1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Damped simple harmonic motion question

  1. Oct 18, 2009 #1
    1. The problem statement, all variables and given/known data

    The frequency fd of a damped oscillator is 100 Hz, and the ratio of the amplitudes of two successive maxima is one half. What is the undamped frequency f0 of this oscillator?


    2. Relevant equations
    this is the equation in my textbook for the position at time t of an underdamped harmonic oscillator:
    x(t) = e^-(yt) * A cos (wd + ϕ)

    where y (really supposed to be gamma) is a constant that affects how quickly the oscillator is damped and w (really supposed to be omega) is the angular velocity and ϕ is just an initial angle

    i chose the underdamped equation because i believe there can not be any maxima if it is overdamped or critically damped.

    Td = 2pi / wd = 2pi / sqrt(Wo^2 - y^2) = 0.01 (Td is the damped period, wd is damped angular velocity, wo is undamped angular velocity)

    and the only other equations i used are T = 2π/w and T = 1/f

    where w is once again the angular velocity

    3. The attempt at a solution
    one maxima occurs at t=0 and the next one at t=T. first i will calculate wd for the position equations.

    Td = 1/fd = 1/100 = 0.01

    Td = 2pi/w
    0.01 = 2pi/wd
    wd = 2pi/0.01 = 200pi

    x(t) = e^-(yt) * A* cos(wd * t)

    x(0) = e^-(y * 0) * A * cos(200pi * 0)) = A

    so the first maxima is just an amplitude, which makes sense. i will now find the height of the next maxima, which occurs when t = T

    x(T) = x(0.01) = e^-(0.01y) * (Acos(200pi)) = A*e^-(0.01y)

    x(T) = A*e^-(0.01y) ( because cos(200pi) = 1)

    x(T) makes sense as a maxima because its just a damped amplitude, there is no cos factor making it smaller. now the question said "the ratio of the amplitudes of two successive maxima is one half" so...

    x(T)/x(0) = A*e^-(0.01y) / A = 1/2

    e^-(0.01y) = 1/2

    -0.01y = ln (1/2)
    y = -100 ln (1/2) = 69.3147181

    now I will solve for Wo, which will give To, which will give Fo. I believe my problem is somewhere in these steps (unless im using some completely wrong equations...)

    Td = 2pi / wd = 2pi / sqrt(Wo^2 - y^2) = 0.01
    0.01 = 2pi / sqrt(wo^2 + 69.3147181^2)
    sqrt(wo^2 + 69.3147181^2) = 200pi
    wo^2 + 69.3147181^2 = sqrt(200pi)
    wo = sqrt(sqrt(200pi) + 69.3147181) = 69.4952979

    so wo is slightly greater than wd, which i expect

    To = 2pi/wo = 0.0904116609
    fo = 1/To = 11.0605202

    now this is what i don't get. why is fo LESS than fd? did i do something wrong?
     
  2. jcsd
  3. Oct 19, 2009 #2

    ehild

    User Avatar
    Homework Helper
    Gold Member

    It should be

    wo^2 + 69.3147181^2 = [(200pi)]^2.


    ehild
     
  4. Oct 19, 2009 #3
    thanks ehild. so now i have

    wo^2 + 69.3147181^2 = 200pi^2
    wo = sqrt((200pi)^2 - 69.3147181^2) = 624.483503

    To = 2pi/wo = 0.0100614112
    Fo = 1/To = 99.3896363

    very good makes sense thanks a lot
     
  5. Oct 19, 2009 #4
    can someone please tell me quickly if it does make sense that fd > fo? is this answer now correct?
     
  6. Oct 19, 2009 #5

    Redbelly98

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    Looks like you switched a +/- sign in your work. The undamped frequency should be higher than the damped frequency; damping will slow down the oscillator.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Damped simple harmonic motion question
Loading...