Hooke's Law and Simple Harmonic Motion

[Soaring Crane]

We did an experiment with a vertical spring-mass system. Here is an example of data collected:
Mass------------T (period)
200 g-----------.3 s
400g------------.5 s

My question is why does the period increase as the mass increases?

I know that the amp. and T are independent of each other, but how do you show this mathematically? I now it’s easy to show visually, but are there any ties in deriving this, along with the relationship that T = 1/sqrt(x)?

Then, my teacher mentioned some of the following derivisions. I understand a majority of it, but I don’t know how these two parts directly correspond with each other. Does this look familiar to anyone? I don’t quite follow the M_efficiency line, but I do get the calculus.
1st Part
F = -kx, where T = 2*pi*sqrt(M/k) and Mg = kL (L=x)
M*[(d^2*x)/(dt^2)] = -(k/M)x (second derivative)
(d^2*x)/(dt^2) = -w^2*x, where w (or omega) = 2*pi*f
x(t) = Asin(wt + phi)
This is to show that T is proportional to k, I guess.



2nd Part
M_efficiency = M + (M_spring/3)

k_spring = (1/2) integrate(dx/dt)^2 dm

Proportion
v/L = (dx/dt)/x
mass density p = dm/dx, where p = M_spring/L, where L is length
k_spring = (1/2)*definite integral(xv/L)^2 pdx L to 0
k_spring = ½*[(pL^3v^2)/(3L^2)] = ½*(M_spring/3)v^2

and somehow KE = ½[M + (M_spring/3)]*v^2. How do you get to this point?

Thanks for ANY help.

 

[futb0l]

I don't know about M efficiency or whatever.
But the period proportional to the mass is pretty simple to explain.

<br /> F = m\ddot{x} = -kx<br />

...

<br /> \ddot{x} + \frac{k}{m}x = 0\ \ \ \ (1)<br />

now you know that x itself is a function and can be modeled as:

<br /> x = A \cos ( \omega t + \phi )<br />

taking the first derivative gives you:

<br /> \dot{x} = -A\omega \cos ( \omega t + \phi )<br />

the second derivative gives you:

<br /> \ddot{x} = -A\omega ^2 cos ( \omega t + \phi )<br />

notice that's also equal to:

<br /> \ddot{x} = - \omega ^2 x<br />

substituting this in equation (1) gives you:

<br /> -\omega ^2x + \frac{k}{m}x = 0<br />

Solving for omega gives you

<br /> \omega = \sqrt{\frac{k}{m}}<br />

so the period will equal to:

<br /> T = 2\pi \sqrt{\frac{m}{k}}<br />

 

[Nylex]

taking the first derivative gives you:

<br /> \dot{x} = -A\omega \cos ( \omega t + \phi )<br />

Should be \dot{x} = -A\omega \sin (\omega t + \phi).

 

[Soaring Crane]

Could you explain the first line before Equation 1? How did you get from that line to Equ(1)?

Thanks.

 

[Pyrrhus]

He got that line by simply applying Newton's 2nd Law (For Constant Mass).

\sum_{i=1}^{n} \vec{F}_{i} = m \vec{a}

 

[Soaring Crane]

Is there a relationship between displacement and period? How do I go about showing this mathematically?

Thanks.