# A couple of 2nd order spring problems are solved for a couple different 
# masses and spring constants.
#
# APPLICATIONS TO HIGHER ORDER DE's-
# SPRING PROBLEMS
# Dr. Kirschner, Math 308
# 
# 
> diffeq:=m*diff(u(t),t$2)+mygamma*diff(u(t),t)+k*u(t)=F;

                       /   2      \
                       |  d       |           /  d      \
           diffeq := m |----- u(t)| + mygamma |---- u(t)| + k u(t) = F
                       |   2      |           \ dt      /
                       \ dt       /
# For example number 5 page 181:
# 
> m:=2/32; mygamma:=0; k:=2/(1/2); F:=0; inits:=u(0)= -1/4,D(u)(0)=0;

                                    m := 1/16

                                  mygamma := 0

                                     k := 4

                                     F := 0

                        inits := u(0) = -1/4, D(u)(0) = 0
> dsolve({diffeq,inits},u(t));

                              u(t) = - 1/4 cos(8 t)
# From the solution, we read off that the amplitude is 1/4, the 
# period is 8, and the frequency is (2*Pi)/8.
> with(plots): plot(rhs("), x=0..10);
Error, (in rhs) invalid arguments

# This is problem number 9, page 188.
# 
# 
> restart;diffeq:=m*diff(u(t),t$2)+mygamma*diff(u(t),t)+k*u(t)=F;

                       /   2      \
                       |  d       |           /  d      \
           diffeq := m |----- u(t)| + mygamma |---- u(t)| + k u(t) = F
                       |   2      |           \ dt      /
                       \ dt       /
> m:=6/32;mygamma:=0;k:=1/(1/2);F:=4*cos(7*t);inits:=u(0)=0,D(u)(0)=0;

                                    m := 3/16

                                  mygamma := 0

                                     k := 2

                                 F := 4 cos(7 t)

                         inits := u(0) = 0, D(u)(0) = 0
> interface(labelling=false); sol:=dsolve({diffeq,inits},u(t)); 

                       32               28             1/2    32
      sol := u(t) = - --- %1 cos(%2) + --- %1 cos(%2) 6    - --- %1 cos(%4)
                      115              115                   115

              28             1/2    32               28             1/2
           - --- %1 cos(%4) 6    - --- %3 sin(%4) - --- %3 sin(%4) 6
             115                   115              115

              32               28             1/2    64
           - --- %3 sin(%2) + --- %3 sin(%2) 6    + --- %1
             115              115                   115

                                        1/2
%1 :=                          cos(4/3 6    t)

                                    1/2
%2 :=                          4/3 6    t + 7 t

                                        1/2
%3 :=                          sin(4/3 6    t)

                                    1/2
%4 :=                          4/3 6    t - 7 t
> plot(rhs(sol),t=0..40);
# NOW, we see one more example, page 190, #22.
# 
# 
> restart;diffeq:=diff(u(t),t$2)+0.125*diff(u(t),t)+u(t)=F; inits:=u(0)=2,D(u)(0)=0;

                        /   2      \
                        |  d       |        /  d      \
              diffeq := |----- u(t)| + .125 |---- u(t)| + u(t) = F
                        |   2      |        \ dt      /
                        \ dt       /

                         inits := u(0) = 2, D(u)(0) = 0
> F:=3*cos(t);

                                  F := 3 cos(t)
> sol:=dsolve({diffeq,inits},u(t));

sol := u(t) = .7514691495 %2 cos(1.998044964 t)

     - .02350638884 %2 sin(1.998044964 t) - .7514691944 %2 cos(.001955036200 t)

     + 24.02350639 %2 sin(.001955036200 t) + 24.02350639 %1 cos(.001955036200 t)

     + .7514691944 %1 sin(.001955036200 t) + .02350638884 %1 cos(1.998044964 t)

     + .7514691495 %1 sin(1.998044964 t) + 2.000000045 exp( - .06250000000 t) %2

     - 23.92176793 exp( - .06250000000 t) %1

%1 :=                         sin(.9980449638 t)

%2 :=                         cos(.9980449638 t)
> trange:=t=0..100: a:=plot(rhs(sol),trange,thickness=3): b:= plot(F,trange): with(plots): display([a,b]);
# As one can see, the amplitude of the response is much larger 
# than the amplitude of the forcing function (The signal was 
# amplified.)
# There is a quarter wavelength or 90 degree lag between the 
# response and the signal (The signal F crosses the x-axis going 
# up, then a quarter of a wavelength later, the response u crosses 
# going up.)
#