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0 0 1 }{CSTYLE "" -1 383 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 384 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 385 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 386 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 1 } {CSTYLE "" -1 387 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 388 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 1 }{PSTYLE "Normal" -1 0 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 }0 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }} {SECT 0 {EXCHG {PARA 0 "" 0 "" {TEXT 256 27 "Coupled spring-mass syste ms" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 658 "Th e spring-mass system under gravity (without and with friction) was dis cussed in the worksheet HOmovie.mws (without and with the application \+ of a driving force). Here we discuss a system with twice the number of degrees of freedom, namely a spring-mass system to which another spri ng-mass system is connected. We do not include gravity, and are intere sted in studying the the added features in the solutions. Martha Abell and James Braselton (Differential Equations using MapleV) obtain the \+ solutions to the problem using the Laplace transform (a technique used particularly by engineering students to solve ODes arising in mechani cs and electric circuits)." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 205 "The set-up of the problem is as follows: we assu me horizontal motion (gravity is balanced by an air track with a varia ble amount of friction; we ignore the friction in the springs). We lab el the masses as " }{TEXT 291 1 "m" }{TEXT -1 7 "1, and " }{TEXT 292 1 "m" }{TEXT -1 27 "2, the spring constants as " }{TEXT 293 1 "k" } {TEXT -1 6 "1 and " }{TEXT 294 1 "k" }{TEXT -1 20 "2, the positions as " }{TEXT 263 1 "x" }{TEXT -1 2 "1(" }{TEXT 264 1 "t" }{TEXT -1 6 ") a nd " }{TEXT 265 1 "x" }{TEXT -1 2 "2(" }{TEXT 266 1 "t" }{TEXT -1 53 " ), etc.. Spring 1 is attached to a fixed point, mass " }{TEXT 290 1 "m " }{TEXT -1 48 "1 to its other end, and spring 2 is attached to " } {TEXT 289 1 "m" }{TEXT -1 9 "1, while " }{TEXT 288 1 "m" }{TEXT -1 54 "2 is attached to the other end of spring2. We measure " }{TEXT 267 1 "x" }{TEXT -1 2 "1(" }{TEXT 268 1 "t" }{TEXT -1 6 ") and " }{TEXT 269 1 "x" }{TEXT -1 2 "2(" }{TEXT 270 1 "t" }{TEXT -1 89 ") from the equil ibrium positions of springs 1 and 2 respectively. This implies that ma ss " }{TEXT 287 1 "m" }{TEXT -1 55 "1 experiences a Hooke force from s pring 1 in the form -" }{TEXT 273 1 "k" }{TEXT -1 2 "1*" }{TEXT 272 1 "x" }{TEXT -1 2 "1(" }{TEXT 271 1 "t" }{TEXT -1 72 "), and from spring 2 (which is attached at the other end) in the form -" }{TEXT 286 1 " k" }{TEXT -1 3 "2*(" }{TEXT 285 1 "x" }{TEXT -1 2 "1(" }{TEXT 284 1 "t " }{TEXT -1 2 ")-" }{TEXT 283 1 "x" }{TEXT -1 2 "2(" }{TEXT 282 1 "t" }{TEXT -1 115 ")). The force for spring 1 is obvious, and the latter i s true since: (i) if the equilibrium position for spring 2, " }{TEXT 274 1 "x" }{TEXT -1 2 "2(" }{TEXT 275 1 "t" }{TEXT -1 61 ") coincides \+ with the equilibrium position of spring 1, i.e., " }{TEXT 281 1 "x" } {TEXT -1 2 "1(" }{TEXT 280 1 "t" }{TEXT -1 33 ") there is no force, an d (ii) if " }{TEXT 279 1 "x" }{TEXT -1 2 "2(" }{TEXT 278 1 "t" }{TEXT -1 17 ") is larger than " }{TEXT 277 1 "x" }{TEXT -1 2 "1(" }{TEXT 276 1 "t" }{TEXT -1 189 ") the second spring tends to balance the effe ct of the first spring, while if it is smaller, it will pull in the sa me direction as spring 1. Thus, we have for spring 1 (including fricti on):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "restart; with(plots ):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 47 "m1:='m1': m2:='m2': k 1:='k1': k2:='k2': b:='b':" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "x1(t):='x1(t)': x2(t):='x2(t)':" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 67 "NE1:=m1*diff(x1(t),t$2)=-k1*x1(t)-k2*(x1(t)-x2(t))-b* diff(x1(t),t);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 