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# viewnote6 - PHYS 302 Unit 6 Viewing Notes – Athabasca...

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Unformatted text preview: PHYS 302: Unit 6 Viewing Notes – Athabasca University PHYS 302: Vibrations and Waves – Unit 6 Viewing Notes In the lecture for Unit 6, Dr. Lewin analyzes the system of the double pendulum, that is, two bobs on one string. French does not include this in the textbook, although the double pendulum system is somewhat similar to the two coupled pendulum case done in the last part of the readings for Unit 5. Dr. Lewin discussed and demonstrated the double pendulum near the end of the video lecture for Unit 5 but did not analyze it. In the small angle approximation, the tension in the upper string supports both (identical) masses, and the tension in the lower string supports just one, so T1=2mg and T2=mg. The restoring force on the lower bob is –T2sinθ2, and sinθ2=(x2-x1)/l. So, NII for this bob is mx T2 sin 2 mg l 2 ( x2 x1 ) . Dividing by m, with 0 g , l 2 2 2 0 x2 0 x1 0 (4:10). Two forces act on the first bob, so NII gives x mx1 T1 sin 1 T2 sin 2 2 mg 2 0 g l , x1 l mg ( x2 x1 ) l (5:30), where we used sinθ1=x1/l. Dividing by m, with 1 3 x x 0 . Checking your work is wise: at this stage we may not intuitively expect the “3” x 2 01 2 02 beside one of the terms (7:15). With x1 C1 cos t and x2 C2 cos t trial solutions, we need to determine three parameters. By taking the derivatives and noting that we do not need to keep the common “cos” terms, there are two equations: from the second DE we get initially get 2 2 C1 2 30 C1 0 C2 0 , and from the first equation we 2 2 2 2 C2 2 C20 C10 0 , then rearrange it to C10 C2 (0 2 ) 0 (9:45). These two equations have three unknowns, but we will be happy just to find the ratio of the Cs and the frequency ω; that is, we can settle for two parameters. At this point, some linear algebra is required. The aim is to be able to solve systems of simultaneous linear equations. The method, which can be generalized to any number of solvable linear equations with the same number of variables as equations, is called Cramer’s Rule. While Cramer’s Rule is a handy way to solve linear systems, it is not considered “elegant.” Since linear algebra is not a prerequisite for PHYS 302, we are happy enough to use it, since it is a recipe that gets the job done. In a system of linear equations with three equations, you may be able to find the values for three (initially unknown) variables x, y, and z. In a well formulated problem with three equations, each will have parameters a, b, and c along with a right-hand side d (each parameter will have a subscript identifying which equation it is in). A solution may be found as a combination of these known items. This solution involves a single number (a scalar), which is a combination of the parameters called the determinant. Computing the determinant involves multiplying the parameters together with appropriate signs. In the simpler case of a system of two equations in two unknowns a1 x b1 y d1 a2 x b2 y d 2 , the determinant is written D a1 a2 b1 and is computed as b2 D a1b2 a2b1 . Note that this formula could come about (non-rigorously) by saying, “take the upper row element, multiply by (-1)1+column , and then multiply by what is not in that row and column.” This recipe can be expanded to larger sets of equations (the set of parameters is, in fact, a matrix) since it will apply even if the “what is not in that row and column” represents another matrix. In this way, we can proceed to calculate three by three determinants (the one above being two by two) to use in Cramer’s Rule as shown in the video lecture. For three a1 x b1 y c1 z d1 a1 equations in three unknowns, a2 x b2 y c2 z d 2 , we have a3 x b3 y c3 z d3 D a1 b2 