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{SECT 0 {EXCHG {PARA 0 "" 0 "" {TEXT -1 10 "Problem #8" }}{PARA 0 "" 
0 "" {TEXT -1 5 "James" }}{PARA 0 "" 0 "" {TEXT -1 6 "Jeremy" }}{PARA 
0 "" 0 "" {TEXT -1 7 "Anthony" }}{PARA 0 "" 0 "" {TEXT -1 7 "John T." 
}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 259 "Question
 #8 States:  A weight hanging on a spring oscillates with a period of \+
3 sec.  After 2 lb is added, the period becomes 4 sec.  Assuming that \+
we can neglect any damping or external forces, determine how much weig
ht was originally attached to the spring." }}{PARA 0 "" 0 "" {TEXT -1 
0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }
}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 98 "The two equations we will need for this p
roblem are:   Omega = 2*pi / P   &    Omega = sqroot(K/m)" }}{PARA 0 "
" 0 "" {TEXT -1 55 "                                                  \+
     " }}{PARA 0 "" 0 "" {TEXT -1 87 "                                \+
                       (we'll use W to symbolize Omega)" }}{PARA 0 "" 
0 "" {TEXT -1 7 "       " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "
" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "
" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 60 "We'll first solve the W
=2*pi/P equation for our two periods." }}{PARA 0 "" 0 "" {TEXT -1 0 "
" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 89 "                             \+
                              Wo = 2*pi / 3 sec = 2.094 /sec" }}{PARA 
0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 89 "               \+
                                             Wf = 2*pi / 4sec = 1.571 \+
/sec" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" 
}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 49 "Now we apply these answers to the other e
quation." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 
79 "                                                            2.094 \+
= sqroot(K/m)" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 85 "                                                         \+
   1.571 = sqroot(K/m+.0625)" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 
0 "" 0 "" {TEXT -1 201 "The \"m+.0625\" comes from the fact that in th
e second equation we have the original mass plus the .0625 slugs that \+
was put on.  Now since you have 2 variables and 2 equations you can so
lve for the mass." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 
-1 38 " After solving you get:  m=.0804 slugs" }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 
-1 45 "  This multiplyed by 32 feet/sec is 2.571 lb." }}{PARA 0 "" 0 "
" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 87 "                         \+
                       So the answer is:     Weight = 2.571 lb" }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 
0 "" 0 "" {TEXT -1 7 "the end" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 16 " It's be
en real." }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 51 "                     \+
                              " }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "
" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}}}{MARK "1 22 0" 0 }
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