1. Limited time only! Sign up for a free 30min personal 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!

Significace of i*sin(wt) in SHM [cos(wt)+i*sin(wt)]

  1. Apr 15, 2008 #1
    confusing title i suspect but didnt know what else to call it.

    so i'm considering a spring and mass with the displacment of the mass being x and the stiffness of the spring being k

    force on the spring = -k*x =m*a

    or rearranged etc

    d2x/dt2 + (k/m)*x = 0 ...............................................[1]

    then i assume a soln of the form e^(at)
    and find a= i*(k/m)^.5 [then let sqrt(k/m)= w)

    so i get x= e^(i*w*t)

    which using the euler identity gives

    x= cos(wt)+i*sin(wt)

    great so i managed to show that the thing will move sinusoidly
    but what about that i*sin(wt) term

    if i ignore it, and let x=cos(wt) this still satisfies [1] above
    so am i justified in ignoring it? or does it have some physical significance?
    or is it something that just gets "thrown up" as a result of me assuming a soln of the form e^at ?

    and more generally
    if i find a soln to a diff eqn, and find that by dropping some terms in the soln that it still satifies the eqn am i still justified in doing that?
    thanks jim
  2. jcsd
  3. Apr 15, 2008 #2


    User Avatar

    You need to simplify a bit further. I can't remember how but eventually the imaginary part disappears when you substitute a certain complex number which IIRC was [tex]\sqrt{\frac{-k}{m}}[/tex].

    You will eventually get to something of the form x = a sin (wt + y). where a is the max displacement, wt is the angular frequency, and y is the phase shift.
    Last edited: Apr 15, 2008
  4. Apr 15, 2008 #3
    I suspect that you'll find that your second order differential equation has two linearly independent solutions. (you somehow threw away the -w root of w^2). So the other solution you discarded was
    [tex]x= cos(wt)-i*sin(wt)[/tex]
    Now, the most general solution to your DE is a linear combination of these solutions, and you can pick the coefficients so that your initial conditions are satisfied. Likely you'll just pick
    [tex]x = e^{i w t} + e^{-i \omega t}[/tex]
    and all your problems will go away.
  5. Apr 15, 2008 #4
    nice one!"
    so i get x(t)=cos(wt)+i*sin(wt)
    and x(t)=cos(wt)- i*sin(wt)

    then adding i get


    and x(t)=cos(wt)

    so i dont have to worry about the complex

    so is this called superposition then?
  6. Apr 15, 2008 #5
    This is not quite right. What you have there are two different solutions, which are independent of each other. So, let's write these like this:

    [tex]x_1(t) = e^{i\omega t}[/tex]
    [tex]x_2(t) = e^{-i\omega t}[/tex]

    However, because these are solutions of a linear differential equation, any linear superposition of the solutions will, itself, be a solution. This superposition will take the form:

    [tex]x(t) = Ax_1(t) + Bx_2(t)[/tex], where A and B are, in general, any complex numbers.

    However, since x_2 is the complex conjugate of x_1, the only way to get real-valued solutions for x(t) is to set [itex]B = A^*[/itex].

    Now, let's let [itex]A = a e^{-i\phi}[/itex].

    In this case:

    [tex]x(t) = a e^{i(\omega t - \phi)} + a e^{-i(\omega t - \phi)}[/tex]

    With a little manipulation, it is clear that this means that [itex]x(t) = 2a\cos (\omega t - \phi)[/itex]. Now we can recognize that 2a will be the largest possible value of x. So:

    [tex] x(t) = x_{max} \cos (\omega t - \phi)[/tex]

    [itex]x_{max}[/itex] are not specified by the original differential equation. This essentially means that the solution can have any amplitude and phase. In other words, the oscillations can have any size and can start at any time.

    If you recall your trig identities, you'll recognize that we could also write this as:

    [tex]x(t) = x_{max}(\cos \phi \cos (\omega t) + \sin \phi \sin (\omega t))[/tex]

    This can also be written:

    [itex]x(t) = a_1 \cos (\omega t) + a_2 \sin (\omega t)[/itex], where we can pick [itex]a_1[/itex] and [itex]a_2[/itex] to have any values we like, including 0. So, in this way, we can see that both [itex]\cos (\omega t)[/itex] and [itex]\sin (\omega t)[/itex] are valid solutions.
  7. Apr 15, 2008 #6
    Yes, you can ignore it. A commonly used trick is to use the identity [itex]cos(x)=Re(e^{ix})[/itex] to convert a trig function problem into an exponential problem.

    The significance is that [itex]sin(\omega t)=Im(e^{i \omega t})[/itex] is also a solution.

    A solution to the equation is a solution to the equation. How you derive it is irrelevant.
    Last edited: Apr 15, 2008
  8. Apr 16, 2008 #7

    Andy Resnick

    User Avatar
    Science Advisor
    Education Advisor

    Others have pointed out that [tex]e^{i \omega t}[/tex] and [tex]e^{-i \omega t}[/tex] are general solutions, the specific form of which can be determined by initial conditions.

    However, for your original question, the term i * sin(wt) does have a physical relevance- in mechanics and electromagnetics, the linear response of a system to a periodic disturbance (a fancy way of thinking about the mass-spring problem) has a component that is in phase to the disturbance (the real cos (wt) part) and a component that is in phase quadrature (i *sin(wt)). In mechanics, the phase quadrature component arises due to viscous or dissipative effects; in electromagnetics, the phase quadrature component corresponds to the reactance, IIRC.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook