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{PARA 258 "" 0 "" {TEXT 257 30 "UNDAMPED MECHANICAL VIBRATIONS" }}
{PARA 259 "" 0 "" {TEXT -1 0 "" }}{PARA 264 "" 0 "" {TEXT -1 9 "Equati
on:" }}{PARA 265 "" 0 "" {TEXT -1 75 "                                \+
                              mu'' + ku = 0" }}{PARA 266 "" 0 "" 
{TEXT -1 2 "or" }}{PARA 267 "" 0 "" {TEXT -1 79 "                     \+
                                          u'' + w0^2 u = 0" }}{PARA 
268 "" 0 "" {TEXT -1 0 "" }}{PARA 269 "" 0 "" {TEXT -1 58 "where w0^2 \+
= k/m is the natural frequency of the vibration" }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart;" }}
}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "MS:=diff(u(t),t$2)+ w0^2*u(
t)=0;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 17 "General solution:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "dsolve(MS,u(t));" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 270 "" 0 "" {TEXT -1 61 "Example
 1: (Similar to Problem 6 on Section 3.8 Boyce's book)" }}{PARA 271 "
" 0 "" {TEXT -1 0 "" }}{PARA 272 "" 0 "" {TEXT -1 289 "A mass of 100 g
 stretches a spring 5 cm. If the mass is set in motion 1 cm below its \+
equilibrium position with a downward velocity of 10cm/sec, and if ther
e is no damping, determine the position u of the mass at any time. Fin
d the frequency, the period, amplitude and phase of the motion. " }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 5 "Answ:" }}
{PARA 0 "" 0 "" {TEXT -1 48 "We need to find \"w0\" or equivalently sq
rt(k/m). " }}{PARA 0 "" 0 "" {TEXT -1 176 "we know mg = kL, then k = m
g/L. Therefore, w0 = sqrt(k/m) = sqrt(g/L). Thus the mass is irrelevan
t in this problem, the important quantities are   g = 980cm/s^2   and \+
L = 5cm. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 31 "g:= 980; L:= 5; w0:= sqrt(g/L);" }}}{EXCHG {PARA 0 
"> " 0 "" {MPLTEXT 1 0 3 "MS;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "
" }}{PARA 0 "" 0 "" {TEXT -1 17 "General solution:" }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "dsolve(MS,u
(t));" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 19 "Initial conditions:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "u0:=1; v0:=10;" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 38 "dsolve(\{MS, u(0)=u0,D(u)(0)=v0\},u
(t));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 46 "S:=rhs(dsolve(\{MS
, u(0)=u0,D(u)(0)=v0\},u(t)));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 17 "u1:=unapply(S,t);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
22 "plot(u1(t), t=0..0.8);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 7 "Remark:" }}{PARA 0 "" 0 "" {TEXT -1 82 "No
tice the difference between the English unit system and the MKS or cgs
 systems. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 
-1 87 "English System:           foot          -   pound      - second
     units of  length - " }{TEXT 275 5 "force" }{TEXT -1 21 " - time, \+
respectively" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 
-1 84 "MKS   System:            meter        -   kilogram  - second   \+
  units of  length - " }{TEXT 276 4 "mass" }{TEXT -1 21 " - time, resp
ectively" }}{PARA 0 "" 0 "" {TEXT -1 86 "cgs      System             c
entimeter -   gram       - second     units of  length - " }{TEXT 277 
4 "mass" }{TEXT -1 21 " - time, respectively" }}{PARA 0 "" 0 "" {TEXT 
-1 0 "" }}{PARA 273 "" 0 "" {TEXT -1 96 "\" The concept of mass and th
e distinction between mass and weigh are often very troublesome ...." 
}}{PARA 274 "" 0 "" {TEXT -1 308 "Confusion over these concepts arise \+
usually from semantics, the use and meaning of words. We have the noun
 weight and the verb to weigh; we have the noun mass but no correspond
ing verb. For this reason one usually uses the word weigh to denote an
 operation designed to obtain either the mass or the weight. \"" }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 73 "From Numb
ers and units for physics of Robert A. Carman. Edit Wiley (1969)" }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 144 "For exam
ple is of common use to write   1 Kg = 2.2 lb. What that usually means
 is that a mass of 1 Kg weighs 2.2 lb on the surface of the earth." }}
}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 267 67 "
Let's find now the amplitude, natural frequency, period, and phase." }
}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 10 "Amplitud
e:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 21 "R:=sqrt(1^2+(5/7)^2);" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 18 "Natural frequency:" }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 3 "w0;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 7 "Period:" }}{PARA 0 "" 
0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "T:= 2*
pi/w0;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 6 "Phase:" }}{PARA 0 "" 0 "" {TEXT -1 46 "To compute the phas
e follow this simple rule: " }}{PARA 0 "" 0 "" {TEXT -1 46 "For A,B >0
 or A>0, B<0     delta = arctan(B/A)" }}{PARA 0 "" 0 "" {TEXT -1 52 "F
or A,B < 0 or A<0, B>0     delta = arctan(B/A) + pi" }}{PARA 0 "" 0 "
" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 29 "In our case:   A=1, B=5
/7 so " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 27 "dleta:= evalf(arctan(5/7));" }}}{EXCHG {PARA 0 "" 0 "
" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 38 "Exercises 1-5 in Boyce's book Sect 3.8" }}{PARA 0 "" 0 "
" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "un1:=4*cos(3*t) - 2*sin(3*t)
;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "delta:=arctan(-2/4);" 
}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "evalf(%);" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "un2:=-2*cos(pi*t) - 3*sin(pi*t);" }
}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "deltaaux:=arctan(3/2);" }}
}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "delta:= pi+evalf(%);" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "un3:=-cos(t) + sqrt(3)*sin(t
);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "deltaaux:=arctan(-sqr
t(3));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "delta:= pi+%;" }}
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