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Damped Simple Harmonic Motion

  1. May 22, 2012 #1
    1. The problem statement, all variables and given/known data

    http://img534.imageshack.us/img534/6503/questiono.jpg [Broken]

    2. Relevant equations

    3. The attempt at a solution

    (a) I'm not sure how to approach this part. I appreciate any help to get me started.

    (b) I know that the period of oscillation is given by

    [itex]T'=\frac{2 \pi}{\omega '}[/itex]

    I can use this later on to express my results in terms of the period T'. The one term in the given expression for x(t) that can turn negative as time increases would be the δ in the cosine. So how do I evaluate the right value for the phase delta?
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. May 22, 2012 #2
    for first one ,differentiate x(t) wrt time and put velocity so obtained=0 at t=0,perhaps that will do.
     
  4. May 22, 2012 #3
    Thanks. I'm still a bit confused. When I differentiated the expression with respect to t I got:

    [itex]x'(t)=\frac{\left( \frac{A_0}{\cos (\delta)}\right) e^{-t/\tau} (\tau \omega ' \sin(\omega' t + \delta)+\cos(\omega' t + \delta))}{\tau}[/itex]

    I've found x'(0):

    [itex]x'(0)=\frac{\left( \frac{A_0}{\cos (\delta)}\right) (\tau \omega ' \sin( \delta)+\cos(\delta))}{\tau}[/itex]


    But I have to show that δ is given by:

    [itex]\tan \delta = - \frac{1}{\omega ' \tau}[/itex]

    So I'm not sure, how can do I use the expression for x'(0) to show this?
     
  5. May 23, 2012 #4
    Did you mean I write down the velocity as:

    [itex]v(t)=\omega ' . \frac{A_0}{\cos (\delta)} e^{-t/\tau} \cos (\omega ' t + \delta)[/itex]

    [itex]\therefore \ v(0) =A_0 \omega '[/itex]

    So, how does this help? :confused:
     
  6. May 23, 2012 #5
    ok.you have done it.equate it to zero(because velocity is initially zero) don't you get what is required.
     
  7. May 23, 2012 #6
    I'm a bit confused. What do I have to do once I equate that expression to zero?
     
  8. May 23, 2012 #7
    by equating you will get,
    w'(tau)sin(delta)=-cos(delta)
    which gives
    tan(delta)=-1/w'(tau)
     
  9. May 23, 2012 #8
    Oh oops I didn't see that part. Thank you so much for the help I really appreciate that!!

    Any ideas how to approach the second part?
     
  10. May 23, 2012 #9
    the best way to it will be to draw(or see elsewhere) the graph and judge when x(t) will be zero for first time or just put it equal to zero and find t for first time.the graph is cosine curve ,shifted leftward and with decreasing amplitude.max. negative displacement will occur after first cycle means after when it crosses first zero and see wikipedia for other assistance.
     
    Last edited: May 23, 2012
  11. May 23, 2012 #10
    If I want to calculate it. We have by setting it equal to zero:

    x(0) = (A0/cos) cos (δ) = 0

    So what value do I use for δ? Do I have to use tan δ = 1/ω'τ and then substitute ω'=2π/T'?
     
  12. May 23, 2012 #11
    I am not saying to put t=0 and then equate x=0.you have to find t by equating it.since
    e(-t/tau) can not be zero.cosine part will be zero find when for the first time this will be zero.
    cos(w't+delta)=0,and choose cos(pi/2) from here you will get t for first zero.max. negative displacement can be obtained at T'/2 time from the beginning means after
    -delta/w'.can you tell the answer so that I can check it.
     
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