Answer:
Spring constant, k = 2.304 N/m
Explanation:
It is given that,
Mass of spring, m = 0.7 kg
Angular frequency, [tex]\omega=2.4\ rad/s[/tex]
We need to find the spring constant of the spring. It is a case of SHM. Its angular frequency is given by :
[tex]\omega=\sqrt{\dfrac{k}{m}}[/tex]
[tex]k=\omega^2\times m[/tex]
[tex]k=(2.4)^2\times 0.4\ kg[/tex]
k = 2.304 N/m
So, the spring constant of the spring is 2.304 N/m. Hence, this is the required solution.
