- #1
brendan3eb
- 54
- 0
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?
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?