1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Harmonic Motion and Springs Question

  1. Jul 22, 2006 #1
    I'm trying to derive the formula for the period T as a function of the mass of the object. Here is my attempt. Note that I cheated and passed the section I had trouble with, without fully understanding it.


    Much of this is quite straightforward for me.

    [tex]y = A sin \alpha[/tex] (basic trigonometry)
    Here is the part I'm not sure I understand fully:

    [tex]\alpha = \omega t[/tex]

    Why does that equation work?

    The angle alpha is replaced by [tex]\psi t[/tex] so the result is:

    [tex]y = A sin \omega t[/tex]

    The speed in the y direction is as follows:

    [tex]v(t) = \frac {dy} {dt} = \omega A cos \omega t[/tex]

    The acceleration in the y direction is:

    [tex]a(t) = \frac {d^2x} {dt^2} = -\omega^2 A sin \omega t[/tex]

    The above is simple calculus.

    When a net force is acting on a body an acceleration will show.

    The force that is acting on the body is (via hookes law):

    [tex]F = -ky[/tex]

    If we replace y with the expression we derived earlier we get

    [tex]F = ma = -m \omega^2 A sin \alpha t[/tex]
    [tex]F = -ky = -k A sin \alpha t[/tex]


    [tex]m \omega^2 = -k \Leftrightarrow \omega = \sqrt{k / m}[/tex]

    Combining the above expression with the commonly known

    [tex]\omega = 2 \pi / T[/tex]

    and you get the final result

    [tex]T = 2 \pi \sqrt {m / k}[/tex]


    My question is:

    Why is [tex]\alpha = \omega t[/tex]?

    Thank you for your time. Have a nice day.
    Last edited: Jul 22, 2006
  2. jcsd
  3. Jul 22, 2006 #2


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    They assume the angular velocity is a constant.
    It probably isn't, so they've hidden away an order of magnitude argument that would show it is a good approximation.

    (That is, they've hidden away everything that physics is about)
  4. Jul 22, 2006 #3
    So it is basically because

    [tex]\alpha = \omega t[/tex]


    rad = rad/s x s

    with the approximation that angular velocity is constant?
  5. Jul 22, 2006 #4


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    Yep, that should be it.
  6. Jul 22, 2006 #5
    I was just wondering why you used psi instead of omega. Nothing wrong with it, but it's not written that way, usually.

    alpaha = omega.t is just the rotational counterpart of s = vt (linear motion with constant velocity).
  7. Jul 22, 2006 #6
    Yes, I noticed that, so I changed it. The "how-to-latex" got me confused for a bit before i realised it.

    Thank you arildno and neutrino! I really should check the units (as in rad = rad/s x s) more often :smile:
  8. Jul 22, 2006 #7


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    The question is not explicitly provided but I was under the impression that that goal was to relate the motion of a mass attached to an ideal spring to circular motion. In that case, using a constant angular velocity is not an approximation or a guess. It follows from the fact that the projection along one of the axis must represent simple harmonic motion. And that implies a constant omega.

    Just a comment.


Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Harmonic Motion and Springs Question
  1. Spring Harmonic Motion (Replies: 1)