15.3 Forced Vibrations From A Steady State Source
Let a harmonic driving force in equation 15.1.10 have a displacement x :
x = xo cos (Ωt)
15.3.1
Which is given for a fixed distance, but a variable frequency. Then the equation for the coupled
seismometer and external driving force is:
2
2
φ + 2βΩ0φ + Ω0 φ = Ω (x0 cos Ωt)
15.3.2
The complete solution of equation 15.3.2 for φ (the displacement of the seismometer mass)will be
the sum of a transient solution, φTR, given by equation 15.2.15, and the steady state solution, φss.
The total solution, φ, is the sum φTR + φss = φ Thus, a seismometer displacement of φ produced
by a driving displacement x = xo cos (Ωt) is the sum of a short term displacement, φTR, and a
steady state long-term displacement, φss .
Now examine the solution of the characteristic equation by letting:
φss = D cos (Ωt-q)
15.3.3
where D is an arbitrary displacement and φss is then:
φ s s = D cos Ωt cos θ + D sin θ sin Ωt.
15.3.4
differentiating:
dφ ss
- (D cos θ ) sin Ω t + Ω (D sin θ ) cos Ω t
dt = Ω
differentiating again:
d 2φ s s
dt 2
2
= - Ω (D cos θ ) cos Ω t − Ω (D sin θ ) sin Ω t
2
11
Then collecting terms and after substituting into 15.3.2:
[
2
2
2
cos Ωt -Ω D cos θ + 2βΩΩ o D sin θ + Ω o D cos θ - Ω x0
[
2
2
]
]
+ sin Ωt -Ω D sin θ - 2βΩΩ0 D cos θ + Ω o D sin θ = 0
For this expression to be satisfied we must set the quantities in the brackets separately equal to
zero and solve for D and tan θ which are:
D=
xo
( )
Ω0
Ω
2
2
−1
2
+ 4β
2Ω 0
Ω
2
2
15.3.5
Which is the displacement term.
tan θ =
2 β Ω Ωo
2
Ω0 − Ω
2
15.3.6
which is the phase angle between the actual driving force, the ground movement, and the seismometer mass movement, the output.
For the transient solution only:
φTR = e−βΩ0t φ0 cos Ωf t +
φ0
Ωf
sin Ωf t
15.3.7
which is associated with the initial displacement, φo, and initial velocity, φ 0 , plus the steadystate solution.
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The steady state solution, φss , is given by substituting 15.3.5 back to 15.3.3:
φs s =
x0
( )
Ω0
Ω
2
2
−1
cos (Ωt - θ )
2
+ 4β
2Ω 0
Ω
2
2
15.3.8
And the total solution is the sum of the steady state and the transient solutions:
φTotal = e−βΩ0t φ0 cos Ωf t +
φ0
Ωf
sin Ωf t +
x0 cos (Ωt - θ)
2
Ω0 − 1
2
Ω
2
+ 2βΩ0
Ω
2
13.3.15
Now plotting x and φ vs time gives the comparison between the ground displacement, x(t), and the
seismometer movement, φ(t). This plot shows two important properties; the inherent magnification of the output, φ(t) compared to the input, x(t), and the time shift or phase lag, θ(t).
θ is the phase lag in radians, but can of course be converted to time given the frequency. Thus,the
ground motion leads the motion recorded on the seismogram by a time equal to θ/Ω. Where Ω is
the driving frequency.
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Extreme conditions of a seismometer reaction
Plotting acceleration against time:.
The velocity is given by:
t
v(t) =
a dt = aot + vo
let vo = 0
o
and the displacement:
t
x=
0
2
v dt = aot + vot + x0 ;
2
let vo = xo = 0
14
Then :
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