Question

A 2.00 kg object is attached to a spring and placed on frictionless, horizontal surface. Ahorizontal force of 18.0 N is required to hold the object at rest when it is pulled 0.250 mfrom its equilibrium position. What is the angular frequency, in rad/s, of the oscillation?

Answer 1

6 rad/s

Step-by-step explanation:

In a spring the angular frequency is calculated as follows:

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

where
`\omega` is the angular frequency,
`m` is the mass of the object in this case
`m=2kg`, and
`k` is the constant of the spring.

To calculate the angular frequency, first we need to find the constant
`k` which is calculated as follows:

`k=(F)/(x)`

Where
`F` is the force:
`F=18N`, and
`x` is the distance from the equilibrium position:
`x=0.25m`.

Thus the spring constant:

`k=(18N)/(0.25m)`

`k=72N/m`

And now we do have everything necessary to calculate the angular frequency:

`\omega=\sqrt{(k)/(m) }=\sqrt{(72N/m)/(2kg) }=√(36)`

`\omega=6rad/s`

the angular frequency of the oscillation is 6 rad/s