Answer:
465.6 N/m
Explanation:
We are given that
F=12 N
[tex]y(t)=4.50cos\left \{(19.5s^{-1}t-\frac{\pi}{8}\right \}[/tex]
We have to find the spring constant of the spring.
[tex]F=mg[/tex]
Where [tex]g=9.8 m/s^2[/tex]
Using the formula
[tex]12=m\times 9.8[/tex]
[tex]m=\frac{12}{9.8}[/tex]kg
Compare the given equation with
[tex]y(t)=Acos(\omega t-\phi)[/tex]
We get [tex]\omega=19.5[/tex]
[tex]k=m\omega^2[/tex]
Using the formula
Spring constant,[tex]k=\frac{12}{9.8}\times (19.5)^2=465.6 N/m[/tex]
