Answer:
200 N/m
20 rad/s
0.31415 seconds
3.18309 Hz
Explanation:
m = Mass of glider = 0.5 kg
x = Displacement of spring
F = Force on spring = 6 N
From Hooke's law we have relation
[tex]F=kx\\\Rightarrow k=\frac{F}{x}\\\Rightarrow k=\frac{6}{0.03}\\\Rightarrow k=200\ N/m[/tex]
The spring constant is 200 N/m
Angular frequency is given by
[tex]\omega=\sqrt{\frac{k}{m}}\\\Rightarrow \omega=\sqrt{\frac{200}{0.5}}\\\Rightarrow \omega=20\ rad/s[/tex]
The angular frequency is 20 rad/s
Frequency is given by
[tex]f=\frac{\omega}{2\pi}\\\Rightarrow f=\frac{20}{2\pi}\\\Rightarrow f=3.18309\ Hz[/tex]
The frequency is 3.18309 Hz
Time period is given by
[tex]T=\frac{1}{f}\\\Rightarrow T=\frac{1}{3.18309}\\\Rightarrow T=0.31415\ s[/tex]
The time period is 0.31415 seconds