c2 b3 c3 b1 a2 c2 a3 c3 c1 a2 b2 a3 b3 b1 c1 D a2 a3 b2 b3 c2 . By the above recipe, c3 , where you should note the sign change in the middle term. This can now be computed by expanding the small determinants using the simple rule for two by two determinants. Note that the variables are arranged by column: x in the first, y in the second, and z in the third. Cramer’s Rule says that we may obtain the values of these variables by taking the determinant from the original parameters, D, and 1 PHYS 302: Unit 6 Viewing Notes – Athabasca University using it as a denominator. The variables are given by fractions, in which the numerators are the determinant arising d1 from replacement of their column by the right hand side (RHS). In this case the RHS is the column d 2 . With d3 d1 a1 b1 D a2 a3 b2 b3 c2 , we have in turn x c3 c1 a1 d1 c1 a1 b1 d1 d2 d3 c1 b1 b2 b3 c2 c3 a2 a3 d2 d3 c2 c3 a2 a3 b2 b3 d2 d3 D ,y D , and z D (10:45). Despite knowing that we will simply get the ratio of the Cs in the end, we start out by trying to solve for them as variables or unknowns. Before going back to the lecture, we will make this more explicit by writing the equations again as 2 (30 2 )C1 2 0 C 2 0 2 2 0 C1 (0 2 )C2 0 . Then D 2 30 2 2 0 2 0 02 2 (11:40), and we can use Cramer’s Rule to get 2 0 0 C1 0 02 2 D 2 30 2 0 2 0 0 and C2 , which are both 0 if D is not 0. These would correspond to trivial D solutions (not wrong, but uninteresting cases of no motion) (13:50). The case in which the determinant D=0 is, in fact, the only case that gives rise to meaningful solutions. Expanding this, we have D0 2 30 2 2 0 2 0 2 0 2 2 2 4 (30 2 ) 0 2 0 , which is just a quadratic in ω2. It is good algebra practice to recall the quadratic formula x b b 2 4ac , which gives the two solutions of 2a ax 2 bx c 0 , one corresponding to the choice of + and the other to –, from the ± symbol. Using these, we expand the above equation for D and solve for the two possible values of ω2. Doing this, one finds that they are 2 2 (2 2)02 and (2 2)02 . Roughly, 0.760 and 1.850 . By messy algebra from the original equations, or possibly by applying l’Hopital’s Rule (which you should have studied in calculus), the C ratio may also be determined. For the lower mode, C2 C1 C2 C1 1 2 2.4 , while for the higher mode 11 2 0.41 (17:30). The whole solution could be obtained from initial conditions. Also, any linear superposition of these solutions is a solution, determined by initial conditions. The previous lecture demonstrated the normal modes and the expected amplitudes. We now proceed to a driven system with driving at the suspension point, with η= η0cosωt, as seen before for a single pendulum. Initially, the changes appear small. They are added in simply in the same equations on the blackboard. The formula for sinθ1, which was sinθ1=x1/l, becomes sinθ1=(x1-η)/l, and is in the DE for x1. This was mx1 T1 sin 1 T2 sin 2 2 mg mx1 2mg ( x1 ) l mg ( x2 x1 ) l x1 l mg ( x2 x1 ) l , which now becomes m cos t , with the first term having an η due to the motion of the 2 00 suspension point and with the addition of a force equivalent to driving based on ω02η0 replacing F0/m as in the third video lecture. Again, with 02 g l , by rearranging terms and dividing through by m, we get 2 PHYS 302: Unit 6 Viewing Notes – Athabasca University 2 2 2 2 1 30 x1 20 0 x2 00 cos t . Putting in η= η0cosωt and moving the η term to the RHS, x 2 1 30 x1 02 x2 020 cos t , which is slightly different from what is given in the lecture (the lecture x misses a 2 on the RHS). ω is now imposed (an important difference). We look for steady state solutions (21:50). Only the amplitudes need to be determined. Subject to checking, we will proceed with the factor 2, to match the lecture. Recall that in the equations for the Cs, the cosωt term was already divided