Question

If the mass is displaced 0.270 m from equilibrium and released, what is its speed when it is 0.130 m from equilibrium?

Answer 1

The speed is 1.52 m.

Step-by-step explanation:

Given that,

Displacement =0.270 m

Distance = 0.130 m

Suppose a 0.321-kg mass is attached to a spring with a force constant of 13.3 N/m.

We need to calculate the angular velocity

Using formula of angular velocity

`\omega=\sqrt{(k)/(m)}`

Put the value into the formula

`\omega=\sqrt{(13.3)/(0.321)}`

`\omega=6.43\ rad/s`

We need to calculate the velocity

Using formula of velocity

`v=\omega√(A^2-x^2)`

Put the value into the formula

`v=6.43*√(0.270^2-0.130^2)`

`v=1.52\ m/s`

Hence, The speed is 1.52 m.