232 "For the second mass we have a simpler force assignment, as it is free at one end; ho wever, the the force from spring 2 acts on it in a way that depends on both positions (similarly, but opposite in sign as in the action on m ass m1):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 58 "NE2:=m2*diff(x2 (t),t$2)=-k2*(x2(t)-x1(t))-b*diff(x2(t),t);" }}}{EXCHG {PARA 0 "" 0 " " {TEXT -1 81 "We assume MKSA (SI) units again, i.e., masses in kg, di stances in m, times in s, " }{TEXT 257 1 "b" }{TEXT -1 10 " in Ns/m, \+ " }{TEXT 258 1 "k" }{TEXT -1 13 " in N/m, etc." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 107 "The initial conditions h ave to be specified in some way. We pick both positions at equilibrium , and masses " }{TEXT 262 1 "m" }{TEXT -1 7 "1 and " }{TEXT 261 1 "m " }{TEXT -1 197 "2 moving with the same velocity. A total of 4 quantit ies have to be specified, as we have two coupled 2nd order ODEs. For a start, we pick the two masses to be equal, and vary the spring consta nts." }}{PARA 0 "" 0 "" {TEXT -1 76 "We can ask ourselves whether we c an set them, e.g., in such a way that mass " }{TEXT 260 1 "m" }{TEXT -1 27 "2 essentially follows mass " }{TEXT 259 1 "m" }{TEXT -1 2 "1." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 62 "First o f all, one has to understand that the location of mass " }{TEXT 309 1 "m" }{TEXT -1 51 "2 with respect to the laboratory frame is given by \+ " }{TEXT 308 1 "x" }{TEXT -1 2 "1(" }{TEXT 307 1 "t" }{TEXT -1 2 ")+" }{TEXT 306 1 "x" }{TEXT -1 2 "2(" }{TEXT 305 1 "t" }{TEXT -1 2 ")." }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 97 "Before be ginning to play seriously with the solutions it helps to figure out so me limiting cases:" }}{PARA 0 "" 0 "" {TEXT -1 12 "(i) suppose " } {TEXT 303 1 "k" }{TEXT -1 22 "1 is much bigger than " }{TEXT 302 1 "k " }{TEXT -1 130 "2, i.e., spring 1 is much stiffer than spring2, and t he motion is started with spring 1 out of equilibrium (position or vel ocity):" }}{PARA 0 "" 0 "" {TEXT -1 68 "in this case equation NE1 corr esponds to an harmonic oscillator for " }{TEXT 304 1 "x" }{TEXT -1 80 "1 with a small perturbation; it makes sense to find an approximate so lution for " }{TEXT 314 1 "x" }{TEXT -1 2 "1(" }{TEXT 313 1 "t" } {TEXT -1 26 ") without the coupling to " }{TEXT 312 1 "x" }{TEXT -1 2 "2(" }{TEXT 311 1 "t" }{TEXT -1 28 "); in the equation for mass " } {TEXT 310 1 "m" }{TEXT -1 37 "2 we can think of the term involving " } {TEXT 316 1 "x" }{TEXT -1 2 "1(" }{TEXT 315 1 "t" }{TEXT -1 51 ") as a predetermined driving force. The motion for " }{TEXT 317 1 "m" } {TEXT -1 153 "2 in the lab frame will be complicated. It is worthwhile to begin the explorations with this case, and to observe whether the \+ expectations are fulfilled." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 134 "(ii) same as in (i), except that spring 2 is s tarted away from equilibrium, while spring 1 starts at rest in the equ ilibrium position " }{TEXT 318 1 "x" }{TEXT -1 7 "1(0)=0:" }}{PARA 0 " " 0 "" {TEXT -1 27 "in this case the motion in " }{TEXT 319 1 "x" } {TEXT -1 22 "1 remains small; mass " }{TEXT 320 1 "m" }{TEXT -1 45 "2 \+ behaves as a perturbed harmonic oscillator." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 15 "(iii) suppose " }{TEXT 322 1 " k" }{TEXT -1 22 "2 is much bigger than " }{TEXT 321 1 "k" }{TEXT -1 2 "1:" }}{PARA 0 "" 0 "" {TEXT -1 126 "irrespective of whether the initi al energy (motion) is in spring 1 or spring 2, the motion corresponds \+ of oscillations around " }{TEXT 324 1 "x" }{TEXT -1 2 "1(" }{TEXT 323 1 "t" }{TEXT -1 2 ")=" }{TEXT 325 1 "x" }{TEXT -1 2 "2(" }{TEXT 326 1 "t" }{TEXT -1 67 "), the net result being damped near-harmonic oscilla tions for mass " }{TEXT 327 1 "m" }{TEXT -1 19 "2 in the lab frame." } }{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 189 "(iv) co mparable spring constants: the motion is complicated with an exchange \+ of energy between the springs happening continuously. This is the resu lt of both oscillators being in resonance." