For the two-dimensional Riemannian space with line element $$(ds)^2 = (a + b \cos φ)^2(dθ)^2 + b^2(dφ)^2 ,\qquad (a, b > 0)\tag{1}$$ Where $φ$ and $\theta$ are functions of a parameter $u$.
I had to find the Euler-Lagrange equations for the geodesics. Hopefully, I have calculated them correctly below:
$$\frac{d}{du}[2(a+b\cosφ)^2\dot\theta]=0\tag{2}$$ $$\frac{d}{du}[2b^2\dotφ]=-2b\sinφ(a+b\cos φ)\dot{\theta}^2.\tag{3}$$
My question concerns when you consider the limit when $|φ|\ll 1$ and $|\theta| \ll 1$, I am supposed to prove that in this limit, $$(a+b)^2\dot{\theta}=k, \qquad k=constant,\tag{4}$$ and $φ(u)$ oscillates at a frequency $$\omega=\sqrt{k^2/b(a+b)^3}.\tag{5}$$
But unfortunately I do not know how to prove that, any help would be greatly appreciated.
Hints:
The Lagrangian reads $$ L(\theta,\varphi,\dot{\theta},\dot{\varphi})~=~(a+b\cos\varphi)^2\dot{\theta}^2+b^2\dot{\varphi}^2. \tag{A}$$
The $\theta$ variable is cyclic, so that the corresponding momentum $$ p_{\theta} ~:=~\frac{\partial L}{\partial \dot{\theta}} ~=~2(a+b\cos\varphi)\dot{\theta}~\approx~2(a+b)\dot{\theta}~\stackrel{(4)}{=:}~\frac{2k}{a+b} \tag{B}$$ is a constant of motion.
The EL eq. for $\varphi$ reads $$ 2b^2\ddot{\varphi}~=~\frac{\partial L}{\partial \varphi} ~=~-2\sin\varphi (a+b\cos\varphi)\dot{\theta}^2~\stackrel{(B)}{=}~-\frac{p_{\theta}^2\sin\varphi}{2(a+b\cos\varphi)}~\stackrel{(B)}{\approx}~-\frac{2k^2\varphi}{(a+b)^3}, \tag{C}$$ which is the eq. for a SHO, so that OP's last eq. (5) follows.