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{SECT 0 {EXCHG {PARA 0 "" 0 "" {TEXT -1 45 "Keith Seal, Sarah Thompson
, Michelle McCrain " }}{PARA 0 "" 0 "" {TEXT -1 9 "p 203 #7 " }}{PARA 
0 "" 0 "" {TEXT -1 292 "There is a 0.5 kg mass hanging on a spring tha
t streches the spring 2 m. The mass is pulled down 0.5 m and is given \+
an upward velocity of 0.5 m/s. What is the equation of simple harmonic
 motion for the spring? How long will it take for the spring to first \+
return to the equilibrium position. " }}{PARA 0 "" 0 "" {TEXT -1 1 " \+
" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 182 "The spring constant for this
 problem would involve using the initial condtion of the spring.The eq
uation used to find the k is mg=kx, where m is 0.5 kg , g=9,8 m/s^2, a
nd x = 0.5 m. " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "restart: \+
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "k:=0.5*9.8/2;" }}{PARA 
11 "" 1 "" {XPPMATH 20 "6#>%\"kG$\"++++]C!\"*" }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 15 "w:=sqrt(k/0.5);" }}{PARA 11 "" 1 "" {XPPMATH 20 
"6#>%\"wG$\"+iVf8A!\"*" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 105 "We obtained k as mg/l  (where l i
s in meters instead of cm, to match with the units of gravity), and th
en" }}{PARA 0 "" 0 "" {TEXT -1 25 "found w=sqrt(k/m).  Thus," }{TEXT 
256 17 "natural frequency" }{TEXT -1 24 " is 2.21/(2Pi) and the  " }
{TEXT 257 6 "period" }{TEXT -1 14 " is 2Pi/2.21  " }}}{EXCHG {PARA 0 "
> " 0 "" {MPLTEXT 1 0 37 "eq1:=diff(x(t),t$2) + w^2 * x(t) = 0;" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$eq1G/,&-%%diffG6$-%\"xG6#%\"tG-%\"$
G6$F-\"\"#\"\"\"F*$\"+********[!\"*\"\"!" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 76 "eq2:=dsolve(\{diff(x(t),t$2) + w^2 * x(t) = 0, x(0)=0
.5, D(x)(0)=0.5\}, x(t));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$eq2G/-
%\"xG6#%\"tG,&-%$sinG6#,$F)$\"+iVf8A!\"*$\"+d(p(eA!#5-%$cosGF-$\"+++++
]F4" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "y1:=unapply(rhs(eq2)
,t);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#y1GR6#%\"tG6\"6$%)operatorG
%&arrowGF(,&-%$sinG6#,$9$$\"+iVf8A!\"*$\"+d(p(eA!#5-%$cosGF/$\"+++++]F
7F(F(F(" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 100 "We obtained the initial conditions in the above using th
at 0.5 m the amount the mass was pulled down" }}{PARA 0 "" 0 "" {TEXT 
-1 103 "below the equilibrium point, which is x(0), and that x'(0)=0.5
 m/s, since we are told that the initial " }}{PARA 0 "" 0 "" {TEXT -1 
91 "velocity is 0.5m/s.  Hence, by looking at the solution, we see tha
t the Amplitude is 0.5 m." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "
" 0 "" {TEXT -1 32 "Here is the plot of the spring. " }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 24 "plot(y1(t),t=0..4*Pi/w);" }}{PARA 13 "" 
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{PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "period:=evalf(2*Pi/w);" }}{PARA 11 
"" 1 "" {XPPMATH 20 "6#>%'periodG$\"+\"z`%QG!\"*" }}}{EXCHG {PARA 0 "
" 0 "" {TEXT -1 148 "The period of the mass would be 2.84 seconds, whi
ch would be the time it takes for the mass to return to the equilbrium
 point after one oscillation." }}}}{MARK "0 0 0" 36 }{VIEWOPTS 1 1 0 
1 1 1803 }