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Deriving angular frequency for simple harmonic motion 
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#1
Mar108, 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 


#2
Mar2508, 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] 


#3
Mar2508, 07:57 PM

HW Helper
P: 6,208

that is in the form [itex]a=\omega^2x[/itex] where [itex]\omega^2=\frac{k}{m}[/itex] 


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