pronounced) lambda (originally, I think, from the Hebrew word for "swan")!

    (and why didn't you just copy-and-paste my λ? :rolleyes:)

    You get λ by using A(t) = A0e-λt, and plugging in the data given to you, and solving.
  6. Apr 26, 2009 #5
    what is the t in e-λt, i would be able to solve for lavender if i had that little t,

    and then once i got lavender i'd just use 1/lavender to solve for the time constant am i correct?

    my first physics teacher was asian so i always thought he was pronouncing it lavender lol, lambda XD lamb duh
    Last edited: Apr 26, 2009
  7. Apr 26, 2009 #6
    w= squareroot spring constant / mass

    w = 5.08 rads^-1

    w = 2pi / t
    2pi / w = t

    3.2 = 7.2 * e^(-x*1.08)
    got lavender being 0.61
    1/0.61 = 1.63, awnser was wrong
  8. Apr 26, 2009 #7
    can someone help please? XD
  9. Apr 26, 2009 #8
    I tried using the energy approach but got an unsolvable awnser


    tried using the energy way, i need to find out how much time it takes for the energy in the system to damp by 37%
    that's what Tau is, t

    E[tex]_{}o[/tex] * e ^ -t / [tex]\tau[/tex] = E(t) = 1/2 k xmax^2 (3.2cm)
  10. Apr 26, 2009 #9
    omfg i did it, I SOLVED IT BOOYAH, my method

    the intital amplitude at t = 0 is A = 7.2 cm
    the time taken to reach 1/2 A is aquired by the following
    w = squareroot k/mass
    = 5.806 rads^-1

    T = 2pi / w
    2pi/5.806 = 1.08 seconds
    and that is for each oscillation

    now to reach 52 oscillations it takes 52 * 1.08 = 56.27 seconds
    and that is the point which it reaches 3.7 cm

    now solving for HALF the amplitude, 7.2cm / 2 = 3.6cm NOT 3.7 cm
    so i have to find out how many oscillations it will have to do for 3.6 cms,
    so if 7.2 - 3.7 = 3.5

    3.5 /52 = 0.06730769231 is the distance reduced by each oscillation
    7.2 - 3.6 = 3.6
    3.6 / 06730769231 = 53.48 = is the number of oscillations needed to reach 3.6 cm

    now i multiply that by the time it takes for each oscillation (from above)
    53.48 * 1.08 = 57.88 seconds is the time needed to dampen to half the amplitude

    from the formulas
    1/2 A = A * e^(-t/2 tau)

    the A's cancel out and tau is the time constant

    1/2 = e^(-t / 2 tau )

    take the logs
    ln(1/2) = ln e ^ (-t / 2tau)

    ln(1/2) = -ln2
    -ln2 = t / 2tau
    -ln2 * 2tau = t
    t/2ln2 = tau

    57.88 / 2ln2 = tau
    41.75!! AND IT WAS CORRECT
    except off my a couple sigfigs so i must've rounded somewhere badlly
    right awnser was 42.3 but it still gave me the correct marking
  11. Apr 27, 2009 #10


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    Hi vorcil! :smile:

    (just got up :zzz: …)

    Well done for perservering, and even more so for getting it right! :biggrin:

    Here's some extra symbols for you to copy-and-paste …

    (and also try using the X2 tag just above the Reply box :wink:)
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