# Period, velocity, acceleration of a spring

1. Dec 3, 2005

### lilkrazyrae

A .390kg mass is suspended from a spring. A force of 1.00N stretches it an additional 35.0mm. Find the force constant of the spring, the period of the motion which results when the mass is released, the maximum acceleration, and the maximum velocity.

Ok so the force constant is F=ky therefore F/y=k and k=1.00/.035=28.6 N/m
The period is T=2pi(sqrt(m/k))
=2pi(sqrt(.390/28.6))
=.734m
That is pretty straightforward but the velocity and the acceleration is what I have a problem with, I don't know if 35.0mm is the amplitude or if you have to find the amplitude.

I used v=(sqrt(k/m))(A)
=(sqrt(28.6/.390))(.0350)
=.300m/s

and for acceleration a=(k/m)(A)
=(28.6/.390)(.0350)
=2.57m/s^2