{VERSION 4 0 "IBM INTEL NT" "4.0" } {USTYLETAB {CSTYLE "Maple Input" -1 0 "Courier" 0 1 255 0 0 1 0 1 0 0 1 0 0 0 0 1 }{PSTYLE "Normal" -1 0 1 {CSTYLE "" -1 -1 "Times" 1 12 0 0 0 1 2 2 2 2 2 2 1 1 1 1 }1 1 0 0 0 0 1 0 1 0 2 2 0 1 }} {SECT 0 {EXCHG {PARA 0 "" 0 "" {TEXT -1 55 "Math 2250 Maple Project 2, March 2003. Tacoma Narrows." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 72 "NAME ____________________________________ ____________ CLASSTIME ______ " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 101 "There are six (6) problems in this proje ct. Please answer the questions A, B, C , ... associated with" }} {PARA 0 "" 0 "" {TEXT -1 97 " each problem. The original worksheet \" project2-spring-2003.mws\" is a template for the solution;" }}{PARA 0 "" 0 "" {TEXT -1 97 " you must fill in the code and all comments. Samp le code can be copied with the mouse. Use pencil" }}{PARA 0 "" 0 "" {TEXT -1 70 " freely to annotate the worksheet and to clarify the code and figures." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 99 "The problem headers for the Spring 2003 revision of David Eyre's project (original was year 2000). " }}{PARA 0 "" 0 "" {TEXT -1 494 " __________ 2.1. OVERDAMPED FREE OSCILLATIONS. \+ The first four problems study the\n __________2.2. UNDERDAMPED FR EE OSCILLATIONS. simplest linear model for x(t).\n _ _________2.3. UNDAMPED FORCED OSCILLATIONS (c=0).\n __________2.4. DAM PED FORCED OSCILLATIONS (c>0).\n __________2.5. LARGE SUSTAINED OSCILL ATIONS. This is the second, nonlinear model.\n _ _________2.6. MCKENNA NON-HOOKES LAW CABLE MODEL. This is the third, \+ nonlinear model.\n" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 44 "2.1. PROBLEM (OVERDAMPED FREE OSCILLATIONS)" }}{PARA 0 " " 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 76 "FREE OSCILLATIONS. Consider the general problem of free linear oscillations\n" }}{PARA 0 "" 0 "" {TEXT -1 63 " m x'' + c x' + k x=0,\n x( 0)=x0, x'(0)=v0," }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 69 " where m=1, k=16 and c is a non-negative constant. T he symbols " }}{PARA 0 "" 0 "" {TEXT -1 76 " x0 and v0 are the ini tial position and initial velocity, respectively.\n" }}{PARA 0 "" 0 " " {TEXT -1 94 " A. Suggest a value for parameter c > 0 so that the free oscillations are overdamped. This" }}{PARA 0 "" 0 "" {TEXT -1 101 " value will be used in item B below. Check your answer by solving the characteristic equation" }}{PARA 0 "" 0 "" {TEXT -1 68 " \+ using Maple's \"solve\" command (example: solve(r^2-1=0,r);). " }}{PARA 0 "" 0 "" {TEXT -1 4 " " }}{PARA 0 "" 0 "" {TEXT -1 100 " B. Use x(0)=1 and x'(0)= -2 for the initial conditions and Maple's \"dsolve\" to find the explicit" }}{PARA 0 "" 0 "" {TEXT -1 95 " \+ real solution x(t). Plot the solution x(t) for t=0 to t=5 using Map le's \"plot\" command." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 121 "EXAMPLE(Wrong parameters! Change it!)\nde:=3*diff(x (t),t,t)+1.5*diff(x(t),t)+4*x(t)=0: # Define the differential equatio n" }}{PARA 0 "" 0 "" {TEXT -1 280 "ic:=x(0)=0,D(x)(0)= -1: \+ # Define the initial conditions\ndsolve(\{de,ic\} ,x(t),method=laplace); # Symbolically solve for x(t)\nX:= unapply(rhs(%),t): # Capture the ds olve symbolic answer as a function X(t)" }}{PARA 0 "" 0 "" {TEXT -1 80 "plot(X(t),t=0..5); # Plo t the solution" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 57 "#2.1-A\n# overdamped means mr^2+cr+k=0 ha s two real roots." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "#2.1-B " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 44 "2.2. PROBLEM (UNDERDAMPED FREE OSCILLATIO NS)" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 68 "FR EE OSCILLATIONS. Consider the problem of free linear oscillations\n" } }{PARA 0 "" 0 "" {TEXT -1 61 " m x'' + c x' + k x=0,\n \+ x(0)=0, x'(0)=1," }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 73 " where m=2, c=4 and k is a non-negative constant . The overdamped case" }}{PARA 0 "" 0 "" {TEXT -1 80 " is studied here , for which the solution x(t) has infinitely many