Deriving angular frequency for simple harmonic motion


by AlexYH
Tags: angular, deriving, frequency, harmonic, motion, simple
AlexYH
AlexYH is offline
#1
Mar1-08, 06:26 PM
P: 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
Phys.Org News Partner Science news on Phys.org
Better thermal-imaging lens from waste sulfur
Hackathon team's GoogolPlex gives Siri extra powers
Bright points in Sun's atmosphere mark patterns deep in its interior
tjr39
tjr39 is offline
#2
Mar25-08, 07:11 PM
P: 12
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]
rock.freak667
rock.freak667 is offline
#3
Mar25-08, 07:57 PM
HW Helper
P: 6,212
Quote Quote by tjr39 View Post
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]
or from here

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

where [itex]\omega^2=\frac{k}{m}[/itex]


Register to reply

Related Discussions
Motion of object (simple harmonic motion?) Introductory Physics Homework 3
Simple Harmonic Motion and Wave Motion General Physics 2
simple harmonic frequency clarification Introductory Physics Homework 3
Angular Simple Harmonic Motion Introductory Physics Homework 3
Simple Harmonic Motion/Wave Motion Introductory Physics Homework 2