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

1.98 rad/s

Step-by-step 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 {(k)/(m)} = \sqrt{(g)/(x_0)}`

`\omega = \sqrt{(g)/(x_0)}\\\\\omega = \sqrt{(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