cipker
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cipker asked in 科學及數學其他 - 科學 · 1 decade ago

Slight damping(SHM)

In slight damping, the amplitude of oscillation decays exponentially with time, i.e. the ratio of amplitudes in successive oscillations is a constant:

A(1)/A(2) = A(2)/A(3) = A(3)/A(4) = .... >1

How to deduce this result?

1 Answer

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  • 1 decade ago
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    Taking a damped spring-and-mass oscillation system as an example with the following parameters:

    x(t): Displacement of the mass with x = 0 at equilibrium position

    μ: Damping coefficient (supposing linear drag)

    k: Spring constant

    m: Mass

    Then the differential equation describing the motion is:

    mx" = -μx' - kx

    mx" + μx' + kx = 0

    With characteristic equation:

    mλ2 + μλ + k = 0 ... (*)

    giving the roots:

    λ = -μ/2m +/- √(μ2/4m2 - k/m)

    = -p +/- q where p = μ/2m and q = √(μ2/4m2 - k/m)

    So in view of the nature of roots of (*), for the case of slight damping, i.e. when μ2/4m2 - k/m < 0

    Then (*) has 2 complex conjugate roots -p +/- iq and therefore x(t) is given by:

    x(t) = e-pt(A cos qt + B sin qt)

    = Ce-pt sin (qt + θ) for some constants C and θ.

    So the amplitude (Ce-pt) decays exponentially and for each successive oscillations, the attenuation factor is Ce-pT where T is the period of each oscillation (being 2π/q from the equation derived)

    So A1/A2 = A2/A3 = A3/A4 = ... = epT > 1 (since p and T are positive, making epT > 1)

    2009-07-26 10:46:18 補充:

    Thanks oldmasterchu

    Source(s): Myself
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