# Why $k/m$ in simple Harmonic motion equal $\omega^2$? [closed]

I've come across this thing in simple harmonic motion but never did I manage to find a reason why $k/m$ should equal $\omega^2$ and the theory behind it. People say it is done for convenience equating dimensional but no actual reason did I find.

• Solve the harmonic oscillator. The solutions involve sines and cosines of frequency $\sqrt{k/m}$. – jinawee Nov 27 '15 at 8:38
• I'm voting to close this question as off-topic because it shows insufficient prior research – John Rennie Nov 27 '15 at 8:50
• You'll find the analysis of the simple harmonic oscillator in any introductory physics book. – John Rennie Nov 27 '15 at 8:51

Let us find the mathematical meaning of $k/m$ .
Here, the equation is $\frac{d^2x}{dt^2}=-\frac{k}{m}x$
suppose $\frac{k}{m}=K$ Then, the solution of this equation is $x=A\sin{\sqrt{K}t}$ quantity inside the sine function is the angle. Thus, $\sqrt{K}$ must be angular velocity. Thus, $$\sqrt{K}=\omega\\ K=\omega^2$$. And this is why we substitute $\frac{k}{m}=\omega^2$