Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Hooke's Law and Simple Harmonic Motion

  1. Dec 11, 2004 #1
    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.
     
  2. jcsd
  3. Dec 11, 2004 #2
    I dont know about M efficiency or whatever.
    But the period proportional to the mass is pretty simple to explain.

    [tex]
    F = m\ddot{x} = -kx
    [/tex]

    ...

    [tex]
    \ddot{x} + \frac{k}{m}x = 0\ \ \ \ (1)
    [/tex]

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

    [tex]
    x = A \cos ( \omega t + \phi )
    [/tex]

    taking the first derivative gives you:

    [tex]
    \dot{x} = -A\omega \cos ( \omega t + \phi )
    [/tex]

    the second derivative gives you:

    [tex]
    \ddot{x} = -A\omega ^2 cos ( \omega t + \phi )
    [/tex]

    notice that's also equal to:

    [tex]
    \ddot{x} = - \omega ^2 x
    [/tex]

    substituting this in equation (1) gives you:

    [tex]
    -\omega ^2x + \frac{k}{m}x = 0
    [/tex]

    Solving for omega gives you

    [tex]
    \omega = \sqrt{\frac{k}{m}}
    [/tex]

    so the period will equal to:

    [tex]
    T = 2\pi \sqrt{\frac{m}{k}}
    [/tex]
     
  4. Dec 12, 2004 #3
    Should be [tex]\dot{x} = -A\omega \sin (\omega t + \phi)[/tex].
     
  5. Dec 12, 2004 #4
    Could you explain the first line before Equation 1? How did you get from that line to Equ(1)?

    Thanks.
     
  6. Dec 12, 2004 #5

    Pyrrhus

    User Avatar
    Homework Helper

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

    [tex] \sum_{i=1}^{n} \vec{F}_{i} = m \vec{a} [/tex]
     
  7. Dec 12, 2004 #6
    Is there a relationship between displacement and period? How do I go about showing this mathematically?

    Thanks.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook