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Damped oscillator

  1. Oct 9, 2008 #1
    1. The problem statement, all variables and given/known data
    A damped oscillator satisfies the equation

    x'' + 2Kx' + [tex]\Omega[/tex]^2 *(x)

    where K and [tex]\Omega[/tex] are positive constants with K < [tex]\Omega[/tex] (underdamping).

    i)At time t =0 the particle is released from rest at the point x=a . Show that the subsequent motion is given by

    x=a*exp(KT)(cos([tex]\Omega[/tex]D*t) +K/([tex]\Omega[/tex]2)*sin([tex]\Omega[/tex]D*t)


    where [tex]\Omega[/tex]D=([tex]\Omega[/tex]^2 - K^2)^1/2.

    ii)Find all the turning points of the function x(t) and show that the rati of successive maximum values of x is e^(-2*[tex]\pi[/tex]*K/([tex]\Omega[/tex]D)

    iii)a certain damped oscillator has mass 10 kg , period 5 seconds and successive maximum values of its displacement are in the ratio 3:1. Find the values of the spring and damping constants [tex]\alpha[/tex] and [tex]\beta[/tex].
    2. Relevant equations



    3. The attempt at a solution

    I had no trouble with part i) so I will skipped directly to part ii and iii.

    ii) Not sure how to calculate the turning points at x(t) and why taking the ratio of those turning points is significant.

    iii) How would knowing that finding the successive maximum values of its displacement are in the ratio 3 :1 aid me in finding the values of the spring and damping constants?
     
  2. jcsd
  3. Oct 9, 2008 #2

    Hootenanny

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    (ii) HINT: Turning points are stationary points.

    (iii) I'm not quite sure which constants you are referring to since there are no alphas or betas in x(t).
     
  4. Oct 9, 2008 #3
    ii) so does that mean x(t) and x'(t) equal zero. What are the turning points supposed to be?
    The problems wants me to find the ratio of successive maximum values of x(t) . When they are talking about successive maximum values at x(t) , do they mean x(t) at certain times and when the velocity is zero?
    iii) I'm not sure what they are talking about either. I think the spring constant is K and I think the omega D constant is the damping constant so I have no idea what alpha and beta are.
     
  5. Oct 9, 2008 #4

    Hootenanny

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    As I said in my previous post, turning points are stationary points, which are maxima, minima and points of inflection. Stationary points are so-called because the function stops increasing/decreasing at this point, in other words the function's derivative vanishes. So you need to find the points were x'(t)=0. Note that it is not necessary that x(t) vanishes.
    As I said above, x'(t) vanishes at maxima and minima. Hence, the points at which x'(t)=0 could either be maxima or minima, you should consider each point and decide which it is.
     
  6. Oct 9, 2008 #5
    For the part 3 of this problem , I know what alpha and Beta are:

    alpha= m*([tex]\Omega[/tex]2)

    beta= 2*m*K

    tau=2*pi/([tex]\Omega[/tex]D)= 2*pi/([tex]\Omega[/tex]2-K2)1/2)

    tau is the period which is 5 seconds
    I know I need to know K and omega in order to find alpha and Beta; I am not sure how to calculate the spring constant or omega.
     
  7. Oct 9, 2008 #6

    Hootenanny

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    HINT: You know that each maxima of x(t) it only one third of the magnitude of the previous maxima.
     
  8. Oct 9, 2008 #7
    So would I write something like:

    the 3:1 ratio of successive maximum displacement values implies:

    3/1 = e^-2pi*3*K/[tex]\Omega[/tex]D/(e^-2pi*K/[tex]\Omega[/tex]D to calculate K? then I could calculate OMEGA since I would now know K and [tex]\Omega[/tex]D
     
  9. Oct 10, 2008 #8

    Hootenanny

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    You're on the right lines. However, notice that the quantity

    [tex]\exp\left(-\frac{2\pi K}{\Omega_D}\right)[/tex]

    Is the ratio of the consecutive maxima of x(t), this what you are asked to show in (ii). The question then tells you that this ratio is equal to 3:1, in other words,

    [tex]\exp\left(-\frac{2\pi K}{\Omega_D}\right) = 3[/tex]

    Do you follow?
     
  10. Oct 10, 2008 #9

    Yes I think so. 2*[tex]\pi[/tex]/([tex]\Omega[/tex]D)= 5 seconds. =[tex]\tau[/tex]

    Therefore , [tex]\exp(-tau*K) = 3[/tex]
     
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