oscillations. " }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 73 " A. F ind a Hooke's constant k > 5 so that the solution x(t) changes" }} {PARA 0 "" 0 "" {TEXT -1 77 " sign infinitely many times and d ecays to zero at t=infinity. Display" }}{PARA 0 "" 0 "" {TEXT -1 58 " \+ the value of k and the exact symbolic solution." }}{PARA 0 " " 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 80 " B. Plot the ex act symbolic solution x(t) from t=0 to t=5. Estimate from the " }} {PARA 0 "" 0 "" {TEXT -1 83 " graph the decimal value of the ps eudoperiod. Display the graphical estimate" }}{PARA 0 "" 0 "" {TEXT -1 81 " and also the exact pseudoperiod 2Pi/w, where w is the natural frequency" }}{PARA 0 "" 0 "" {TEXT -1 77 " of the trig onometric term in the solution x(t) found in item 2.2.A." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 88 "Maple tip: Click \+ with the mouse on the graphic to print the cursor location (left upper " }}{PARA 0 "" 0 "" {TEXT -1 85 "corner of the maple window). The coo rdinates printed are of the form (x,y). From this" }}{PARA 0 "" 0 "" {TEXT -1 66 " coordinate information, a simple subtraction estimates t he period" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 38 "#2.2-A Define k, then solve and plot." }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 7 "#2.2-B " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 49 "2.3. \+ PROBLEM (UNDAMPED FORCED OSCILLATIONS (c=0))" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 57 "FORCED LINEAR OSCILLATIONS. Co nsider the forced problem\n" }}{PARA 0 "" 0 "" {TEXT -1 103 " x'' + 1.5625 x = 10 cos(wt),\n x(0)=0, x'(0)=0,\n\n where w i s a non-negative constant." }}{PARA 0 "" 0 "" {TEXT -1 2 " " }} {PARA 0 "" 0 "" {TEXT -1 76 " A. Choose w=3.75, so that the forc ing frequency w is 3 times larger " }}{PARA 0 "" 0 "" {TEXT -1 76 " \+ than the natural frequency w0=1.25. Solve for x(t) using dsolv e()." }}{PARA 0 "" 0 "" {TEXT -1 77 " Plot the solution x(t) \+ on a suitable interval in order to show the " }}{PARA 0 "" 0 "" {TEXT -1 48 " global behavior of the solution x(t)." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 84 " B. The solution \+ x(t) is the sum of two functions, one of period 2Pi/w and the " }} {PARA 0 "" 0 "" {TEXT -1 82 " other of period 2Pi/w0. Approxi mate the period of x(t) by examining the " }}{PARA 0 "" 0 "" {TEXT -1 88 " graph in 2.3.A. Display the estimate for the period and a lso the exact period, " }}{PARA 0 "" 0 "" {TEXT -1 87 " as cal culated from the solution formula for x(t) -- see page 341 for detail s " }}{PARA 0 "" 0 "" {TEXT -1 49 " about how to calculate the exact period." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 85 " C. Suggest a value for the forcing frequency w so t hat the oscillations exhibit" }}{PARA 0 "" 0 "" {TEXT -1 55 " \+ resonance. Show resonant behavior on a graph." }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }{TEXT -1 2 " " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 7 "#2.3-A " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 6 "# 2.3-B" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 6 "#2.3-C" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 47 "2.4. PROBLEM (DAMPED FORCED OSCILLATIONS (c>0))" }}{PARA 0 "" 0 "" {TEXT -1 31 " Consider the forced problem\n" }}{PARA 0 "" 0 "" {TEXT -1 61 " x'' + c x' + 21 x = 5cos(w t),\n x(0)=0, x'(0 )=0,\n" }}{PARA 0 "" 0 "" {TEXT -1 71 " A. Consider the damping con stants c=2, c=1 and c=1/2. Compute the" }}{PARA 0 "" 0 "" {TEXT -1 81 " amplitude function C(w) [page 346] for these three equat ions, then plot" }}{PARA 0 "" 0 "" {TEXT -1 75 " for w=0 to w=2 0 the three amplitude graphs on a single set of axes." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 72 " B. For each case \+ c=2, c=1, c=1/2, print the values w*, C* where " }}{PARA 0 "" 0 "" {TEXT -1 73 " C*=C(w*)=max \{C(w) : 0 <= w <= 20\}. The three \+ data pairs should " }}{PARA 0 "" 0 "" {TEXT -1 57 " show that \+ C* becomes larger as c tends to zero. " }}{PARA 0 "" 0 "" {TEXT -1 40 " SAVE YOUR MAPLE FILE FREQUENTLY" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 75 " Maple Hint: Use Maple's m ouse interface on the graphic of Part C. " }}{PARA 0 "" 0 "" {TEXT -1 79 " Specifically, click on a possible maximum (horizontal tan gent) in the " }}{PARA 0 "" 0 "" {TEXT -1 85 " graph to displa y the values w*, C* on the screen. Copy the values on paper." }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 66 "EXAMPLE(B eware! Wrong values!)