out, so we can also divide it out when we add the terms due to forcing: 2 200 C1 2 (30 2 )C1 2 2 0 C2 200 2 2 0 C1 (0 2 )C2 0 . Now in Cramer’s rule, 2 0 02 2 0 D 2 2 30 2 200 2 0 0 and C2 C2 . We can do these determinants to get C1 D 2 2 200 (0 2 ) (25:40) and 2 2 ( 2 )( 2 ) 2 (26:10). You should verify by substitution that the denominator is D, but is written in 2 ( )( ) 4 00 2 2 2 a different form, since we noted above that the formula for D was a quadratic in ω2, denoting the solutions ω-2 and ω+2. Finding these solutions amounts to factoring D into the product shown as the denominator. These functions of ω are obviously quite complicated, although they clearly blow up at frequencies ω- and ω+. A discussion of the graph of the Cs, normalized by dividing by η0, follows. Not surprisingly, this graph resembles Fig. 5-8 in the textbook somewhat, colour coded with the objects of both curves plotted on one axis instead of two. In the low frequency limit, the whole double pendulum is simply translated slowly back and forth; essentially, it does not swing. In this ω=0 case, C1=C2= η0. Since the C curves are normalized to η0, they both start at 1 on the graph. Approaching the first resonance, C2> C1, approaching C2=2.4 C1 as expected, but they are in phase. This is the lower mode, in which both bobs are swinging, with the lower (2) bob having higher amplitude. At the point where the frequency is exactly that of a single pendulum, ω=ω0, C1=0, and C2= 2η0. Amazingly, the first bob is stationary and the lower one moves (36:00). The demo that follows includes even this strange situation (42:00). Break Following the break, Dr. Lewin considers the three oscillator problem with oscillating cars coupled by springs. In the setup, assume that x1>η, x2> x1, and x3> x2. We can then quickly write the DE for the first object: mx1 k ( x1 ) k ( x2 x1 ) (the DE will work even if the assumption we made is not true). Similar force analysis gives mx2 k ( x2 x1 ) k ( x3 x2 ) and mx3 k ( x3 x2 ) kx3 . Imposing ω, we put x1 C1 cos t , x2 C2 cos t , and x3 C3 cos t . Doing algebraic grinding (with s2 2 2s2 D s2 2 0 2 s 2 2 s 0 2 s s2 2 2 . Using Cramer’s Rule after putting the 2 s 3 k m ) we should get PHYS 302: Unit 6 Viewing Notes – Athabasca University s20 2 2s2 s20 C2 s2 0 0 2 0 proposed solutions into the DE, we get C1 s2 s2 s2 0 2 2 0 2 s s2 , 2 2s2 2 2s2 0 0 2 s2 s20 s2 2 2s2 0 0 0 , and C3 2 s 2 s (57:20). We could now solve for the resonance or normal mode frequencies by putting D=0. This corresponds to no driving and 0 determinants on top, since η0 would be 0. These frequencies turn out to be and 2 2 (2 2)s2 , 2s2 2 (2 2)s2 (59:30). Another multiple curve graph shows these results well. The 0 frequency case is not really surprising—all three springs are stretched. At ωs (called ω0 on the graph), the first car stands still while the two adjacent cars move with the amplitude of and in antiphase to the driver. The demo clearly shows this after the transient dies out (1:08:30). Next, near one of the resonances, the middle car should move only a little while the outer cars move a lot in antiphase. D is also 0 at this point, which gives C2 a finite amplitude. Next, Dr. Lewin examines the triple pendulum with plots and demonstrations. Obviously (now), the 0 frequency case has in-phase motion with the amplitude of the driver. Near the first resonance, the bobs swing in phase, with larger amplitude toward the bottom. At a certain frequency (maybe 0.77ω0) the top one will stand still and the others will move. This is too hard to demonstrate. But at ω0, the middle one will stand still, the top one will move at about 0.7η0, and the bottom one will move at about -1.57η0. This can be demonstrated, since ω0 can be “burned into the chips.” 4 ...
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