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Hey guys. This was a homework problem (I could never figure it out) and I asked my instructor for help as well as researching on the Internet, but still no dice. It seems basic enough but it bothers me that I still don't know how to solve it.
Show that the period of the simple harmonic motion of a mass hanging from a spring is [tex]2 \pi \sqrt{l/g}[/tex], where [tex]l[/tex] denotes the distance beyond the spring's natural length when the mass is at equilibrium.
I figured the differential equation we should solve is a simple mass-on-a-spring case but with the external force being mass-gravity. That is:
[tex]my'' + ky = mg[/tex].
The general solution looks something like:
[tex]c_1 cos(\sqrt{\frac{k}{m}}t) + c_2 sin(\sqrt{\frac{k}{m}}t) + y_p[/tex]
For now, let's say:
[tex]\omega := \sqrt{\frac{k}{m}}[/tex]
Now we need to solve for [tex]y_p[/tex].
I did this in the two ways I know how, that being method of undetermined coefficients and variation of parameters. Both of them ended in some pretty nonsensical solutions (0 = mg??).
For the sake of comparison, after using the method of variation of parameters, I ended up with this particular solution and resulting position function:
[tex]y_p = \frac{g}{\omega(1+\omega)}[/tex]
(I know that can't be right since it's a constant.)
[tex]y(t) = (L - \frac{g}{\omega(1+\omega)}) cos(\omega t) + \frac{g}{\omega(1+\omega)}[/tex]
Where L is the natural length of the spring (I used the initial conditions y(0) = L, y'(0) = 0).
Anyone want to take a whack at it or point me to where someone already has?
Homework Statement
Show that the period of the simple harmonic motion of a mass hanging from a spring is [tex]2 \pi \sqrt{l/g}[/tex], where [tex]l[/tex] denotes the distance beyond the spring's natural length when the mass is at equilibrium.
Homework Equations
I figured the differential equation we should solve is a simple mass-on-a-spring case but with the external force being mass-gravity. That is:
[tex]my'' + ky = mg[/tex].
The Attempt at a Solution
The general solution looks something like:
[tex]c_1 cos(\sqrt{\frac{k}{m}}t) + c_2 sin(\sqrt{\frac{k}{m}}t) + y_p[/tex]
For now, let's say:
[tex]\omega := \sqrt{\frac{k}{m}}[/tex]
Now we need to solve for [tex]y_p[/tex].
I did this in the two ways I know how, that being method of undetermined coefficients and variation of parameters. Both of them ended in some pretty nonsensical solutions (0 = mg??).
For the sake of comparison, after using the method of variation of parameters, I ended up with this particular solution and resulting position function:
[tex]y_p = \frac{g}{\omega(1+\omega)}[/tex]
(I know that can't be right since it's a constant.)
[tex]y(t) = (L - \frac{g}{\omega(1+\omega)}) cos(\omega t) + \frac{g}{\omega(1+\omega)}[/tex]
Where L is the natural length of the spring (I used the initial conditions y(0) = L, y'(0) = 0).
Anyone want to take a whack at it or point me to where someone already has?