\nF:=15: m:=1: k:=25: c:='c': w:='w':" }}{PARA 0 "" 0 "" {TEXT -1 88 "C:=(w,c)->F/sqrt((k-m*w*w)^2+(c*w)^2):\nplot(\{C( w,4),C(w,3),C(w,2)\},w=0..15,color=black);" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 51 "#2.4-A \+ Plot C(w), three graphics on one set of axes" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 42 "#2.4-B Table of six data values for w*, C*" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 43 "2.5. PROBLEM (LARGE SUSTAINED OSCILLATIONS)" }}{PARA 0 " " 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 121 "NONLINEAR MODEL W ITH GEOMETRY INCLUDED.\nConsider the nonlinear, forced, damped oscilla tor equation for torsional motion, " }}{PARA 0 "" 0 "" {TEXT -1 240 "w ith bridge geometry included,\n\n x'' + 0.05 x' + 2.4 sin(x)cos(x) \+ = 0.06 cos (12 t/10) ,\n x(0) = x0, x'(0) = v0\n\nand its correspo nding linearized equation\n\n x'' + 0.05 x' + 2.4 x = 0.06 cos (12 t /10) , \n x(0) = x0, x'(0) = v0.\n" }}{PARA 0 "" 0 "" {TEXT -1 84 "The spring-mass system parameters are m=1, c = 0.05, k = 2.4, w = 1 .2 , F = 0.06." }}{PARA 0 "" 0 "" {TEXT -1 155 "Maple code used to so lve and plot the solutions appears below.\n\n # WARNING: set the par ameters on the second line! Use \"copy as maple text\" for maple 6+." }}{PARA 0 "" 0 "" {TEXT -1 58 " m:=1: F := 0.06: w := 1.2: m:=1: c := 0.05: k:= 2.4: " }}{PARA 0 "" 0 "" {TEXT -1 29 " x0:=0: v0:=0: a :=0: b:=50:" }}{PARA 0 "" 0 "" {TEXT -1 198 " deNonLinear:= m*diff(x (t),t,t) + c*diff(x(t),t) + k*sin(x(t))*cos(x(t)) = F*cos(w*t):\n de Linear:= m*diff(x(t),t,t) + c*diff(x(t),t) + k*x(t) = F*cos(w*t):\n \+ with(DEtools): opts:=stepsize=0.1:" }}{PARA 0 "" 0 "" {TEXT -1 158 " \+ DEplot(deNonLinear,x(t),t=a..b,[[x(0)=x0,D(x)(0)=v0]],opts,title='No nLinear');\n DEplot(deLinear,x(t),t=a..b,[[x(0)=x0,D(x)(0)=v0]],opts ,title='Linear');\n" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 " " {TEXT -1 103 " A. Let x0=0, v0=0. Plot the solutions of the li near and nonlinear equations from t=160 to t=260." }}{PARA 0 "" 0 "" {TEXT -1 79 " These plots represent the steady state solutions of the two equations." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 103 " B. Let x0=1.2, v0=0. Plot the solutions of the linear and nonlinear equations from t=220 to t=320." }}{PARA 0 "" 0 " " {TEXT -1 108 " These plots represent the steady state solution s of the two equation, with new starting value x0=1.2." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 90 " C. Argue in a s entence why the two linear plots have to be identical, based upon the \+ " }}{PARA 0 "" 0 "" {TEXT -1 88 " superposition formula x(t)= xh(t)+xss(t), even though the homogeneous solution " }}{PARA 0 "" 0 " " {TEXT -1 103 " xh(t) is different for the two plots. Please include a discussion of the amplitude of xh(t) " }}{PARA 0 "" 0 "" {TEXT -1 44 " on the corresponding t-interval. " }}{PARA 0 " " 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 94 " D. Determine \+ the ratio of the apparent amplitudes (a number > 1) for the nonlinear \+ plots." }}{PARA 0 "" 0 "" {TEXT -1 88 " Explain why \"large su stained oscillations\" is an appropriate description of the" }}{PARA 0 "" 0 "" {TEXT -1 41 " nonlinear steady-state behavior." }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }} }{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 6 "#2.5-A" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 6 "#2.5-B" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 45 "#2.5-C\n# The plots are identical because ...." