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Homework Help: Maximum speed of a mass on a spring

  1. May 16, 2010 #1
    1. The problem statement, all variables and given/known data

    Prove that the maximum speed (Vmax) of a mass on a spring is given by 2PifA


    2. Relevant equations



    ac=4 Pi^2 r / T^2

    T=2Pi sqrt(r/ac) , T=2Pi sqrt(A/ac)


    T=2Pi sqrt(-x/ac) , T=2Pi sqrt(m/k)

    f=1/2Pi sqrt(k/m) , f=1/2Pi sqrt(a/-x)

    3. The attempt at a solution

    Ok so after much trial and failure and producing much gibberish full of variables..... This is what i know so far that is right, is the mass is going at max speed then the total energy which is made of potential energy plus kinetic energy, is all made of just kinetic energy because potential energy should equal 0 at max speed, well i think so....


    and anything after that doesn't make sense whatever route i try.

    I am taking a correspondence course by ILC ( the independent learning centre at Ontario) , and its been 6 years since i graduated from high school so my grade 12 physics its not very good.

    Thank you for all the help !!
  2. jcsd
  3. May 16, 2010 #2

    Doc Al

    User Avatar

    Staff: Mentor

    Good thinking.

    Hint: Try eliminating k/m using one of your equations. (First rewrite this equation in terms of k/m.)

    For some reason, your list of relevant equations has stuff about centripetal acceleration. That's not relevant here.
  4. May 17, 2010 #3
    o the thing about the centripetal equations is that the lesson was using them for explaining on how to obtain certain variations with T and f equations, so i tried everything but to no avail.

    Dude i could kiss you!! well metaphorically speaking hahaha, here is my work and well i hope this can be used for other people.




    sqrt (k/m) = v/A

    v=A sqrt(v/A) i pause here and i work on another equation to insert on this one

    T=2Pi sqrt(m/k)



    2Pi/T=sqrt(k/m) now i insert on the previous equation and i get this

    v=A (2Pi/T)

    v= A2Pi/ T

    v=A 2Pi f

    got the result!!!! thanks for the hint!!!
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