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{SECT 0 {EXCHG {PARA 0 "" 0 "" {TEXT -1 77 "redoctbr.mws  26 Aug 95   \+
 An array of hydrophones tracks a russian submarine" }}{PARA 0 "" 0 "
" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 230 "Red October submarine \+
(S) is at (8000,0) m when t=0. Its velocity in the y-direction is 10 k
nots. She has a problem and is radiating sound at a frequency of 3 khz
. Your boat has 15 detectors, spaced 0.25 m apart in the y-direction" 
}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 68 "a) Writ
e an expression for the location of S as a function of time. " }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 88 "b) Determ
ine the distance L[i] from the moving source S to each detector, as a \+
function " }}{PARA 0 "" 0 "" {TEXT -1 13 "of the time. " }}{PARA 0 "" 
0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 89 "c) Show that at t=0 \+
the distances are the same to a very small fraction of a wavelength. \+
" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 77 "d) Fi
nd the resultant wave R from all 15 detectors as a function of the tim
e " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 75 "e) \+
Plot the amplitude of R as a function of time from t=-300 s to t=+300s
. " }}{PARA 0 "" 0 "" {TEXT -1 54 "Note the time when the detector sig
nal falls to zero. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "> " 0 
"" {MPLTEXT 1 0 8 "restart;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
29 "assume(k>0,wavelength>0,t>0);" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 22 "wavelength:=1500/3000;" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 20 "k:=2*Pi/wavelength; " }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "X:=8000;
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "Y:=t*10*2000*36*100/(39
37*3600);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 6 "N:=15;" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "m:=trunc(N/2);" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 69 "Determine
 the distances L[i] from the source to each detector element" }}{PARA 
0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "for i fro
m 1 to N  do" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 47 "L[i]:=sqrt( (Y-spac
ing*(N-i-m))^2 + X^2);   od:" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 68 "C
heck that distances are all 'same' well within a wavelength at t=0." }
}{PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "for i from 1 to N  do" }}{PARA 0 "
> " 0 "" {MPLTEXT 1 0 41 "evalf(subs(spacing=1/4,t=0,L[i]) );   od;" }
}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "R_x:=0:  R_y:=0:" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 102 "Let t=0 and get magnitude of resu
ltant; first obtain x and y components of the resultant, then square.
" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 21 
"for i from 1 to N  do" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "R_x:=R_x+
A*cos(k*L[i]);" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "R_y:=R_y+A*sin(k*
L[i]);       od:" }}}{EXCHG {PARA 2 "" 0 "" {TEXT -1 1 "\n" }}{PARA 0 
"> " 0 "" {MPLTEXT 1 0 22 "Rsquared:=R_x^2+R_y^2:" }}}{EXCHG {PARA 0 "
> " 0 "" {MPLTEXT 1 0 26 "values:=\{A=1,spacing=1/4\};" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 104 "Substitu
te problem values into the resultant vector squared. This will depend \+
on time since theta does. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 
"> " 0 "" {MPLTEXT 1 0 26 "g:=subs(values,Rsquared): " }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "plot(g,t=-100..300,numpoints=200);
" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 8 "Remarks." }}{PARA 0 "" 0 "" 
{TEXT -1 64 "The signal is a max at t=0 and fades to zero in around 20
0 sec. " }}}}{MARK "17 1 0" 2 }{VIEWOPTS 1 1 0 1 1 1803 }