Why does $k/m=\omega^2$ for harmonic motion? Can anyone please give me a proof for $k/m=w^2$ in simple harmonic motion?
I have tried energy conservation and Newton's laws as follows :
In the case of a mass-spring system,
$$F=ma =-kx\\
F=ma = mr\omega^2 $$
hence 
$$\frac km  = \omega^2$$
Or 
$$\frac12 mv^2 = \frac12kx^2$$
$$\frac km  =  \frac{V^2}{x^2}   =   \omega^2$$
therefore $k/m  =  \omega^2$
Are these valid and correct? I have been stuck in this for the  entire day.
 A: The simple harmonic oscillator is governed by Hooke's Law, $F=-kx$. Since Newton's laws tell us that $F=ma=m\ddot x$ (where $\ddot x$ is the second derivative of $x$ with respect to time), we have a second-order differential equation
$$
\ddot x = -\frac kmx.
$$
We want to find the position as a function of time, $x(t)$, which solves this equation. A function which is proportional to its derivative usually involves an exponential, so let's guess
\begin{align}
x(t) &= e^{\beta t} \\
\dot x &= \beta x \\
\ddot x &= \beta^2 x
\end{align}
This guess does in fact solve our differential equation if and only if
$$
\beta^2 = -\frac km,
$$
which is permitted if $\beta$ is imaginary. There are two solutions, then: 
$\beta = \pm i\omega$, where $\omega$ is real and has units of $\mathrm s^{-1}$.  This gives you $\omega^2 = +\frac km$, as you asked.
We can construct purely real solutions thanks to the Euler identity:
\begin{align}
x_\text{even}(t) &= \frac{e^{i\omega t}+e^{-i\omega t}}{2} = \cos \omega t \\
x_\text{even}(t) &= \frac{e^{i\omega t}-e^{-i\omega t}}{2i} = \sin \omega t \\
\end{align}
This construction makes the interpretation of $\omega$ as angular frequency more obvious: the position repeats whenever the dimensionless product $\omega t$ increases by $2\pi$, so we can think of $\omega t$ as a "phase angle" (though no physical angle need be involved) measured in radians.  The number of cycles that have occurred up to time $t$ is $\omega t/2\pi$, so the clock frequency is $f=\omega/2\pi$.
