Question

a 0.400-kg mass is attached to a spring and set into simple harmonic motion. if its period is 1.7 seconds, what is the spring constant?

Answer 1

The spring constant is 5.46N/m.

Step-by-step explanation:

The period
`T` of the mass in SHM is related to the spring constant
`k` by the equation

`T = 2\pi \sqrt{(m)/(k) },`

using this we solve for
`k` to get:

`k = (4\pi ^2m)/(T^2).`

For our case we put in
`m = 0.400kg` and
`T = 1.7s`:

`k = (4\pi ^2(0.400kg))/((1.7s)^2).`

`\boxed{k = 5.46N/m}`

which is our spring constant.