$$\begin{gathered} At a general point \\ -2T\sin \theta = ma (\theta is measured from horizontal) \\ \Rightarrow a = -2Mg\theta /m = -2Mgy/ml \\ compare with a = -{\omega }^{2}y \\ \omega =\sqrt{\frac{2Mg}{ml}} \left(1\right) \\ Now for max displacement of small m \\ mg(h+y) = 2Mgx and x = y^2/2l \\ on solving this quadratic eq. for y/l we get \\ \frac{y}{l}-\frac{m}{2M}=\sqrt{\frac{m}{M}\frac{h}{l}}\Rightarrow \frac{y}{l}=\sqrt{\frac{m}{M}\frac{h}{l}} (given that \frac{m}{M} <<\frac{h}{l}) \\ so x = \frac{m}{2M}h \left(2\right) \\ so v_{max} = x\omega \\ now solve \end{gathered}$$