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 36 "m1:=1; m2:=1; k1: =4; k2:=5; b:=1/10;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 44 "IC:= x1(0)=0,D(x1)(0)=1/2,x2(0)=0,D(x2)(0)=0;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 66 "This initial condition implies that both masses move in u nison at " }{TEXT 295 1 "t" }{TEXT -1 85 "=0, i.e., there is no relati ve displacement, and no relative velocity between masses " }{TEXT 297 1 "m" }{TEXT -1 6 "2 and " }{TEXT 296 1 "m" }{TEXT -1 2 "1." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 158 "In Maple 5.01 \+ one needs to specify the Laplace method to obtain a solution. In Maple 6.0 a solution is obtained without this hint (and by a different meth od)." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 55 "sol:=dsolve(\{NE1,N E2,IC\},\{x1(t),x2(t)\},method=laplace);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "assign(sol);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 54 " Now that the solution is assigned, we can simply type:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 6 "x1(t);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 8 "but not:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 6 "x1(1 );" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "simplify(%);" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 380 "Therefore, we define mappings bas ed on x1(t) and x2(t). Note that assigning a solution has the drawback , that one has to unassign the dependent variables before re-using it. This was the reason for statements of the type: x1(t):='x1(t)': befor e defining the Newton equation. It has to be executed before re-using \+ the assignment with a new choice of parameters in the same session." } }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 51 "X1:=unapply(Re(x1(t)),t): \+ X2:=unapply(Re(x2(t)),t):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 63 "plot([X1(t),X2(t),X1(t)+X2(t)],t=0..40,color=[red,blue,green]);" } }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 74 "From the graph of the solutions \+ for both positions we note several things:" }}{PARA 0 "" 0 "" {TEXT -1 9 "(i) mass " }{TEXT 300 1 "m" }{TEXT -1 17 "2 indeed follows " } {TEXT 301 1 "m" }{TEXT -1 25 "1. The interpretation of " }{TEXT 329 1 "x" }{TEXT -1 2 "2(" }{TEXT 328 1 "t" }{TEXT -1 91 ") is that the seco nd spring-mass system does carry out oscillations around its equilibri um;" }}{PARA 0 "" 0 "" {TEXT -1 285 "(ii) the solutions imply that we \+ are running a simulation of some kind of superspring: the springs can \+ move to both sides of the equilibrium position (this means that theu a re pre-stretched), but more importantly, the two masses can switch pos itions (they can pass through each other)." }}{PARA 0 "" 0 "" {TEXT -1 29 "(iii) for the motion of mass " }{TEXT 330 1 "m" }{TEXT -1 94 "2 in the lab frame we obtain deformed damped harmonic oscillations for \+ many intial conditions." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 379 "The next step in understanding the solut ion is to generate an animation as in HOmovie.mws. We draw the system \+ horizontally to stress that gravity is not involved (apart from provid ing the friction, which is reduced by the airtrack). For the drawing w e will choose the masses to be thin vertical slabs. As before, we pick a width for the springs, as well as their number of turns." }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "w:=0.2: nT:=10:" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 145 "Given a position for the mass y(t) we di vide this length into 2 nT segments, and generate a list of points whi ch alternate on the left and right." