Question about simple harmonic motion 
A massive bird lands on a taut light horizontal wire. You observe that after landing the bird oscillates up and down with approximately simple harmonic motion with period $T$.
You also notice that during the initial oscillations the maximum height the bird reaches above its position at landing is equal to the distance below its landing position at which the bird finally comes to rest, when all oscillations have stopped.
In terms of only $T$ and $g$, the acceleration due to gravity, estimate the magnitude of the vertical velocity $v_0$ that the bird had at the instant of landing on the wire.

So, to attack this problem by energy conservation I use
\begin{equation}
\frac{1}{2}mv_0^2=mgA
\end{equation}
Where now $v_0 = \omega A$. So
\begin{equation}
\frac{1}{2}\omega^2 A^2=gA \Rightarrow A=\frac{2g}{\omega^2}=\frac{2g}{\frac{4\pi^2}{T^2}}=\frac{2gT^2}{4\pi^2}
\end{equation}
So
\begin{equation}
v_0 =  \omega A = \frac{2\pi}{T} \frac{2gT^2}{4\pi^2} = \frac{gT}{\pi}
\end{equation}
But the answer is
\begin{equation}
v_0 = \frac{\sqrt{3}gT}{2\pi}
\end{equation}
What am I doing wrong?
 A: The second line is the clincher in the problem. Once you understand its significance, the problem becomes very easy. When the bird comes to rest permanently on the wire, it reaches the equilibrium position which is the mean position of SHM. If this is A units below landing position, and the maximum height reached is A units above landing position, then amplitude is 2A. This means that at the landing position the displacement is half of maximum amplitude and hence, the velocity at this point( which is required in the problem) is  √3/2 times the maximum velocity.( This comes from the fact that velocity and amplitude are cosine and sine functions respectively in case of SHM).
Now, you can get the maximum velocity by equating $mω^2A=mg$ since the bird is in equilibrium at A units below the landing position. Remember, the amplitude is 2A. The maximum velocity that you get from this is the same as the answer that you have obtained in your own attempt, $gT/π$. The initial velocity occurs at half amplitude and is √3/2 times the maximum velocity i.e √3/2 $gT/π$.
A: Using conservation of energy comparing the point that the point that the bird arrives and the maximum potential energy stored in the wire:
1/2 * m * v^2 + m * g * (3h) = 1/2 * k * (3h)^2.
Solving with mg = kh with the force relation and the period T = 2 * pi * sqrt(m/k) would give you the correct answer.
It helps to draw a diagram for the conservation of energy. The equilibrium is 2h below the maximum height, and the amplitude is 2h, as the answer above indicates too. I did an extension 2h downwards from equilibrium as my 0 potential energy level. Then the potential at the landing point is mg(3h). The stored energy in the wire uses the displacement from the 0 point to the landing point, which is a length of 3h, so the potential is 1/2k(3h)^2.
