Problem with a physical equation When I solved a physics problem, I found a little problem of math calculation at :
$$E=E_{c}+E_{p}= \frac{ms'^2}{2}+ \frac{mgs^2}{8R} ~~~~\text{(1)}$$
(this is the equation where I met the problem for solving it). The problem said that are no neconservative force, so our $E$ will be zero. 
I derivate the equation $\text{(1)}$ and I obtained:
$$s''+ \frac{gs}{4R} =0 ~~~~\text{(2)} $$ 
Now, the problem asks for pulsation. I don't know how to find angular-frequency [rad/s] beginning from the equation $\text{(2)}$. Any help or hint will be receive very well !!
 A: Just from glancing at your problem, it looks like the differential equation will have periodic solutions of the form 
$$
s (t) = A \sin \sqrt{ \frac{g}{4R} } t + B \cos \sqrt{ \frac{g}{4R} } t .
$$
In which case, the frequency of oscillation is given by the quantity $\sqrt{ \frac{g}{4R} }$. I don't know if that answers your question.
A: Hint:
Any equation of type $y''+a^2y=0$ can be solved by testing $y=e^{rx}$ as a solution which will result in $r^2+a^2=0$. 
$s''+ {\sqrt{\frac{g}{4R}}}^2s =s''+ a^2s=0\implies s=C_1exp(iat)+C_1exp(-iat)\implies s (t) = A \cos \sqrt{ \frac{g}{4R} } t + B \sin \sqrt{ \frac{g}{4R} } t $.
angular-frequency [rad/s] means $a$ in $s''+ a^2s=0$ which is $\sqrt \frac{g}{4R}$. You can check its dimensional also.    
A: Let $ \theta = \dfrac {s}{R},$ be the angle of pendulum thread with respect to vertical direction.
$$s''+ \frac{gs}{4R} =0 ~~~~\text{(2)} $$ 
$$\theta ''+ \frac{g \theta }{4R} =0 ~~~~\text{(3)} $$ 
This is simple harmonic angular motion in place of the linear motion. (It is a linear vibration model which approximates $\sin\theta $ as $\theta $)
An odd solution is 
$$ \theta = \alpha \sin ( \omega \,\theta) ~~~~\text{(4)} $$ where
$$ \alpha = \theta_{max} ,\omega^2 = \frac{g}{4 R} , \omega = 2 \pi f = \frac{2 \pi}{T}, ~~~~\text{(5)}  $$
where $ T$ is the time period, $f$ is frequency and $\omega$, the angular velocity just as it is with rectilinear motion.