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 70 "DL:=proc(t) local i,L,d1,d2,le1,le2,t0,x10; global w, nT; t0:=evalf(t);" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 68 "d1:=evalf(X1(t 0)/(2*nT+1)); le1:=1; le2:=1; L:=[[0.,0.],[d1*0.5,w]]:" }}{PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 488 "for i from 1 to nT do: L:=[op(L),[d1*(2*i-0.5 ),-w],[d1*(2*i+0.5),w]]; od: L:=[op(L),[X1(t0),0.],[X1(t0)-w/80,1.1*w] ,[X1(t0)+w/80,1.1*w],[X1(t0)+w/80,-1.1*w],[X1(t0)-w/80,-1.1*w],[X1(t0) ,0.]]; d2:=evalf(X2(t0)/(2*nT+1)); x10:=X1(t0); L:=[op(L),[x10+d1*0.5, w]]: for i from 1 to nT do: L:=[op(L),[x10+d2*(2*i-0.5),-w],[x10+d2*(2 *i+0.5),w]]; od: L:=[op(L),[x10+X2(t0),0.],[x10+X2(t0)-w/80,1.1*w],[x1 0+X2(t0)+w/80,1.1*w],[x10+X2(t0)+w/80,-1.1*w],[x10+X2(t0)-w/80,-1.1*w] ,[x10+X2(t0),0.]]; end:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 37 " PL:=seq(plot(DL(j_t/10)),j_t=1..320):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 39 "display(PL,insequence=true,axes=boxed);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 76 "An interesting graph is obtained by plott ing parametrically the position of " }{TEXT 298 1 "m" }{TEXT -1 21 "2 \+ vs the position of " }{TEXT 299 1 "m" }{TEXT -1 90 "1: if we choose bo th masses to be at equilibrium initially, the plot starts at the origi n." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "plot([X1(t),X2(t),t=0..10]);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 140 "We proceed by performing a simulation of a system that is more easily set up on an air track. Three springs (e .g., of equal spring constant " }{TEXT 331 1 "k" }{TEXT -1 32 ") hold \+ two (e.g., equal) masses " }{TEXT 332 1 "m" }{TEXT -1 6 "1 and " } {TEXT 334 1 "m" }{TEXT -1 145 "2 between themselves. The springs are p re-stretched by this set-up, so that they all can be compressed and st retched by the motion of the masses." }}{PARA 0 "" 0 "" {TEXT -1 5 "Ma ss " }{TEXT 338 1 "m" }{TEXT -1 56 "1 is attached on one side to sprin g 1 (characterized by " }{TEXT 337 1 "k" }{TEXT -1 9 "1), mass " } {TEXT 336 1 "m" }{TEXT -1 12 "2 to spring " }{TEXT 335 1 "k" }{TEXT -1 36 "2, and both are connected by spring " }{TEXT 333 1 "k" }{TEXT -1 2 "3." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 262 "It is best to use a coordinate system that uses the equilibrium p ositions for springs 1 and 2 as the coordinates and of opposite sign t o each other. That way the equaations of motion are cast into symmetri cal form. The middle spring's compression is measured by " }{TEXT 341 1 "x" }{TEXT -1 2 "1+" }{TEXT 340 1 "x" }{TEXT -1 2 "2." }}{PARA 0 "" 0 "" {TEXT -1 44 "For this choice of coordinates the force on " } {TEXT 339 1 "m" }{TEXT -1 16 "1 consists of -" }{TEXT 345 1 "k" } {TEXT -1 11 "1*x1, and -" }{TEXT 344 1 "k" }{TEXT -1 3 "3*(" }{TEXT 342 1 "x" }{TEXT -1 2 "1+" }{TEXT 343 1 "x" }{TEXT -1 19 "2). Similarl y, for " }{TEXT 346 1 "m" }{TEXT -1 11 "2 we have -" }{TEXT 350 1 "k" }{TEXT -1 11 "2*x2, and -" }{TEXT 349 1 "k" }{TEXT -1 3 "3*(" }{TEXT 347 1 "x" }{TEXT -1 2 "1+" }{TEXT 348 1 "x" }{TEXT -1 2 "2)" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "restart; with(plots):" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 57 "m1:='m1': m2:='m2': k1:='k1' : k2:='k2': k3:='k3': b:='b':" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "x1(t):='x1(t)': x2(t):='x2(t)':" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 67 "NE1:=m1*diff(x1(t),t$2)=-k1*x1(t)-k3*(x1(t)+x2(t))-b* diff(x1(t),t);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 67 "NE2:=m2*d iff(x2(t),t$2)=-k2*x2(t)-k3*(x1(t)+x2(t))-b*diff(x2(t),t);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 99 "It helps to choose a weak coupling betwee n the two oscillators, i.e., a weak spring between masses " }{TEXT 352 1 "m" }{TEXT -1 6 "1 and " }{TEXT 351 1 "m" }{TEXT -1 3 "2.:" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 46 "m1:=1; m2:=1; k1:=1; k2:=1; \+ k3:=1/10; b:=1/50;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 42 "IC:=x 1(0)=0,D(x1)(0)=1,x2(0)=0,D(x2)(0)=0;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 55 "sol:=dsolve(\{NE1,NE2,IC\},\{x1(t),x2(t)\},method=lap