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 6 "#2.5-D" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 55 "2.6. PROBLEM . ( MCKENNA'S NON-HOOKE'S LAW CABLE MODEL)" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 68 " MCKENNA'S NON-HOOKE'S LAW CABL E MODEL FOR THE TACOMA NARROWS BRIDGE" }}{PARA 0 "" 0 "" {TEXT -1 0 " " }}{PARA 0 "" 0 "" {TEXT -1 73 "The model of McKenna studies the brid ge with a nonlinear, forced, damped " }}{PARA 0 "" 0 "" {TEXT -1 78 "o scillator equation for torsional motion that accounts for the non-Hook e's law" }}{PARA 0 "" 0 "" {TEXT -1 75 "cables coupled to the equation s for vertical motion. The equations in this" }}{PARA 0 "" 0 "" {TEXT -1 78 "case couple the torsional motion with the vertical motion . The equations are:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 63 " x'' + c x' - k G(x,y) = F sin wt, x(0) = x0, x' (0) = x1," }}{PARA 0 "" 0 "" {TEXT -1 63 " y'' + c y' + (k/3) H(x,y ) = g , y(0) = y0, y'(0) = y1," }}{PARA 0 "" 0 "" {TEXT -1 0 "" } }{PARA 0 "" 0 "" {TEXT -1 81 "where x(t) is the torsional motion and y (t) is the vertical motion. The functions" }}{PARA 0 "" 0 "" {TEXT -1 74 "G(x,y) and H(x,y) are the models of the force generated by the cab le when " }}{PARA 0 "" 0 "" {TEXT -1 81 "it is contracted and stretche d. Below is sample code for writing the differential" }}{PARA 0 "" 0 " " {TEXT -1 77 "equations and for plotting the solutions. It is ready t o copy with the mouse." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 276 "with(DEtools): \+ \nw := 1.3: F := 0.05: f(t) := F*sin(w*t): \+ \nc := 0.01: k1 := 0.2: k2 := 0.4: g := 9. 8: L := 6: \nSTEP:=x->piecewise(x<0,0,1):\nfp(t) := \+ y(t)+(L*sin(x(t))):" }}{PARA 0 "" 0 "" {TEXT -1 384 "fm(t) := y(t)-(L* sin(x(t))):\nSm(t) := STEP(fm(t))*fm(t):\nSp(t) := STEP(fp(t))*fp(t): \nsys := \{ \n diff(x(t),t,t) + c*diff(x(t),t) - k1*co s(x(t))*(Sm(t)-Sp(t))=f(t),\n diff(y(t),t,t) + c*diff(y(t),t) + k2 *(Sm(t)+Sp(t)) = g\}:\nic := [[x(0)=0, D(x)(0)=0, y(0)=27.25, D(y)(0)= 0]]:\nvars:=[x(t),y(t)]: \nopts:=stepsize=0.1: \nDEplot(sys,vars,t= 0..300,ic,opts,scene=[t,x]);\n" }}{PARA 0 "" 0 "" {TEXT -1 96 "The ama zing thing that happens in this simulation is that the large vertical \+ oscillations take " }}{PARA 0 "" 0 "" {TEXT -1 80 "all the tension ou t of the springs and they induce large torsional oscillations." }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 86 " A. TORSI ONAL OSCILLATION PLOT. Get the sample code above to produce the plot o f x(t) " }}{PARA 0 "" 0 "" {TEXT -1 39 " [that's what scene=[t,x] means]. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 97 " B. Estimate the number of degrees the roadway oscillates based on the plot; recall that x in the" }}{PARA 0 "" 0 "" {TEXT -1 95 " p lot is reported in radians. Comment on the agreement of this result wi th historical data." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 " " {TEXT -1 94 " Tip: Average the five largest amplitudes in the pl ot to find an average maximum amplitude" }}{PARA 0 "" 0 "" {TEXT -1 75 " for t=0 to t=300. Convert to degrees using Pi radians = 180 degrees." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 81 " C. VERTICAL OSCILLATION PLOT. Modify the DEplot code to scene=[t, y] and plot the" }}{PARA 0 "" 0 "" {TEXT -1 98 " oscillation y(t) on t=0 to t=300. The plot is supposed to show 30-foot vertical oscill ations" }}{PARA 0 "" 0 "" {TEXT -1 92 " that dampen to 7-foot ver tical oscillations after 300 seconds.Comment on the agreement" }} {PARA 0 "" 0 "" {TEXT -1 98 " between these oscillation results a nd the historical data for Tacoma Narrows, especially the" }}{PARA 0 " " 0 "" {TEXT -1 66 " visual data present in the film clip of the \+ bridge disaster." }}{PARA 0 "" 0 "" {TEXT -1 6 " " }}{PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 32 " #2.6-A Torsional plot t-versus-x" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 59 "#2.6-B Roadway oscillation estimate in degrees + com ments." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 43 "#2.6-C Vertical p lot t-versus-y + comments." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{MARK "26 0 0" 43 }{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 } {PAGENUMBERS 0 1 2 33 1 1 }