Answer:
0.75 kg
Explanation:
c = Damping coefficient = 3 kg/s
x = Displacement of spring = 0.5 m
F = Force = 1.5 N
From Hooke's law we get
[tex]F=kx\\\Rightarrow k=\dfrac{F}{x}\\\Rightarrow k=\dfrac{1.5}{0.5}\\\Rightarrow k=3\ \text{N/m}[/tex]
In the case of critical damping we have the relation
[tex]c^2-4mk=0\\\Rightarrow m=\dfrac{c^2}{4k}\\\Rightarrow m=\dfrac{3^2}{4\times 3}\\\Rightarrow m=0.75\ \text{kg}[/tex]
Mass that would produce critical damping is 0.75 kg.
