Question

5. A spring of spring constant = 8.75/ is hung vertically from a rigid support. A mass
0.500 is placed on the end of the spring and supported by hand at a point so that the
displacement of the spring is 0.250 . The mass is suddenly released and allowed to fall. At
the lowest position of the mass what is the displacement of the spring from its equilibrium
position? (Hint—Apply Equation 5 with 1 = 0.250 and 2 the unknown. This will lead to
a quadratic equation with one of the solutions the unknown 2, and the other solution the
original 0.250 displacement.) Show your work

Answer 1

Answer:

0.87 m

Explanation:

Energy is conserved, so the change in potential energy must equal the change in elastic energy, resulting in Equation 5:

mg (x₂ − x₁) = ½ k (x₂² − x₁²)

Given that m = 0.500, g = 9.8, x₁ = 0.250, and k = 8.75:

(0.500) (9.8) (x₂ − 0.250) = ½ (8.75) (x₂² − (0.250)²)

4.9 (x₂ − 0.250) = 4.375 (x₂² − 0.0625)

4.9 x₂ − 1.225 = 4.375 x₂² − 0.273

0 = 4.375 x₂² − 4.9 x₂ + 0.952

Solve with quadratic formula:

x = [ -b ± √(b² − 4ac) ] / 2a

x₂ = [ -(-4.9) ± √((-4.9)² − 4(4.375)(0.952)) ] / 2(4.375)

x₂ = (4.9 ± 2.71) / 8.75

x₂ = 0.25 or 0.87

At the lowest position of the mass, its displacement from the equilibrium position is 0.87 m.