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!

Deriving angular frequency for simple harmonic motion

  1. Mar 1, 2008 #1
    1. The problem statement, all variables and given/known data
    Derive the equation for angular frequency for simple harmonic motion of a spring.


    2. Relevant equations
    Derive omega = sqrt(k/m) from F = -kx
    (sorry i don't know how to use notation)


    3. The attempt at a solution
    I asked my teacher how to do this, and he used some crazy math I didn't learn yet, including Euler's identity and differential equations. I'm in an AP calculus bc class, and i understand differential equations, just not some aspects. Does anyone know a simple solution for this? Thanks in advance
     
  2. jcsd
  3. Mar 25, 2008 #2
    If [tex]F=ma[/tex] and [tex]F=-kx[/tex]

    Then [tex]ma=-kx[/tex] (by equating the forces.)

    Which can be also written as [tex]ma+kx=0[/tex]

    or [tex]a+\frac{k}{m}x=0[/tex]

    Now if x is displacement, differentiating once with respect to time will give you velocity of the spring and then differentiating again with respect to time will give acceleration.

    Displacement of a spring can be given by

    [tex]x=A * Cos (\omega t)[/tex]

    where A is the Amplitude of motion and [tex]\omega [/tex] is the angular frequency

    Now Differenting once will give velocity;

    [tex]v=-A\omega Sin(\omega t)[/tex]

    and again to give acceleration

    [tex]a=-A \omega^{2} Cos(\omega t)[/tex]

    Now substituting our formula for Acceleration and displacement into our equation of motion

    [tex]a+\frac{k}{m}x=0[/tex]

    Gives [tex]-A \omega^{2} Cos(\omega t) +\frac{k}{m}A Cos (\omega t)=0[/tex]

    Which can be rearranged to;

    [tex]A(-\omega^{2} +\frac{k}{m})Cos(\omega t)=0[/tex]

    Can get rid of the [tex]A[/tex] and [tex]Cos(\omega t)[/tex]

    which leaves [tex]-\omega^{2} +\frac{k}{m}=0[/tex]

    which can be rearranged to [tex]\omega=\sqrt{\frac{k}{m}}[/tex]
     
  4. Mar 25, 2008 #3

    rock.freak667

    User Avatar
    Homework Helper

    or from here

    that is in the form [itex]a=-\omega^2x[/itex]

    where [itex]\omega^2=\frac{k}{m}[/itex]
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Deriving angular frequency for simple harmonic motion
Loading...