Trick question.
Assuming that when stretched/compressed the spring is not let to it's own agenda and being held stretched/compressed by an outside force then there is no acceleration in any of the three states. There is potential energy, but if the spring is not dynamic in those states then there can't be acceleration.
EDIT: Rereading, the question asks "state(s)" and "maximum acceleration." With that wording, I would think the compressed and extended options are correct because there can not be any acceleration when the spring is at x=0m.
Stubbee wrote:Oscillating spring ?
The relaxed state is the fastest acceleration point.
Force is applied from either the extended or compressed states and is continuously applied until the spring reached the relaxed state. At which time the force from the opposite state starts to slow it down.
So from x meters until the relaxed state there is a continuous force i.e. continuous increase in acceleration until it passes the relaxed state where the opposite force starts to decelerate.Agreed, though under different circumstances. If after being compressed the spring is let go, F=-kx says yes the force exerted is constant therefor in a closed system acceleration is constant as well.
However, your interpretation of that confuses me. At 0m for x, there would be no force and no acceleration.
Is there anything more to this problem, a diagram or further explanation? If not I can't get much more out of it.EDIT2: Thinking about it again, assume the compression of a spring of a spring would be -x. Under compression, -k*-x=positive force, positive acceleration. -k*x=negative acceleration. At 0x, there would be no force. Under that system, the only positive acceleration would be when the spring is under compression. There would be acceleration when it is expanded however a positive acceleration is greater then a negative acceleration because acceleration is a vector.
Last edited by Defiance (2007-12-09 23:59:50)
