Question

A spring stretches 0.1 m when a 0.5 kg mass is hung from it. The spring is now oriented horizontally with the mass attached to one end. If it is stretched 0.2 m from the equilibrium length, determine the maximum velocity of the mass.

Answer 1

Answer:

1.98 rad/s

Explanation:

Given that a spring stretches at a distance $x_0$  0.1 m when a mass of 0.5 kg is attached to it.

The angular velocity $w = \sqrt {\frac{k}{m}} = \sqrt{\frac{g}{x_0}}$

$\omega = \sqrt{\frac{g}{x_0}}\\\\\omega = \sqrt{\frac{9.81}{0.1}}\\\\\omega = 9.9 \ rad/s$

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