Forget about the motion for a moment. First just consider a pendulum of length $\ell$ at an angle $\vartheta$ from rest. If $(0,0)=0\mathbf{e}_x+0\mathbf{e}_y$ is where the pendulum is attached, then the end of it is at
$$
\bigl(\ell\cdot\sin(\vartheta), -\ell\cdot\cos(\vartheta)\bigr)
$$
So, now let's describe the motion in terms of these variables. $\ell$ would usually be constant, so we don't need to care about this. It's $\vartheta$ that varies in time, so we get
$$
y(t) = -\ell\cdot\cos(\vartheta(t)),
$$
$$
x(t) = \ell\cdot\sin(\vartheta(t)),
$$
Unfortunatly, we don't know $\vartheta(t)$ yet.
We can either
- Assume that it does oscillate harmonically, that is, $\vartheta(t)=A\cdot\sin(\omega t)$ for some $A$ and $\omega$, which is often a good approximation.
- Decide we can do better, which is necessary if you consider big angles like $30^\circ$.
Lets stick to small angles here. $\vartheta(t)=A\cdot\sin(\omega t)$ with $A\ll1$. We can then try to further simplify the cartesian coordinate representations: in
$$
x(t) = \ell\cdot\sin(\vartheta(t)) = \ell\cdot\sin\bigl(A\cdot\sin(\omega t)\bigr),
$$
we can use that $A\cdot\sin(\omega t)\ll 1$ and $\sin(x)\approx x$ for $x\ll 1$, therefore
$$
x(t) \approx \ell\cdot A\cdot\sin(\omega t).
$$
In
$$
y(t) = -\ell\cdot\cos(\vartheta(t)) = -\ell\cdot\cos\bigl(A\cdot\sin(\omega t)\bigr),
$$
we can use that $\cos(x)\approx 1-\tfrac{x^2}2$ for $x\ll 1$, therefore
$$
y(t) \approx \ell\cdot \bigl(\tfrac{A^2}2\cdot\sin^2(\omega t)-1\bigr)
$$
which, due to $\sin^2(x)=\tfrac12-\tfrac12\cos(2x)$, can be further simplyfied to
$$
y(t) \approx \tfrac{A^2}{4}\ell\cdot\cos(2\omega t) - \tfrac\ell2.
$$
As you see, two harmonic oscillators is in fact not such a bad assumption, and the important thing is that $y$ must oscillate at twice the frequency of $x$.
Sorry, I forgot about this question...
When I say $\sin(x)$ is approximately $x$, I mean just what I say: if you look at the sine function, and "zoom in" very close to 0 (that is, $x\ll1$, for instance $x=.001$), then this function looks very much like the simple $f(x)=x$.
http://www.wolframalpha.com/input/?i=plot+sin%28x%29+from+-.01+to+.01
It's just when you look very carefully that you notice it's in fact bent a little bit.
It's been awhile since I've studied QM, but I recall it often being stated that classical mechanics was the limit of quantum mechanics as $\hbar \to 0$. Yes, the Hamiltonian approach to mechanics is often used in both. The Poisson bracket of classical Hamiltonian mechanics has it's quantum mechanical analog in the quantum mechanical commutator. IIRC, this is one way to quantize classical theories. For classical or quantum field theory, one typically works in the Lagrangian framework, where one tries to minimize some action.
Best Answer
When moving up you are pushing yourself in the opposite direction of the force of gravity. Therefore you do a positive work which is approximately $mgh$ ($h$ is the height of the step). While coming down gravity will do the same work for you.