Spring hanging from ceiling

1. Jul 19, 2012

EBBAzores

I have a question that's being bugging me around. This might be simple but I can't figure it out. If there's a spring hanging from the ceiling and we want to prove that there's a Simple Harmonic Oscilation then why don't we account for the gravitational force?

The sum of forces equals to mass times aceleration (ma) according to newton's 2nd law, so the sum of forces is the gravitional force plus the force by the spring (there should be vectors above the forces of course).

This is a differential equation since aceleration equals the second derivitive of position and to prove that there is a S.H.O. the equation must be
(d2x)/(dt2) + (K/m)x = 0

But in fact what I actually get is
(d2x)/(dt2) + (K/m)x + g = 0

I know that in this equation g is a constant but my question is is g relevant if we want to prove that there is a S.H.O?

And following that could anybody tell me how do we demonstrate that the period is in fact
T=2π√(m/K)

Thanks for your time

Last edited: Jul 19, 2012
2. Jul 19, 2012

tiny-tim

Welcome to PF!

Hi EBBAzores! Welcome to PF!

(have a pi: π and a square-root: √ and try using the X2 button just above the Reply box )
Hint: put y = x + gm/K

3. Jul 19, 2012

Staff: Mentor

Just do a change of variables with x=X-gm/k

tiny-tim got me!

4. Jul 19, 2012

EBBAzores

Ok one problem down, so i'll assume that g isn't relevant since I change variables for the S.H.O. but now how do we prove that the period is in fact the above one?

P.S.
Thanks for welcoming me in here =)

5. Jul 19, 2012

tiny-tim

uhh?

solve the equation!

6. Jul 19, 2012

EBBAzores

I've solved it. Now I get the actual S.H.O. equation wich is now
(d2y)/(dt2) + (K/m)y = 0

That part I understood but now I want to know why the period of oscillation is T=2π√(m/k)
how do we demonstrate that?

I can't figure it out sorry for wasting your time with this, maybe I'm really tired and thats why I cant figure it out

7. Jul 19, 2012

tiny-tim

ok, so y'' = -(K/m)y …

the solutions are … ?

8. Jul 19, 2012

EBBAzores

This might be a REALLY wild guess but where it goes xD

(d2y)/(dt2) = -(K/m)y

so that meand that

dt2 =(d2)y/(-k/m)y the y's dissapear so I'll assume now that dt is an aproximation to the period T and that the minus sign before k doesn't really interest us so we'll exclude it therefore

T2=(d2)/(k/m) and now

T=1/√(k/m) and to put it in angular references T=2π/√(k/m)

I don't know if this is right our am I getting dumber xp problaby I need some sleep

9. Jul 19, 2012

tiny-tim

(you're confusing the X2 button with the X2 button )
no they don't disappear!!

you can't possibly do that!
yes, get some sleep :zzz:

(then think about what solutions of shm you know)

10. Jul 19, 2012

HallsofIvy

You have been told repeatedly "solve the equation" but have not. Do you not know the general solution to y''= ay?

11. Jul 19, 2012

EBBAzores

Ok so here I am again ready to finish it already.

Since we know that y=Acos(ωt+$\phi$) so (d2y)/(dt2)=-Aω2cos(ωt+$\phi$) we can easily get that

-Aω2cos(ωt+$\phi$)= -(k/m)(Acos(ωt+$\phi$))

from that expression we obtain that ω2=k/m concluding that ω=√(k/m)

we also know that T=2π/ω=2π√(m/k)

I think that this time its really finished!