# Deriving angular frequency for simple harmonic motion

 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
 P: 12 If $$F=ma$$ and $$F=-kx$$ Then $$ma=-kx$$ (by equating the forces.) Which can be also written as $$ma+kx=0$$ or $$a+\frac{k}{m}x=0$$ 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 $$x=A * Cos (\omega t)$$ where A is the Amplitude of motion and $$\omega$$ is the angular frequency Now Differenting once will give velocity; $$v=-A\omega Sin(\omega t)$$ and again to give acceleration $$a=-A \omega^{2} Cos(\omega t)$$ Now substituting our formula for Acceleration and displacement into our equation of motion $$a+\frac{k}{m}x=0$$ Gives $$-A \omega^{2} Cos(\omega t) +\frac{k}{m}A Cos (\omega t)=0$$ Which can be rearranged to; $$A(-\omega^{2} +\frac{k}{m})Cos(\omega t)=0$$ Can get rid of the $$A$$ and $$Cos(\omega t)$$ which leaves $$-\omega^{2} +\frac{k}{m}=0$$ which can be rearranged to $$\omega=\sqrt{\frac{k}{m}}$$
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P: 6,208
 Quote by tjr39 If $$F=ma$$ and $$F=-kx$$ Then $$ma=-kx$$ (by equating the forces.) Which can be also written as $$ma+kx=0$$ or $$a+\frac{k}{m}x=0$$
or from here

that is in the form $a=-\omega^2x$

where $\omega^2=\frac{k}{m}$

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