Quiz 11
Sample Quiz 11, Problem 1. Vertical Motion Seismoscope
The 1875 horizontal motion seismoscope of F. Cecchi (1822-1887) reacted to an earthquake.
It started a clock, and then it started motion of a recording surface, which ran at a speed of 1 cm
per second for 20 seconds. The clock provided the observer with the earthquake hit time.
A Simplistic Vertical Motion Seismoscope
The apparently stationary heavy mass on a spring writes
with the attached stylus onto a rotating drum, as the
ground moves up. The generated trace is x(t).
The motion of the heavy mass m in the figure can be modeled initially by a forced spring-mass
system with damping. The initial model has the form
mx00 + cx0 + kx = f (t)
where f (t) is the vertical ground force due to the earthquake. In terms of the vertical ground
motion u(t), we write via Newton’s second law the force equation f (t) = −mu00 (t) (compare to
falling body −mg). The final model for the motion of the mass is then


x00 (t) + 2βΩ0 x0 (t) + Ω20 x(t) = −u00 (t),





k

 c = 2βΩ0 ,
= Ω20 ,
m
(1)
m



x(t) = center of mass position measured from equilibrium,





u(t) = vertical ground motion due to the earthquake.
Terms seismoscope, seismograph, seismometer refer to the device in the figure. Some observations:
Slow ground movement means x0 ≈ 0 and x00 ≈ 0, then (1) implies Ω20 x(t) = −u00 (t). The
seismometer records ground acceleration.
Fast ground movement means x ≈ 0 and x0 ≈ 0, then (1) implies x00 (t) = −u00 (t). The
seismometer records ground displacement.
A release test begins by starting a vibration with u identically zero. Two successive maxima
(t1 , x1 ), (t2 , x2 ) are recorded. This experiment determines constants β, Ω0 .
The objective of (1) is to determine u(t), by knowing x(t) from the seismograph.
The Problem.
Assume the seismograph trace can be modeled at time t = 0 (a time after the earthquake
struck) by x(t) = 10 cos(3t). Assume a release test determined 2βΩ = 16 and Ω20 = 80.
Explain how to find a formula for the ground motion u(t), then provide details for the
answer u(t) = 710
cos(3t) − 160
sin(3t).
9
3
Solution.
Background:
A release test is an experiment which provides initial data x(0) > 0, x0 (0) = 0 to the
seismoscope mass. The model is x00 + 2βΩ0 x0 + Ω20 x = 0 (ground motion zero). The
recorder graphs x(t) during the experiment, until two successive maxima (t1 , x1 ), (t2 , x2 )
appear in the graph. This is enough information to find values for β, Ω0 .
In an under-damped oscillation, the characteristic equation is (r + p)2 + ω 2 = 0 corresponding to complex conjugate roots −p ± ωi. The phase-amplitude form is x(t) =
Ce−pt cos(ωt − α), with period 2π/ω.
The equation x00 + 2βΩ0 x0 + Ω20 x = 0 has characteristic equation (r + β)2 + Ω20 = 0.
Therefore x(t) = Ce−βt cos(Ω0 t − α).
The period is t2 − t1 = 2π/Ω0 . Therefore, Ω0 is known. The maxima occur when the
cosine factor is 1, therefore
Ce−βt2 · 1
x2
=
= e−β(t2 −t1 ) .
x1
Ce−βt1 · 1
This equation determines β.
Details. The equation −u00 (t) = x00 (t) + 2βΩ0 x0 (t) + Ω20 x(t) (the model written backwards)
determines u(t) in terms of x(t). If x(t) is known, then this is a quadrature equation for the
ground motion u(t).
For the example x(t) = 10 cos(3t), 2βΩ0 = 16, Ω20 = 80, the quadrature equation is
−u00 (t) = x00 (t) + 16x0 (t) + 80x(t).
After substitution of x(t), the equation becomes
−u00 (t) = 710 cos(3t) − 480 sin(3t),
which can be integrated twice by hand. All integration constants will be assumed zero. The
answer:
710
160
u(t) =
cos(3t) −
sin(3t).
9
3
A Maple answer check has this brief input:
de:=-diff(u(t),t,t) = diff(x(t),t,t) + 16*diff(x(t),t) + 80* x(t);
x:=t->10*cos(3*t);
dsolve(de,u(t));subs(_C1=0,_C2=0,%);
