How Does Gravity Affect Equations in Vertical SHM Problems?

[brendan3eb]

I have been doing quite a few SHM problems, and I just have a few questions in general. A lot of questions evolved from one particular problem type: A mass attached to the end of a vertical spring of spring constant k.

My questions:
1. How can we prove that we can use the equation w=(k/m)^(1/2) for this problem. Normally, you can just go:
ma=-ky
a=-k/m * y
a=-w^2*y
y's cancel out
w=(k/m)^(1/2)
but in this case you should have to account for the mg force, but in most solutions, I do not seem mg accounted for?

In one problem, I was asked to solve for the maximum amplitude the shm could have in order to not surpass a certain acceleration. Once again, all answers were along the lines:
ma=-kA
mg=-kA
A=-mg/k
Once again, how can you neglect the mg force?

My only idea is that since we determine the equilibrium point for most of these problems at the beginning - the point where the spring force matches the gravitational force - that they treat this equilibrium point like the spring's equilibrium point and can somehow, magically, neglect the spring force?

 

[brendan3eb]

I have been doing problems for the last two hours, and still haven't really gotten much further on figuring this out..

 

[kamerling]

you didn't prove that you could use w=(k/m)^(1/2). you have a second order differential equation y'' = -k/m * y
To solve this just try y = a sin (b *t) as a solution and then find out what a and b are.

if you would include an mg force then your new differential equation would become

y'' = -k/m * y - mg. try to prove that if y=F(t) is a solution of the first differential equation, that y = F(t) - mg/k is a solution of the second one.

 

[rock.freak667]

When the spring is hanging vertically in equilibrium: Tension =Weight i.e ke=mg

When you displace it a small distance,x, Resultant force,F =mg-k(e+x)...use F=ma now.