Period, velocity, acceleration of a spring

[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

Are these methods right or should a different amplitude be used? Please help asap!

 

[Doc Al]

I didn't check your arithmetic, but your methods look good to me. And yes, you are using the correct amplitude. (When the mass is hanging undisturbed, it rests at the equilibrium position. When the mass is then given a displacement and released, it will oscillate about the equilibrium position with an amplitude equal to the initial displacement.)

 

[quicknote]

Your calculations for the force constant and period appear to be correct. For the velocity and acceleration, the amplitude should be the maximum displacement from equilibrium, which in this case is 35.0mm. So your calculations for velocity and acceleration are correct. The maximum velocity would occur when the mass is at its equilibrium position and the maximum acceleration would occur at the extremes of the motion. It is important to note that these values are only valid for the simple harmonic motion of the mass-spring system and may change if other external forces are present.