Quiz 11, Problem 2. Laplace Theory
Laplace theory implements the method of quadrature for higher order differential equations, linear systems of differential equations, and
certain partial differential equations.
Laplace’s method solves differential equations.
The Problem. Solve by table methods or Laplace’s method.
(a) Forward table. Find L(f (t)) for f (t) = 3(t + 1)2 e2t + 2et sin(3t).
(b) Backward table. Find f (t) for
L(f (t)) =
4s
s−1
+
.
s2 + 4 s2 − 2s + 5
(c) Solve the initial value problem x00 (t) + 2x0 (t) + 5x(t) = et , x(0) = 0, x0 (0) = 1.
Solution (a).
L(f (t)) = L(3(t + 1)2 e2t + 2et sin(3t))
t
= 3 L((t + 1)2 e2t
) + 2 L(e sin(3t))
2
= 3 L((t + 1) ) s=s−2
+ 2 L(sin(3t))|s=s−1
2
= 3 L(t + 2t +1) s=s−2 + 2 L(sin(3t))|
s=s−1
2
1
3 2
+ 2 s2 +9 = 3 s3 + s2 + s = 3
2
(s−2)3
+
s=s−2
2
1
+
2
s−2
(s−2)
+
s=s−1
6
(s−1)2 +9
Linearity
Shift rule twice
Expand (t + 1)2
Forward table
Apply shifts
Solution (b).
s−1
L(f (t)) = s24s+4 + s2 −2s+5
s
s−1
= 4 s2 +4 + (s−1)
2 +4
s−1
= 4 L(cos 2t) + (s−1)
2 +4
f (t)
=
=
=
=
=
Prep for backward table
backward table
4 L(cos 2t) + s2s+4 s = s − 1
4 L(cos 2t) + L(cos 2t)| s = s − 1
4 L(cos 2t) + L(et cos 2t)
L(4 cos 2t) + et cos 2t)
4 cos 2t + et cos 2t
shift rule
backward table
shift rule
Linearity
Lerch’s cancel rule
Solution (c).
L(x00 (t) + 2x0 (t) + 5x(t))
s L(x0 ) − x0 (0) + 2s L(x) − 2x(0) + 5 L(x)
s(s L(x) − x(0)) − x0 (0) + 2s L(x) − 2x(0) + 5 L(x)
s2 L(x) − 1 + s L(x) + 5 L(x)
(s2 + 2s + 5) L(x)
=
=
=
=
=
L(et )
L(et )
L(et )
L(et )
1 + L(et )
L acts like matrix mult
Parts rule
Parts rule
Use x(0) = 0, x0 (0) = 1
Collect L(x) left
L(x) =
s+L(et )
(s2 +2s+5)
Isolate L(x) left
L(x) =
s+1/(s−1)
(s2 +2s+5)
Forward table
L(x) =
s2 −s+1
(s−1)(s2 +2s+5)
Algebra
L(x) =
A
s−1
+
Bs+C
s2 +2s+5
Partial fractions
L(x) =
A
s−1
+
B(s+1)+C−B
(s+1)2 +4
Prepare for shift
L(x) =
B(s)+C−B A
+
s−1
(s)2 +4 s=s+1
Shift rule
L(x) = A L(et ) + B L(cos 2t) +
C−B
2
L(x) = A L(et ) + B L(e−t cos 2t) +
L(x) = L(Aet + Be−t cos 2t +
x(t)
= Aet + Be−t cos 2t +
L(sin 2t) C−B
2
s=s+1
L(e−t sin 2t)
C−B −t
sin 2t)
2 e
C−B −t
sin 2t
2 e
Backward table
Shift rule
Linearity
Lerch’s rule
The partial fraction problem remains:
A
Bs + C
s2 − s + 1
= + 2
(s − 1)(s2 + 2s + 5)
s
s + 2s + 5
This problem is solved by clearing the fractions, then swapping sides of the equation, to obtain
A(s2 + 2s + 5) + (Bs + C)(s − 1) = s2 − s + 1.
Substitute three values for s to find 3 equations in 3 unknowns A, B, C:
s=0
5A − C
= 1
s=1
8A
= 1
s = −1 4A + 2B − 2C = 3
Then A = 1/8, B = 7/8, C = −3/8 and finally
x(t) = Aet + Be−t cos 2t +
C − B −t
1
1
3
e sin 2t = et − e−t cos 2t + e−t sin 2t.
2
8
8
8
Answer Checks
# Quiz 11
# Problem 2
# answer check problem 2(a)
f:=3*(t+1)^2*exp(2*t)+2*exp(t)*sin(3*t);
with(inttrans): # load laplace package
laplace(f,t,s);
convert(%,parfrac,s);
# answer check problem 2(b)
F:=4*s/(s^2+4)+(s-1)/(s^2-2*s+5);
invlaplace(F,s,t);
# answer check problem 2(c)
x:=’x’:de:=diff(x(t),t,t)+2*diff(x(t),t)+5*x(t)=exp(t);
ic:=x(0)=0,D(x)(0)=1;
dsolve([de,ic],x(t));
# answer check problem 2(c), partial fractions
convert((s^2-s+1)/((s-1)*(s^2+2*s+5)),parfrac,s);
> # Quiz 11 solutions
> # Problem 1
> de:=­diff(u(t),t,t) = diff(X(t),t,t) + 16*diff(X(t),t) + 80* X(t)
;
> X:=t­>10*cos(3*t);
> dsolve(de,u(t));subs(_C1=0,_C2=0,%);
(1)
> > > > # answer check problem 2(a)
f:=3*(t+1)^2*exp(2*t)+2*exp(t)*sin(3*t);
with(inttrans): # load laplace package
laplace(f,t,s);
(2)
> convert(%,parfrac,s);
(3)
> # answer check problem 2(b)
> F:=4*s/(s^2+4)+(s­1)/(s^2­2*s+5);
> invlaplace(F,s,t);
(4)
> # answer check problem 2(c)
> x:='x':de:=diff(x(t),t,t)+2*diff(x(t),t)+5*x(t)=exp(t);
ic:=x(0)=0,D(x)(0)=1;
> dsolve([de,ic],x(t));
> # answer check problem 2(c), partial fractions
(5)