Answer to Question #69979 in Mechanics | Relativity for Devi

Question

A spiral spring of natural length 400mm is suspended vertically from a fixed point with its upper end fixed. A mass of 0.2kg is suspended at rest from the lower end of the spring to increase its length to 460mm. The mass is then pulled down a further distance of 40mm and released from rest so it oscillates about the equilibrium.
a) calculate the stiffness constant
b)calculate the maximum tension in the spring

Answer 1

a) For the equilibrium:
kΔx=mg
The stiffness constant is
k=mg/Δx=((0.2)(9.8))/(0.46-0.4)=33 N/m.
b) The maximum tension in the spring is
T=kΔx^'=33(0.06+0.04)=3.3 N.