Question

A 10.00 kg mass is attached to a 250N/m spring and set into vertical oscillation. When the mass is 0.50m above the equilibrium it is moving at speed 2.4 m/s. Find the amplitude of the oscillation

Answer 1

assuming the reference line to measure the height for gravitational potential energy lying at the equilibrium position

m = mass attached to the spring = 10.00 kg

k = spring constant of the spring = 250 N/m

h = height of the mass above the reference line or equilibrium position = 0.50 m

x = compression of the spring = 0.50 m

v = speed of mass = 2.4 m/s

A = maximum amplitude of the oscillation

v' = speed of mass at the maximum amplitude location = 0 m/s

using conservation of energy between the point where the speed is 2.4 m/s  and the highest point at which displacement is maximum from equilibrium

kinetic energy + spring potential energy + gravitational potential energy = kinetic energy at maximum amplitude + spring potential energy at maximum amplitude  + gravitational potential energy at maximum amplitude

(0.5) m v² + m g h + (0.5) k x² = (0.5) m v'² + m g A + (0.5) k A²

inserting the values

(0.5) (10) (2.4)² + (10) (9.8) (0.50) + (0.5) (250) (0.50)² = (0.5) (10) (0)² + (10) (9.8) A + (0.5) (250) A²

109.05 = (98) A + (125) A²

A = 0.62 m