lace);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "assign(sol);" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 51 "X1:=unapply(Re(x1(t)),t): X2 :=unapply(Re(x2(t)),t):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 45 " plot([X1(t),X2(t)],t=0..60,color=[red,blue]);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 158 "Interestingly enough, if we look at the difference of the solutions, we find that it satisfies simple, damped harmonic moti on (subtract the Newton equations):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 37 "plot(X1(t)-X2(t),t=0..60,color=blue);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 111 "We observe the so-called beats phenomeno n: the masses oscillate alternatingly. The opposite sign definition of " }{TEXT 354 1 "x" }{TEXT -1 6 "1 and " }{TEXT 353 1 "x" }{TEXT -1 58 "2 is responsible for the apparent intial movement of mass " } {TEXT 356 1 "m" }{TEXT -1 50 "2 in the opposite direction of the oscil lation of " }{TEXT 355 1 "m" }{TEXT -1 2 "1." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 144 "Now we need to re-write our an imation program. While it was straightforward to understand the beat p henomenon based on the deviations of masses " }{TEXT 358 1 "m" }{TEXT -1 6 "1 and " }{TEXT 357 1 "m" }{TEXT -1 114 "2 from the equilibrium p ositions, our graph of the entire system requires a translation into t he laboratory frame." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 " " {TEXT -1 130 "We omit the outer two springs in the drawing, and choo se an arbitrary length (2*Le1) for the initial length of the middle s pring." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "w:=0.2: nT:=10:" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 145 "Given a position for the mass \+ y(t) we divide this length into 2 nT segments, and generate a list of \+ points which alternate on the left and right." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 70 "DL:=proc(t) local i,L,d1,d2,t0,x10,x20,Le1; glob al w,nT; t0:=evalf(t);" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 211 "Le1:=1; \+ d1:=evalf((2*Le1-X1(t0)-X2(t0))/(2*nT+1)); x10:=-Le1+X1(t0); x20:=Le1- X2(t0); L:=[[x10,0.],[x10-w/80,1.1*w],[x10+w/80,1.1*w],[x10+w/80,-1.1* w],[x10-w/80,-1.1*w],[x10-w/80,1.1*w],[x10,0.],[x10+d1*0.5,w]]:" }} {PARA 0 "> " 0 "" {MPLTEXT 1 0 186 "for i from 1 to nT do: L:=[op(L),[ x10+d1*(2*i-0.5),-w],[x10+d1*(2*i+0.5),w]]; od: L:=[op(L),[x20,0.],[x2 0-w/80,1.1*w],[x20+w/80,1.1*w],[x20+w/80,-1.1*w],[x20-w/80,-1.1*w],[x2 0,0.]]; end:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 39 "PL:=seq(plo t(DL(2*j_t/10)),j_t=0..350):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 39 "display(PL,insequence=true,axes=boxed);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 215 "If we extend the number of frames such that at the en d the solution almost connects to the intial condition (for small enou gh friction this is achievable), then the continuous simulation in HTM L will look realistic." }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 79 "One sho uld try initial conditions that correspond to the two fundamental mode s:" }}{PARA 0 "" 0 "" {TEXT -1 72 "(i) the masses moving together (no \+ compression of the spring connecting " }{TEXT 365 1 "m" }{TEXT -1 6 "1 and " }{TEXT 364 1 "m" }{TEXT -1 4 "2); " }}{PARA 0 "" 0 "" {TEXT -1 77 "this can be achieved by picking opposite, equal in magnitude veloc ities for d" }{TEXT 360 1 "x" }{TEXT -1 3 "1/d" }{TEXT 361 1 "t" } {TEXT -1 6 " and d" }{TEXT 363 1 "x" }{TEXT -1 3 "2/d" }{TEXT 362 1 "t " }{TEXT -1 4 " at " }{TEXT 359 1 "t" }{TEXT -1 7 "=0 (as " }{TEXT 367 1 "x" }{TEXT -1 6 "1 and " }{TEXT 366 1 "x" }{TEXT -1 22 "2 have o pposite sign)." }}{PARA 0 "" 0 "" {TEXT -1 42 "(ii) the masses moving \+ against each other." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 " " {TEXT -1 95 "These two modes correspond to motions described by a si ngle Newton equation in the coordinates " }{TEXT 368 1 "q" }{TEXT -1 2 "1=" }{TEXT 369 1 "x" }{TEXT -1 2 "1+" }{TEXT 370 1 "x" }{TEXT -1 7 "2, and " }{TEXT 371 1 "q" }{TEXT -1 2 "2=" }{TEXT 372 1 "x" }{TEXT -1 2 "1-" }{TEXT 373 1 "x" }{TEXT -1 132 "2 respectively.Verify that t he two Newton equations decouple in these two cases, and that the osci llation frequencies are different." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 117 "There are mathematical ways to understan d the appearance of so-called normal modes, i.e., modes where the moti ons of " }{TEXT 375 1 "m" }{TEXT -1 6 "1 and " }{TEXT 374 1 "m" } {TEXT -1 234 "2 reduce to a single degree of freedom. One technique is to write the system of two second-order equations as a system of four coupled first-order equations, and to diagonalize the coefficient mat rix. The transformed variables become " }{TEXT 377 1 "q" }{TEXT -1 6 " 1 and " }{TEXT 376 1 "q" }{TEXT -1 2 "2." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 304 "Let us see whether the the soluti on for one of the masses (in the general, coupled case, i.e., for an i ntial condition where the beat phenomenon occurs) knows about the freq uencies associated with the normal modes. A Fourier transform from the time to the frequency domain should reveal this information." }} {PARA 0 "" 0 "" {TEXT -1 199 "A look at the solution shows indeed that we have a superposition of two solutions with particular frequencies. The symbolic solution was given above. We read out that both frequenc ies are very close:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "eval f(X1(t));" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 316 "For the beat phenom enon to be prominent we need, in fact, the closeness of the two freque ncies. Instrument tuners (and orchestras) know that two strings are ou t of tune if the wobbling sound associated with the beat phenomenon is heard. The wobble disappears only if the two frequencies get very clo se to each other." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 67 "Fourier analysis can be performed analytically on simple \+ functions." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "with(inttrans );" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "fourier(sin(t),t,f); " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "fourier(sin(t)+cos(t/2) ,t,f);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 155 "Note that the cosine c omponents acquire real amplitudes, while the sine components have imag inary amplitudes, and that the frequency spectrum is symmetric." }} {PARA 0 "" 0 "" {TEXT -1 300 "As we have an initial-value problem, and the damping would result in a blow-up for negative times, we are inte rested in a transform that integrates over positive times only. Obviou s candidates are fourier cosine, and fourier sine transforms. The four ier sine transform gives a very reasonable result." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "fouriersin(x1(t),t,f);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "plot(%^2,f=0.9..1.2);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 326 "We \+ find that the frequency spectrum has a finite width about the natural \+ frequencies. We leave it as an exercise to vary the damping constant a nd to verify that the spectrum approaches that of delta functions at t he natural frequencies for the case of weaker damping, and vice versa \+ that it broadens, if the friction constant " }{TEXT 378 1 "b" }{TEXT -1 14 " is increased." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 104 "For the fourier cosine transform we find that the squ ared amplitude vanishes at the natural frequencies." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "fouriercos(x2(t),t,f);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "plot(%^2,f=0.9..1.2);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 108 "For more complicated time signals one performs d iscrete Fourier transforms to unravel the frequency content." