Oscillation of an object suspended on an unstretched horizontal elastic string The physics concepts here are: small oscillations around equilibrium, elastic strings.
Let me show you my approach to the following problem. But first the problem description from the problem book.
Unstretched elastic string is suspended horizontally. Its stiffness is $k=16N/m$. At the centre of the string, a weight of mass $m=0.024kg$ is attached and then lightly pulled downwards. Find the period of small oscillations, given that angle between horizontal and the string is small.

Before the weight is pulled downward, let $\alpha$ be the angle from the string to the horizontal and x the total elongation of the string. From force equilibrium I got: $mg=2kx\cdot sin(\alpha)$.
After the motion has begun, I got: $ma=mg-2kX\cdot sin(\beta)$ (positive downwards). Where $X$ is the new total elongation of the string and $\beta$ is the new angle.
Let $L$ be the unstretched length of the string. This allows to write two equations:
$sin(\alpha)=\frac{2y}{\sqrt{4y^2+L^2}}$ and
$sin(\beta)=\frac{2Y}{\sqrt{4Y^2+L^2}}$
Where $y$ is the distance by which weight drops from the horizontal line in the equilibrium. Furthermore, $Y$ is the distance from the horizontal line during oscillation.
One can also say that: $y^2+\frac{L^2}{4}=\frac{(L+x)^2}{4}$ and $Y^2+\frac{L^2}{4}=\frac{(L+X)^2}{4}$
I then got rid of $x, X, sin(\alpha)$ and $sin(\beta)$ from first two equations.
Combining these two gives:
$ma=4k\cdot[y-Y+L(\frac{Y}{\sqrt{4Y^2+L^2}}-\frac{y}{\sqrt{4y^2+L^2}})]$
The $4k\cdot (y-Y)$ part looks good! Some signs of SHM equation appear!
Linearization around $Y-y=0$ gives:
$ma=4k(y-Y)\cdot[1-\frac{L^3}{\sqrt{4y^2+L^2}^3}]$
Well, I showed that ma is proportional to $(y-Y)$ or $(Y-y)$. But this doesn't give me the answer from the author (given only using $m$ and $k$...). Which I give below. Can you help me with identifying what went wrong? Or maybe should I use a different approach?
I couldn't find similar problems elsewhere, I only found a Polish forum thread from 15 years ago when some physicists tried to solved it with no success.

 A: 
I got this solution :
the kinetic energy is
$$T=\frac 12\,m\,v^2\quad,
 v=\dot y=\frac L2\frac{d}{dt}\,\tan(\alpha)$$
the potential energy is:
$$U=m\,g\,y+k\,s_l^2=m\,g\,\frac L2\,\tan(\alpha)+
k\,s_l^2$$
where $~s_l~$  is the "spring length"
$$s_l^2=\left(\frac L2\right)^2+y^2= 
\left(\frac L2\right)^2+\left(\frac L2\,\tan(\alpha)\right)^2$$
from here with EL the equation of motion
$$\ddot\alpha+f(\alpha~,~\dot\alpha)=0\tag 1$$
linearized Eq.1 $~\alpha \ll~$
$$\ddot\alpha+2\,{\frac {g}{L}}+2\,{\frac {k\alpha}{m}}+2\,\alpha\,{\dot\alpha }^{2}
=0$$
thus $~\omega^2=\frac {2\,k}{m}~$
A: Use symmetry to consider only half the problem: a mass $\frac{m}{2}$ constrained to move vertically (with slight deflection $y$ measured downward), suspended from a nearly horizontal string of length $L$ and at slight downward angle $\theta$ (thus, $\frac{y}{L}=\tan\theta\approx\theta$).
The stretching of the string is
$$\Delta L=\sqrt{L^2+y^2}-L\approx \frac{y^2}{2L},$$
and so the axial string force is
$$k\Delta L=\frac{ky^2}{2L}$$
with upward component multiplied by $\sin\theta$, or
$$\frac{ky^2}{2L}\sin\theta\approx \frac{ky^2\theta}{2L}\approx \frac{ky^3}{2L^2}.$$
Linearized for small displacements around position $y_0$, this is
$$\frac{ky^3}{2L^2}\approx \frac{ky_0^3}{2L^2}+3\frac{ky_0^2}{2L^2}y,$$
so the equation of motion, summing upward and downward forces, is
$$\frac{m}{2}\ddot y=-\frac{ky_0^3}{2L^2}-\frac{3ky_0^2}{2L^2}y+\frac{mg}{2}.$$
But the equilibrium position $y_0$ is where the weight counteracts the string tension at rest, or
$$\frac{ky_0^3}{2L^2}=\frac{mg}{2},$$
so the linearized equation of motion is
$$\frac{m}{2}\ddot y=-\frac{3ky_0^2}{2L^2}y,$$
corresponding to a natural frequency of $\omega=\sqrt{\frac{3ky_0^2}{mL^2}}$ and a period of $$T=2\pi\sqrt{\frac{mL^2}{3ky_0^2}}=2\pi\sqrt{\frac{y_0}{3g}}=2\pi\left(\frac{mL^2}{27kg^2}\right)^{1/6}.$$
I'm not seeing how to usefully simplify this further.
