Order of terms & harmonic approximation I have a question regarding some basics about determining the order of a term in the context of the harmonic approximation.
I am regarding the equation
\begin{align}
2ml\dot{l}\dot{\phi} + ml^2\ddot{\phi} + mgl\sin(\phi) = 0
\end{align}
and I want to simplify it using $\sin(\phi) \approx \phi$ and the harmonic approximaton (which means that I can ignore terms of order greater than $2$).
The textbook states that $2ml\dot{l}\dot{\phi}$ is a term of order $3$, hence the simplification looks as follows:
\begin{align}
ml^2\ddot{\phi} + mgl\phi = 0.
\end{align}
My questions are:

*

*Why is $2ml\dot{l}\dot{\phi}$ a term of order $3$?

*Why is $ml^2\ddot{\phi}$ a term of order $\leq 2$?

*Maybe the most important question (and more general): How can I quickly determine the order of terms that look like the ones above?

 A: Since the derivation from your question is a bit rough, I take the liberty to elaborate on it a little more. The first equation from your question can be written in the form
$$\frac{d}{dt}\left( ml^2 \dot{\phi}\right) + mgl\sin(\phi)=0$$
Trying to find the corresponding Lagrangian, we can further write this as
$$\frac{d}{dt}\frac{\partial}{\partial \dot \phi}\left( \frac{1}{2}ml^2 \dot{\phi}^2\right)-\frac{\partial}{\partial \phi}\left( mgl\cos(\phi)\right)=0$$
Hence, concluding from the Euler-Lagrange-equations, the Lagrangian must have the form
$$L=\frac{1}{2}ml^2 \dot{\phi}^2 + mgl\cos(\phi)+L_2(l,\dot l)=T_1(\phi,\dot \phi, l, \dot l)-V_1(l,\phi)+L_2(l,\dot l)$$
Obviously this is a kind of pendulum problem where $l$ is the variable length (otherwise its time derivative would not appear) of the pendulum and $\phi$ is the angle of rotation.
The harmonic oscillator approximation generally comprises two things:

*

*there is an equilibrium state where $V$ is minimal. Independent of how the system in our case behaves with respect to $l$, $V_1$ is minimal for $\phi = n\cdot 2\pi$, where $n$ is an integer. For the pendulum this is clearly one of the equivalent bottom configurations distinguished only by the "winding number" $n$. Without loss of generality we might choose $\phi_0=0$ as the equilibrium point.

*the Taylor expansion of the Lagrangian around the equilibrium point up to second order terms. Given that the forces are zero (or potential energy is minimal) in equilibrium, first order terms disappear, so only second order terms stay.

Suppose there is also an equilibrium point for $l$ given by $l_0$, then the harmonic oscillator approximation is obtained by expressing the Lagrangian in the new equilibrium coordinates
$$\phi\to \phi_0+\phi \qquad \mbox{and} \qquad l\to l_0+l$$
where now $\phi$ and $l$ denote the small deviations from the equilibrium point (it would be clearer to denote them by $\delta \phi$ and $\delta l$, but I want to avoid notational clutter). Then due to $\phi_0=0$ we obtain
$$L=\frac{1}{2}m(l_0+l)^2 \dot{\phi}^2 + mg(l_0+l)\cos(\phi)+L_2(l_0+l,\dot l_0+\dot l)$$
Taylor expansion in the small deviations $\phi$ and $l$ up to second order yields
$$L\approx \frac{1}{2}ml_0^2 \dot{\phi}^2 + mg(l_0+l) - \frac{1}{2}mgl_0\phi^2+(\dots \mbox{terms from $L_2$} \dots)$$
Especially note, that in the Taylor expansion, cubic (i.e. order 3) terms in $l\dot \phi^2$ and $l\phi^2$, quartic (order 4) terms in $l^2\dot \phi^2$, and even higher orders (from the cosine) have been neglected, as it is required by the harmonic oscillator approximation.
The above Lagrangian leads, with respect to $\phi$, to the second equation from your question, except that it denotes by $l_0$ the equilibrium length of the pendulum, which is, in your version, still denoted by $l$. However, the version from your "textbook" is misleading since $l_0$ is actually a constant, while $l$ seems to suggest that it is still a variable length.
You could also directly introduce the transformation to the equilibrium variables into the equation of motion. Then you would take the Taylor expansion only up to first order (as stated by TBissinger's answer, and this is why in Eli's answer it is called linearization, order 1 means linear). However, as stated above, if you do not take account of the equilibrium position for $l$, you will not understand what term is of what order. So answering your numbered questions makes no sense because they ignore this fact.
Of course, the above derivation is a bit more lengthy than yours, but at least you understand what is happening. Usually, if you have done such calculations a few times, you don't have to go through all the details anymore, and may relapse into more hand-waving "calculations".
A: 
The equations of motion are:
$${\ddot l}-{\dot\varphi }^{2}l-g\cos \left( \varphi  \right)+\frac km\,l=0\\
\ddot\varphi +2\,{\frac {{\dot l}\,\dot\varphi }{l}}+{\frac {g\sin
 \left( \varphi  \right) }{l}}=0
$$
Linearization
$$\cos(\varphi)=1-\frac 12\,\varphi^2
+\frac{1}{24}\varphi^4+\ldots\approx 1\\
\sin(\varphi)=\varphi-\frac 16 \varphi^3+\frac{1}{120}
\varphi^5+\ldots\approx \varphi \\
\dot{\varphi}^2=0~,\text{second order }\\
\dot l\,\dot\varphi=0 ~,\text{second order }$$
$\Rightarrow$
$${\ddot l}-g+\frac km\,l=0\\
l\,\ddot\varphi +g\,\varphi
  =0
$$
A: The order depends on where you drop the terms. You have two levels of description, the Lagrangian or Hamiltonian of your system and the equation of motion derived from it.

*

*The harmonic approximation at the level of the Lagrangian (or Hamiltonian) is exactly that: dropping all terms beyond quadratic order. So you have a second order polynomial as your Lagrangian/Hamiltonian after the approximation. Basically a particle in a square well potential.

*To get equations of motion, you have to use derivation on the Lagrangian/Hamiltonian. Deriving a second order polynomial once leads to a first order polynomial as your equation of motion. Basically Hooke's law.