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 1197 "We repeat and strengthen a statement made before about the natural frequencies. One might wonder why two different frequencies appear for the apparently \+ symmetrical system of two identical masses. We have chosen identical s prings to attach the masses to their respective sides and a weak sprin g to couple them together. The student is encouraged to re-run the sim ulations, and the frequency analysis with a stronger (and weaker) coup ling between the two masses. The two different natural modes arose obv iously for the symmetrical set-up as those where the masses do not com press the spring which couples them (presumably the lower-energy mode) , and the one where in addition to compressing/expanding their support springs the masses compress/expand the coupling spring (the higher-fr equency mode). It is for weak coupling that the frequencies become nea rly degenerate (almost equal), and the beat phenomenon occurs (which i s even more fascinating to watch in real life than on the computer scr een; try to put together your own pendula coupled by a very weak rubbe r band). Which of the two motions has the higher/lower frequency is de monstrated by the decoupling of the equations carried out below." }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 117 "We compl ete the calculations by carrying out the transformation of the Newton \+ equations for our choice of parameters." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "x1(t):='x1(t)': x2(t):='x2(t)':" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 46 "m1:=1; m2:=1; k1:=1; k2:=1; k3:=1/10; b:=1/50 ;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 67 "NE1:=m1*diff(x1(t),t$2 )=-k1*x1(t)-k3*(x1(t)+x2(t))-b*diff(x1(t),t);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 67 "NE2:=m2*diff(x2(t),t$2)=-k2*x2(t)-k3*(x1(t)+x2(t ))-b*diff(x2(t),t);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 52 "NE1p :=subs(x1(t)=q1(t)-q2(t),x2(t)=q1(t)+q2(t),NE1);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 52 "NE2p:=subs(x1(t)=q1(t)-q2(t),x2(t)=q1(t)+q2(t ),NE2);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "NE1pp:=NE1p+NE2p ;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "NE2pp:=NE1p-NE2p;" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 76 "We have decoupled the two differen tial equations. Let us find the solutions." }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 26 "sol1:=dsolve(NE1pp,q1(t));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "evalf(rhs(sol1));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "sol2:=dsolve(NE2pp,q2(t));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "evalf(rhs(sol2));" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 57 "Now, the lower frequency occurs in the mode described by \+ " }{TEXT 380 1 "q" }{TEXT -1 2 "2(" }{TEXT 379 1 "t" }{TEXT -1 6 ") = \+ 2(" }{TEXT 381 1 "x" }{TEXT -1 2 "2(" }{TEXT 382 1 "t" }{TEXT -1 4 ") \+ - " }{TEXT 383 1 "x" }{TEXT -1 2 "1(" }{TEXT 384 1 "t" }{TEXT -1 101 " )). This mode is the one where the masses run in parallel (remember th at according to our convention " }{TEXT 386 1 "x" }{TEXT -1 6 "1 and \+ " }{TEXT 385 1 "x" }{TEXT -1 31 "2 have opposite sign, and that " } {TEXT 387 1 "x" }{TEXT -1 2 "1/" }{TEXT 388 1 "x" }{TEXT -1 136 "2 mea sures the displacement of the respective mass from the equlibrium poin t of the spring which connects it to its respective support)." }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 373 "For the \+ case of arbitrary masses and spring constants one would have to find t he right linear combinations to perform the decoupling. It is for this case that one prefers the mathematical machinery of decoupling system s of first-order differential equations with constant coefficients by \+ means of a similarity transformation (diagonalization of the matrix of coefficients)." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}} {MARK "0 2 0" 51 }{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 }{PAGENUMBERS 0 1 2 33 1 1 }