Deriving angular frequency for simple harmonic motion

[AlexYH]

Homework Statement

Derive the equation for angular frequency for simple harmonic motion of a spring.

Homework Equations

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

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

 

[tjr39]

If $$F=ma$$ and $$F=-kx$$

Then $$ma=-kx$$ (by equating the forces.)

Which can be also written as $$ma+kx=0$$

or $$a+\frac{k}{m}x=0$$

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 $$\omega=\sqrt{\frac{k}{m}}$$

 

[rock.freak667]

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