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Homework Help: Simple Harmonic Motion Question

  1. Oct 22, 2008 #1
    1. The problem statement, all variables and given/known data

    A weight of mass m is placed on a string. as a result it exerts a force equal to xk that tends to reduce x, where k is the spring constand and x is the distance from the equilibrium position of the spring.

    1. Confirm that x = Asin(wt+ f) is indeed a solution of theequation of SHM. (given below)
    2. Write down the relationship between k, m, and w^2.

    2. Relevant equations

    d^2 / dt ^ 2 = -w^2 * x




    3. The attempt at a solution

    1. Well i got this part by differentiating x = Asin(wt+f) and indeed it is a solution
    2.

    I have no idea how to do this part...
    All i can think of is that at the peaks and troughs of oscillation a is zero so the forces mg and xk must be equal but that doesn't really help me relate them in general
     
  2. jcsd
  3. Oct 22, 2008 #2
    Think in terms of Newton's second law of motion.
     
  4. Oct 22, 2008 #3
    Well i'm tryin to...but not sure how....
    Fnet = mg + xk = ma

    (ma - mg) / k = x


    well this is tthe relationship between m and k and... I doubt its even right... and i have no idea how to go to relate to w^2 of all things
     
  5. Oct 22, 2008 #4
    How did the mg come in?

    Anyway, Fnet is also equal to m(d2x/dt2).
     
  6. Oct 22, 2008 #5
    d^x / dt^2 = a(t) right?
    so would it be f = ma
    so Fnet = m * d^2/dt^2?

    crap, i misread the first time lol
     
    Last edited: Oct 22, 2008
  7. Oct 22, 2008 #6
    It is not mentioned that the spring is hanging vertically. You must therefore assume that is placed on a smooth horizontal surface. In that case, the force due to gravity and the normal force due to the surface cancel each other.


    Please look at the equation I typed once more. If it is still not clear, read your textbook (on Newton's second law).

    No.
     
  8. Oct 22, 2008 #7
    OK so then for a horizontal string :
    kx = -mw^2x
    so they are related by

    k = -mw^2 (i think)

    But can you help me figure out what the answer would be with a vertical spring as well cause now im curious...
    Would it be mg + xk = -mw^2x?
     
  9. Oct 22, 2008 #8
    There is a rather important detail missing.


    Again, the same detail is missing here. But as it turns out, the relation between these three quantities (omega, k and m) is the same no matter how the spring is oriented. This of course assumes ideal conditions, such as a negligible mass for the spring, etc.
     
  10. Oct 22, 2008 #9
    is it that its a restorive force so it shud be k = mw^2?
     
  11. Oct 22, 2008 #10
    Exactly.
     
  12. Oct 22, 2008 #11
    ahhh excellent thanks a lot for all the help
     
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