Answer:
1.98 rad/s
Explanation:
Given that a spring stretches at a distance [tex]x_0[/tex] 0.1 m when a mass of 0.5 kg is attached to it.
The angular velocity [tex]w = \sqrt {\frac{k}{m}} = \sqrt{\frac{g}{x_0}}[/tex]
[tex]\omega = \sqrt{\frac{g}{x_0}}\\\\\omega = \sqrt{\frac{9.81}{0.1}}\\\\\omega = 9.9 \ rad/s[/tex]
If it is stretched 0.2 m from the equilibrium length; we have:
The maximum velocity v = 0.2 × 9.9
v = 1.98 rad/s
Therefore; the maximum velocity of the mass = 1.98 rad/s
