Period of a pendulum hanging from point attached to two horizontal, identical springs Let's suppose there is a rigid pendulum attached to a two-spring system like in the image. Both springs are identical, and their elastic constant is $k$. The pendulum has length $l$ and there is a point mass $m$ attached to its free end. The point P can oscillate back and forth in a straight line, but the sum of the length of both springs at each moment must be constant, as is depicted. The gravity points downwards.

I want to work out the equations of motion for the mass $m$ using the Lagrangian formalism. First, I have described the kinetic energy of the system as the sum of the energy associated with the translation of the point $P$ and the angular movement of the mass $m$. Let $\theta$ be the angle between the pendulum and the vertical line arising from P and $x(t)$ the position of $P$ at any given point.
$$T = \frac{1}{2} m \dot{\theta}^2 l^2 + \frac{1}{2}m\dot{x}^2$$
The potential energy is equal to that stored in the springs plus the gravitational energy arising from the movement of $m$. If $L_0$ is the natural length of the springs, then the potential elastic energy equals (if I am not mistaken):
$$V_{elastic} = \frac{1}{2}k (L_0 + x)^2 + \frac{1}{2}k (L_0 - x)^2 = (…) = kL^2_0 + kx^2$$
And, because of the angular movement of $m$, the gravitational potential energy is: $$V_{gravitational} = mgl \sin{\theta}$$
And therefore the lagrangian of the system is:
$$\mathcal{L} = T - V = \frac{1}{2}m\dot{\theta}^2 l^2 + \frac{1}{2}m \dot{x}^2 - mgl\sin\theta -kL^2_0 - kx^2$$
Applying the Euler-Lagrange equation, I arrive at the equations of motion:
$$\ddot{\theta} = -\frac{g}{l}\cos\theta$$
$$m\ddot{x} = -2kx$$
But I don't know 1) whether or not I got the right result and 2) how to find out, for small oscillations, the period of the pendulum. I have to prove that the period is: $$\tau=2 \pi \sqrt{\frac{m}{2 k}+\frac{g}{l}}$$
 A: 
First, I have described the kinetic energy of the system as the sum of the energy associated with the translation of the point P and the angular movement of the mass m

You are not allowed to do that!  It would be permitted if point $P$ were the center of mass of the system, but it isn't.
To determine kinetic energy, express the particle's position in Cartesian coordinates:  $\vec r = [x + l \sin\theta, l \cos\theta]$ (where $\hat j$ is down and $\theta$ is counter-clockwise from  $\hat j$).  Then $\vec v  = [\dot x + l \dot \theta \cos\theta, -l\dot \theta \sin \theta]$ and you can determine $T$ from that.
You defined $\theta$ to be the angle with the vertical, in which case your expression for $V_{\text{gravitational}}$  is incorrect.  Please review it.  $V$ should be symmetrical around the vertical $V(\theta) = V(-\theta)$ and it should be the lowest when $\theta=0$.
To solve for period of small oscillations, retain the first significant  term w.r.t. displacement in both kinetic and potential energy. Hint: The resulting lagrangian would also describe the motion of a particle attached to 2 springs connected in series with spring constants $2k$ and $mg/l$.
Finally, the expression you wrote down for the period is not dimensionally consistent. While $m \over 2k$ is $s^2$, ${g \over l}$ is $s^{-2}$. Either you copied it incorrectly or the problem statement has a typo. $l$ and $g$ need to be switched.
