Given : A = 0.25m, g = π2 = 10 m s2
During vertical oscillations, the acceleration is maximum at the turning points at the top and bottom. The block will just lose contact with the piston when its apparent weight is zero at the top,
i.e., when its acceleration is amax = g, downwards.
|amax| = ω2A = 4π2 η2max A = g
∴ η2max = \(\sqrt{\frac{g}{4π^2A}}\) = \(\sqrt{\frac{10}{4×10×0.25}}\) = 1 Hz
This gives the required frequency of the piston.
E = ½ mω2A2 = ½ m(4π2 η2)A2
∴ E/m = 2p2 h2A2 = 2 x 10 x 12 x (1/4)2 = 1.25 J/kg