## Trending News

# 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

- 六呎將軍Lv 71 decade agoFavorite Answer
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