A pendulum clock is mounted on an elevator, which starts from rest and goes up with constant acceleration $a_0$, where $\frac{a_0}{g}=\frac{2}{5}$. At a height h, the acceleration of the elevator reverses in direction, its magnitude remaining constant. After a time $\frac{\psi }{5}\sqrt{\frac{2h}{a_0}}$, from the start the clock will shown right time again. Find the value of $\psi$. (Given, $\sqrt{7}=2.6,\sqrt{3}=1.7and\sqrt{5}=2.2$)