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{SECT 0 {EXCHG {PARA 0 "" 0 "" {TEXT -1 24 "Rebekah, Melinda, Andrew" 
}}{PARA 0 "" 0 "" {TEXT -1 18 "Page 203 number 3." }}{PARA 0 "> " 0 "
" {MPLTEXT 1 0 20 "restart:with(plots):" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 319 "A mass weighing 8 lb stretches a spring 2 ft on coming t
o rest at equilibrium.  The mass is then lifted up 6 in. above the equ
ilibrium point and given a downward velocity of 1 ft/sec.  Determine t
he simple harmonic motion of the mass.  How fast and in what direction
 will the mass be moving 10 sec after being released?" }}{PARA 0 "" 0 
"" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 178 "Since k=m*g/l, we kno
w mass is 8 lbs and  l = 2ft.  But since we are dealing with feet, we \+
must convert gravity from m/s^2 to ft/s^2, which is 32 ft/sec^2.  Henc
e k=128 lb/sec^2." }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 10 "k:=8*32/2;" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"kG\"$G\"" }}}{EXCHG {PARA 0 "" 0 "
" {TEXT -1 216 "To find omega, our angular frequency, for our d.e., we
 take k and divide it by the mass of our weight, then we take the squa
re root of that answer.    Hence our natural frequency is w/(2*Pi) and
 our period is 2*Pi/w." }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "w:=sqrt(k
/8);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"wG\"\"%" }}}{EXCHG {PARA 
0 "" 0 "" {TEXT -1 135 "This is our solved d.e. before applying the in
itial conditions to it.  Notice that it is 2nd order homogeneous and t
he angle is 4*time." }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 39 "dsolve(diff(
x(t),t$2)+w^2*x(t)=0,x(t));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%\"xG
6#%\"tG,&*&%$_C1G\"\"\"-%$sinG6#,$F'\"\"%F+F+*&%$_C2GF+-%$cosGF.F+F+" 
}}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 607 "Let's consider our initial con
ditions.  The first initial condition is x ( 0 ) = - .5.  The reason f
or this is because we started out at 2 ft and we raised it up 1/2 ft. \+
 We consider the pull downward our positive direction thus to raise th
e spring makes the sign negative.  The .5 is for the 1/2 ft we raised \+
our spring.  Our second initial condition is the derivative, or initia
l velocity, of our spring.  Thus D(x)(0) = 1 because we gave it a down
ward velocity of 1 ft/sec.  Since the velocity is downward, we have a \+
positive velocity.  Our answer is the simple harmonic motion of the ma
ss on the spring." }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 65 "eq1:=dsolve(\{
diff(x(t),t$2)+w^2*x(t)=0,x(0)=-.5,D(x)(0)=1\},x(t));" }}{PARA 11 "" 
1 "" {XPPMATH 20 "6#>%$eq1G/-%\"xG6#%\"tG,&-%$sinG6#,$F)\"\"%$\"+++++D
!#5-%$cosGF-$!+++++]F2" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 58 "To make
 life easier, we unapplied the rhs of our equation." }}{PARA 0 "> " 0 
"" {MPLTEXT 1 0 24 "y1:=unapply(rhs(eq1),t);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%#y1GR6#%\"tG6\"6$%)operatorG%&arrowGF(,&-%$sinG6#,$9$
\"\"%$\"+++++D!#5-%$cosGF/$!+++++]F5F(F(F(" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 51 "To find the period, we plugged in for w and solved." }}
{PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "evalf(2*Pi/4);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#$\"+Fjzq:!\"*" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 340 "
To see the motion of our d.e. we plotted using t = 0 to roughly 12.56.
  Since our problem asks us to look at 10 sec., we wanted the graph to
 show us up to and beyond 10 seconds.  To find our t coordinates, we s
tarted at t = 0 and then using our period, Pi/2, we found at 7*Pi/2, w
e had enough oscillations to evaluate graphically at t = 10." }}{PARA 
0 "> " 0 "" {MPLTEXT 1 0 24 "plot(y1(t),t=0..7*Pi/2);" }}{PARA 13 "" 
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