Calculate small small oscillations of a pendulum The system is setup as follows:
A point $O_1$ moves along the $x$ axis with it's $x$ coordinate being $a\sin(\omega t)$ and $\omega\ne\sqrt{\frac{g}{l}}$.
There's a pendulum attached to $O_1$ of length $l$ and a weight of mass $m$.
I'm asked to determine the small oscillations of the pendulum if at $t=0$ the pendulum was still and it's angular displacement was $\varphi_0$.
What I attempted is to get the Euler-Lagrange equations for the system. My velocity was $v=l\varphi'+x'$ where $\varphi$ is the angular displacement of the pendulum and $x$ is the location of the point $O_1$.
The final equation, assuming I didn't make a mistake is
$$2l\varphi''-2a\omega^2\sin(\omega t)+g\varphi'\cos(\varphi)=0$$
If this was in fact what I needed to do, what now?
How do I get the actual answer to the problem?
I admit I'm unclear on what it means to "determine small oscillations".
Any help is appreciated.
 A: This is the picture of your problem 

So point M movement is described by $x_M = a \sin(\omega t)$. This is a system with one degree of freedom and for the coordinate that completely describes this system we will use angle $\theta$. Let us describe $x$ and $y$ position of a pendulum at any moment $$x_m = x_M + r \sin(\theta)$$ and $$y_m = r \cos(\theta)$$. Also let $x=0$ be referent level where gravitational potential energy will be zero. Now kinetic and potential energy will be: 
\begin{align}
T&=\frac{1}{2} m (\dot{x}_m^2 + \dot{y}_m^2) \\
&=\frac{1}{2} m (a^2 \omega^2 \cos^2(\omega t) + 2a\omega r \dot{\theta} \cos(wt) \cos(\theta) + r^2 \dot{\theta}^2 \cos^2(\theta) + r^2\dot{\theta}^2 \sin^2(\theta))\\
&=\frac{1}{2} m (a^2 \omega^2 \cos^2(\omega t) + 2a\omega r \dot{\theta} \cos(wt) \cos(\theta) + r^2 \dot{\theta}^2)
\end{align}
\begin{align}
U&=-mgy_m \\
 &=-mgr\cos(\theta)
\end{align}
Now you form Lagrangian $L = T-U$
$$L =\frac{1}{2} m (a^2 \omega^2 \cos^2(\omega t) + 2a\omega r \dot{\theta} \cos(wt) \cos(\theta) + r^2 \dot{\theta}^2) + mgr \cos(\theta)$$
Now you write Euler-Lagrange equations
$$\frac{d}{dt}\left (\frac{\partial L}{\partial \dot{\theta}} \right) - \frac{\partial L}{\partial \theta}  = 0$$
and you get
$$mr^2\ddot{\theta} - ma\omega^2 r \sin(wt) \cos\theta  + mg r \sin\theta = 0$$
Now you have an equation of motion of this system for some general angle $\theta$. Small oscillations are occurring when you displace a system from stable equilibrium for some small amount, after that system will tend to go back to the stable equilibrium and and therefore oscillate around it. Around stable equilibrium if you displace $\theta$ for some small value $\eta$ your angle will be $\theta = \theta_0 + \eta$ ($\theta_0$ is equilibrium position, $\dot{\theta}=\dot{\eta}$ and $\ddot{\theta} = \ddot{\eta}$) where $\sin \eta \sim \eta$  and $\cos \eta \sim 1$. With this approximations you can simplify this differential equation that we got and find final solutions.
A: The final equation turns out to be:
$$ l \ddot{\phi} - a \omega^2 \sin{\omega t} \cos\phi + g\sin\phi = 0$$
Now, as Aleksandar has suggested, all that remains is to put in the condition of small oscillations i.e. small values of $\phi$. Therefore
\begin{align}
\sin\phi &\approx \phi \\
\cos\phi &\approx 1
\end{align}
Your equation reduces to
$$ l\ddot{\phi} = - g\phi + a \omega^2 \sin{\omega t}$$
This is a typical driven harmonic oscillator. The steady state solution will finally be a sinusoidal with an angular frequency $\omega$. You can find the solution by substituting $\phi$ for the following generic form of solution:
$$ \phi = A\sin(\omega t)$$
where a complex $A$ can account for the phase lag that may be present (will be clear after you work it out).
