Answer:
0.04865 m
Explanation:
k = Spring Constant
m = Mass
d = Distance
g = Acceleration due to gravity = 9.81 m/s²
Angular frequency is given by
[tex]\omega=\sqrt{\dfrac{k}{m}}\\\Rightarrow \dfrac{k}{m}=\omega^2\\\Rightarrow \dfrac{k}{m}=14.2^2[/tex]
At equilibrium we have
[tex]kd=mg\\\Rightarrow d=\dfrac{mg}{k}\\\Rightarrow d=\dfrac{g}{\omega^2}\\\Rightarrow d=\dfrac{9.81}{14.2^2}\\\Rightarrow d=0.04865\ m[/tex]
The distance by which the spring stretches from its unstrained length is 0.04865 m
