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Vibrations `

  1. Jul 19, 2005 #1
    How does phase shifting occur?

    All i know is that when i you the regular equation

    mx + cx + kx Damping bring phase shift (excuse my derivatives)

    mx + kx No phase shift

    Can someone explain this... :zzz:
  2. jcsd
  3. Jul 20, 2005 #2


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    [itex]x=Acos(\omega t) [/itex] is clearly a solution for the undamped oscillator equation, [itex]\ddot{x} = -(k/m)x [/itex] and clearly, NOT a solution for the damped oscillator equation, [itex]\ddot{x} = -(c/m) \dot{x} -(k/m)x [/itex]. For this latter equation, [itex]x=Ae^{kt}cos(\omega t + \phi ) [/itex] is a solution.
    Last edited: Jul 21, 2005
  4. Jul 21, 2005 #3
    In general, the solution to an underdamped system is

    e^(-squiggle*wn*t)[A*cos(wd*t) + B*sin(wd*t)]

    wd = damped frequency (rad/s)
    squiggle = damping ratio (dimensionless, always between 0 and 1, normally
    represented with a lower-case greek xi)
    wn = natural frequency (rad/s)
    t = time (s)

    How do i get this junk? well

    x'' + c/m*x' + k/m*x = 0 for a system without any driving forces

    let c/m = 2*squiggle*wn
    let k/m = wn^2

    the characteristic equation becomes

    r^2 + 2*squiggle*wn*r + wn^2 = 0

    applying the quadratic formula and simplifying gives

    r = -squiggle*wn +- wn*sqrt(squiggle^2 - 1)

    lets define
    wd = wn*sqrt(1 - squiggle^2)

    because sqrt(squiggle^2 - 1) = imaginary as squiggle < 1
    thus in the characteristic equation the imaginary part is the cause of the cosine and sine terms in the general solution at the very top.

    From the big ugly thing at the top...

    A and B are arbitrary constants which depend on the initial conditions

    we must solve a 2nd order system such that
    x(0) and x'(0) are satisfied with a single set of values for A and B

    If A and B are non zero then we have a phase shift, why?

    If B = 0 then we are left with


    Same can be said if A = 0

    So lets just look at the stuff inside the brackets as the e^junk only causes the vibrations to decay and does not effect the frequency or the phase.
    A*cos(wd*t) + B*sin(wd*t)

    This looks something like sin(wd*t + C)
    (look at some of your trig identites such as the sum of angles)

    C is the phase is this case and is found by taking the inverse tangent of the cosine term divided by the sine term. YES THE COSINE DIVIDED BY THE SINE. thats the way the math works out so in this case the complete solution when only using a sine to represent the behaviour of the system is

    R*e^(-squiggle*wn*t)*sin(wd*t + phi)

    R = sqrt(A^2 + B^2)
    phi = atan(A/B)

    Also, the statment that mx'' + kx = 0 has no phase shift is incorrect
    This system is undamped and has the general form of

    A*cos(wn*t) + B*sin(wn*t)

    Which if put in terms of only a sine function we have

    R*sin(wn*t + phi)

    R = sqrt(A^2 + B^2)
    phi = atan(A/B)
    wn = sqrt(k/m)

    Pretty much as long as your system has an initial displacement and some non-zero initial velocity there will just about always be a phase shift. Actuall it turns out in the undamped case that...

    A = x(0)
    B = x'(0)/wn

    phi = atan[x(0)*wn/x'(0)]

    as long as phi is not zero or a multiple of pi/2 there is a phase shift.

    so as long as x(0) is not zero and x'(0) is not zero and the spring constant and masses are not zero. Then we have a phase shift.

    In general, its safer to say that phase shifts depend on the initial conditions and not the physical constants of the system.
    Last edited: Jul 21, 2005
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