Spring system with oscillation: two different displacements

[awertag]

Homework Statement

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/s2

Homework Equations

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 cm


does 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!

 

[Rick88]

It's you again :P
I recognised your "delta" !For part a), how about using energies?

Set the initial potential energy, with h=\Deltax_extra and let mgh=1/2 mv².

R.

 

[awertag]

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

 

[Rick88]

do you know what the answer should be?

 

[awertag]

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

 

[Rick88]

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 \Deltax² = mgx + 1/2mv²

However, the velocity is maximum when the potential energy is 0.

(\Deltax is the displacement, x is the position along the axis. they are not the same thing)

 

[awertag]

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 :)

 

[Rick88]

It's a pleasure :)