Question

A block with mass m =6.5 kg is hung from a vertical spring. When the mass hangs in equilibrium, the spring stretches x = 0.21 m. While at this equilibrium position, the mass is then given an initial push downward at v = 4 m/s. The block oscillates on the spring without friction. After t = 0.37 s what is the speed of the block?

Answer 1

Sought speed $v = v_0cos((\frac{g}{x})^{0.5}\tau) = 4cos2.55 = - 3.3 (m/s).$