# spring system with oscillation: two different displacements

1. The problem statement, all variables and given/known data
A 0.77 kg mass is attached to a vertical spring and is lowered until it reaches equilibrium at a distance x. The force constant of the spring is 220 N/m. The mass is then further displaced and released causing an oscillation with a maximum speed of 0.40 m/s. Find the following quantities related to the motion of the mass.
(c) the amplitude cm (d) The actual total force in the spring at the lowest position N (e) the maximum magnitude of the acceleration m/s22. Relevant equations
3. The attempt at a solution
I found the initial stretch distance x to be .0343m
and the period to be .3717 seconds
for the amplitude i wanted to do Fextra=k/\xextra where the k=spring constant;Fextra=force applied to further displace it;/\xextra=amplitude
however, i had too many unknowns and didn't know what to do next.
so then i tried doing average v = .2 m/s^2(avg v)(T) = d then d/4 because there are 4 amplitudes per period and get amp = .01858 m or 1.858 cmdoes it have to do with the trick of turning the spring horizontal and setting equilibrium as relaxed length w/zero spring energy? any good help will be greatly appreciated!!
It's you again
I recognised your quot;deltaquot; !For part a), how about using energies?
Set the initial potential energy, with h=xextra and let mgh=1/2 mv2.
R.
haha yess it's me again!! sorry i don't know how to do those fancy deltas..
well i had actually tried what you said as well, but it wasn't right either so i don't know, i'm out of ideas
do you know what the answer should be?
no unfortunately it's on webassign (i don't know if you're familiar?) but i know when i'm wrong, but do not know the actual answer
Ah, of course.
You do need to use energy conservation, but I told you the wrong thing.
total energy of system = potential energy + kinetic energy
1/2 k x2 = mgx + 1/2mv2
However, the velocity is maximum when the potential energy is 0.
(x is the displacement, x is the position along the axis. they are not the same thing)
ok thanks so much again!! i had actually done that for the initial stretch at first, but i didn't think to try it for the amplitude. you're saving my life here
It's a pleasure |