# Problem using lagrangian

I wanted to approach this problem analytically, but i got stuck midway.

PROBLEM: Consider the rod-spring mechanism shown in the figure. The rods are massless, and are hinged at the vertices. A mass M is hung from the lowest vertex, and a spring of spring constant k is attached across the horizontal line BD. In equilibrium, the angle between spring and each rod is 45 deg. If the length of each rod is b,find the angular frequency of small oscillation of the system.

MY ATTEMPT: Let the angle between the diagonal through the lowest vertex and the corresponding rod be $\phi$. If the distance moved down by the mass is $x$ , then clearly: $$x = (2b\cos(\phi) - \sqrt2 b)\tag{1}$$

$$\dot x = -2b\sin(\phi)\tag{2}$$

writing the lagrangian $L = T- V$ (kinetic - potential):

$l$ = equilibrium spring length.

$$L = \frac{1}{2}M(4b^2\sin^2(\phi)) - \frac{1}{2}k(2b\sin(\phi)-l)^2 - mg(2b\cos(\phi) - \sqrt2b)$$

Using the Euler lagrange equation, we have directly:

$$m(4b^2\sin^2(\phi))\ddot\phi = -k(2b\cos(\phi))(2b\sin(\phi) - l) + mg2b\sin(\phi)$$

This is where I'm stuck. I cannot turn this completely into $x$ and its derivatives, to derive the frequency. Even on using equations (1) and (2), the result is no better. Any help regarding the last bits of simplifications would be appreciated.

• Check your equations, (2) in particular. The angle $\phi$ is not time-independent. – Troy Oct 9 '16 at 14:23

• As pointed out in another answer, $\dot{x}=-2b \sin \phi \dot{\phi}$, but you probably just forgot to write a piece since you correctly have $\ddot{\phi}$ in the equations of motion.
• Since at equilibrium $\phi=\pi/4$, one has $l=b\sqrt 2$
• Since you need to find small oscillations you can Taylor expand around the equilibrium position $\phi=\pi/4$. This gives $$\cos\phi\approx{\sqrt2\over2}(1-\epsilon+\dots)$$ $$\sin\phi\approx{\sqrt2\over2}(1+\epsilon+\dots)$$ Note in particular how this simplifies your expressions for the displacements, since the constant parts cancel each other.