Queries on Damped Harmonic Motion

  1. So we know that SHM can be described as:
    x(t) = Acos(ωt + ϕ)
    v(t) = -Aω sin(ωt + ϕ)
    a(t) = -Aω^2 cos(ωt + ϕ)

    it can be easily said that the max acceleration (in terms of magnitude) a SHM system can achieve is Aω^2

    In Damped Harmonic Motion we know that:
    x(t) = (A)(e^(-bt/2m))cos(ωt + ϕ)

    given that:
    A' = (A)(e^(-bt/2m))
    ω' = sqrt( (ω^2) - (b/2m)^2 )

    Is it true that the max acceleration at any given time is (A')(ω')^2?

    My intuition tells me that the above statement is not true =/
    because differentiating the function x(t) = (A)(e^(-bt/2m))cos(ωt + ϕ) gives me a complex function (which has sine & cosine in it) & it doesn't really give me anything close to the (A')(ω')^2 term
     
    Last edited: Dec 5, 2013
  2. jcsd
  3. AlephZero

    AlephZero 7,300
    Science Advisor
    Homework Helper

    This is easier using complex numbers.

    For damped motion, ##x(t)## = the real part of ##Ae^{(-s + i\omega')t}## where ##s## is your ##b/2m##. Note, ##A## is a complex constant (to account for your phase angle ##\varphi##) and of course ##e^{i\theta} = \cos\theta + i\sin\theta##.

    So ##a(t)## = the real part of ##(-s + i\omega')^2 x(t)##

    Your intuition is right, but if the damping is small, ##(-s + i\omega')^2## is close to ##-\omega^2##.
     
    1 person likes this.
  4. I'm not really familiar with complex numbers (other than i^2 = -1) but your explanation does makes some sense to me. Thank You!! =)
     
Know someone interested in this topic? Share a link to this question via email, Google+, Twitter, or Facebook

Have something to add?

0
Draft saved Draft deleted