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{SECT 0 {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "restart:with(plots):
" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 61 "      p. 203, #6.      Calman
 Lankton, Jon Conowall, Katie M." }}{PARA 0 "" 0 "" {TEXT -1 68 "     \+
 A  mass of 5kg is attached to a spring hanging from a ceiling," }}
{PARA 0 "" 0 "" {TEXT -1 70 "therby stretching the spring 2m on coming
 to rest at equilibrium.  The" }}{PARA 0 "" 0 "" {TEXT -1 66 "mass is \+
then lifted up 1m above the equilibrium point and given an" }}{PARA 0 
"" 0 "" {TEXT -1 68 "upward velocity of 1/3 m/sec.  Determine the equa
tion for the simple" }}{PARA 0 "" 0 "" {TEXT -1 64 "harmonic motion of
 the mass.  When will the mass first reach its" }}{PARA 0 "" 0 "" 
{TEXT -1 41 "maximum height after being set in motion." }}{PARA 0 "" 
0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 65 "     First, we will \+
define the spring constant k, and the angular" }}{PARA 0 "" 0 "" 
{TEXT -1 63 "velocity, w.  k is equal to the mass times the force of g
ravity" }}{PARA 0 "" 0 "" {TEXT -1 70 "divided by the amount of spring
 stretch at equilibrium.  w is equal to" }}{PARA 0 "" 0 "" {TEXT -1 
41 "the square root of k divided by the mass." }}{PARA 0 "> " 0 "" 
{MPLTEXT 1 0 11 "k:=5*9.8/2;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"kG
$\"++++]C!\")" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "w:=sqrt(k/
5);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"wG$\"+iVf8A!\"*" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 4 "    " }}{PARA 0 "" 0 "" {TEXT -1 64 " Now,
 using the defined equation for the undamped spring with no" }}{PARA 
0 "" 0 "" {TEXT -1 70 "external forces, we can write our de.  For the \+
purposes of simplicity," }}{PARA 0 "" 0 "" {TEXT -1 70 "we will let th
e upwards direction be positive.  Thus our value of x at" }}{PARA 0 "
" 0 "" {TEXT -1 68 "time 0 is 1m, and our value of dx/dt, which is equ
al to velocity, at" }}{PARA 0 "" 0 "" {TEXT -1 68 "time 0 is 1/3 m/s. \+
 With these initial conditions, Maple immediately" }}{PARA 0 "" 0 "" 
{TEXT -1 67 "gives us the equation for the simple harmonic motion of t
he mass.  " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "> " 0 "" 
{MPLTEXT 1 0 67 "eq1:=dsolve(\{diff(x(t),t$2) + w^2 * x(t) = 0, x(0)=1
, D(x)(0)=1/3\}," }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 6 "x(t));" }}{PARA 
11 "" 1 "" {XPPMATH 20 "6#>%$eq1G/-%\"xG6#%\"tG,&-%$sinG6#,$F)$\"+iVf8
A!\"*$\"+0l%e]\"!#5-%$cosGF-$\"\"\"\"\"!" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 24 "y1:=unapply(rhs(eq1),t);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%#y1GR6#%\"tG6\"6$%)operatorG%&arrowGF(,&-%$sinG6#,$9$
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{PARA 0 "" 0 "" {TEXT -1 3 "   " }}{PARA 0 "" 0 "" {TEXT -1 69 "Now we
 can plot the function to get a general idea when the mass will" }}
{PARA 0 "" 0 "" {TEXT -1 32 "first reach it maximum height.  " }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "plo
t(y1(t),t=0..4*Pi/2.213594362);" }}{PARA 13 "" 1 "" {GLPLOT2D 400 300 
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45.000000 0 }}}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 3 "   " }}{PARA 0 "" 
0 "" {TEXT -1 66 "     As you can see from the plot, the mass will rea
ch its maximum" }}{PARA 0 "" 0 "" {TEXT -1 66 "height almost immediate
ly after it is given its initial velocity. " }}{PARA 0 "" 0 "" {TEXT 
-1 68 "This makes sense as the mass was given an initial upwards veloc
ity. " }}{PARA 0 "" 0 "" {TEXT -1 69 "To determine this value numerica
l, we must simply take the derivative" }}{PARA 0 "" 0 "" {TEXT -1 70 "
of our SHM equation with respect to t, and set it equal to zero.  Then
" }}{PARA 0 "" 0 "" {TEXT -1 69 "we can solve for t to find the time w
hen the mass will first reach it" }}{PARA 0 "" 0 "" {TEXT -1 15 "maxim
um height." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "> " 0 "" 
{MPLTEXT 1 0 19 "m:=(diff(eq1,t)=0);" }}{PARA 11 "" 1 "" {XPPMATH 20 "
6#>%\"mG//-%%diffG6$-%\"xG6#%\"tGF-,&-%$cosG6#,$F-$\"+iVf8A!\"*$\"+MLL
LL!#5-%$sinGF1$!+iVf8AF5\"\"!" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 70 "solve(.3333333334*cos(2.213594362*t)-2.213594362*sin(2.213594362
*t)=0," }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 3 "t);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#$\"+G1*>v'!#6" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 3 "  \+
 " }}{PARA 0 "" 0 "" {TEXT -1 67 "     Thus, our time when the mass fi
rst reaches a maximum height is" }}{PARA 0 "" 0 "" {TEXT -1 70 "0.0675
199 seconds.  This value agrees with our plot as it is very soon" }}
{PARA 0 "" 0 "" {TEXT -1 27 "after the motion was begun." }}{PARA 0 ">
 " 0 "" {MPLTEXT 1 0 0 "" }}}}{MARK "1 0 0" 61 }{VIEWOPTS 1 1 0 1 1 
1803 }