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6 Pages

### sol6

Course: PHYSICS 507, Fall 2008
School: Rutgers
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Word Count: 1107

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507 Physics Homework Solutions#6 Due: late Oct. 23, 2008 1.) With xi measured from the left wall, and with 1 = x1 a, 2 = x2 3a the displacements from equilibrium, the kinetic and potential energies are T = V 1 1 m(x2 + x2 ) = m(1 + 2 ) 1 2 2 2 2 2 1 = k[(x1 a)2 + 2(x2 x1 2a)2 + (x2 3a)2 ] 2 1 = k[(1 )2 + (2 )2 + 2(2 1 )2 ] 2 1 = k[3(1 )2 + 3(2 )2 41 2 ]. 2 The kinetic energy is already in the form where...

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507 Physics Homework Solutions#6 Due: late Oct. 23, 2008 1.) With xi measured from the left wall, and with 1 = x1 a, 2 = x2 3a the displacements from equilibrium, the kinetic and potential energies are T = V 1 1 m(x2 + x2 ) = m(1 + 2 ) 1 2 2 2 2 2 1 = k[(x1 a)2 + 2(x2 x1 2a)2 + (x2 3a)2 ] 2 1 = k[(1 )2 + (2 )2 + 2(2 1 )2 ] 2 1 = k[3(1 )2 + 3(2 )2 41 2 ]. 2 The kinetic energy is already in the form where the mass matrix is a constant multiple of the identity, so all that is necessary is to diagonalize the potential Bjk = k where O is a rotation by /4, 1 O= 2 1 1 , 1 1 C=k 1 0 . 0 5 3 2 2 3 = O1 CO, rst of these corresponds to = O1 1 = (1, 1) Re (Aj ei k/mt ), with both masses moving in the same direction. This corresponds to a body of mass 2m and a spring constant 2k from the two springs in parallel. The second mode is (1, 1) Re (Aj ei 5k/mt ), with each mass eectively coupled to a xed point with one spring with constant k and another (half of the middle spring) with constant 4k. Thus the eigenfrequencies are 1 = k/m and 2 = 5k/m, with corresponding normal modes j = Re (Aj eij t ). For the displacements j , the 2.) The rst mass has coordinates x1 y1 x2 y2 = = = = L sin 1 L cos 1 L (sin 1 + sin 2 ) L (cos 1 + cos 2 ) L 1 - m 1 2 so v1 = L2 2 and 2 v2 = L2 [(1 cos 1 + 2 cos 2 )2 +(1 sin 1 + 2 sin 2 )2 ] 2 2 2 = L [ + + 21 2 cos(1 2 )]. 1 2 2 The potential energy is U = m1 gy1 + m2 gy2 = gL (m1 cos 1 + m2 (cos 1 + cos 2 )) . T = L m 2 For shorthand, let r = m1 /m2 . Then the Lagrangian is L = T U , where 1 2 2 m2 L2 (1 + r)1 + 2 + 21 2 cos(1 2 ) , 2 U = m2 L [(1 + r) cos 1 + cos 2 ] b) Equilbrium is at 1 = 2 = 0, so to second order in the s and s, 1 2 2 m2 L2 (1 + r)1 + 2 + 21 2 , 2 1 2 2 U = m2 L (1 + r)1 + 2 , 2 where an irrelevant constant term in the potential energy has been dropped. In matrix form this gives 1 1 T = m2 L2 T M , U = m2 gLT A , 2 2 with 1+r 1 1+r 0 M= , A= . 1 1 0 1 T = c) As U is already in diagonal form, it is easier if we reverse the steps of the procedure in the book. With y1 = 1 + r1 and y2 = 2 , we have U = 1 m2 gLy T y, and T = 1 y T m y with 2 2 m= 1 (1 + r)1/2 (1 + r)1/2 . 1 This has equal diagonal elements, so we know the diagonalizing rotation is through 45 , or u1 = (y1 + y2 )/ 2 = ( 1 + r1 + 2 )/ 2 u2 = (y1 y2 )/ 2 = ( 1 + r1 2 )/ 2, so 1 = (u1 u2 )/ 2, 2 = (u1 + u2 )/ 2(1 + r) Then U = 1 m2 gL(u2 + u2 ) and 1 2 2 T = 1 1 m2 L2 (u1 + u2 )2 + 2 2 1 1 = m 2 L2 1 + 2 1+r 1 1 (u1 u2 )2 + (u2 u2 ) 2 2 1+r 1 1 u2 + 1 1 u2 ) . 2 1+r We now have independent oscillators, with angular frequencies given by the square root of the ratio of the coecient in the potential to that in the kinetic energy. Thus the two frequencies are f= 1 2 1 g 1 L 1+r 1/2 = 1 2 g L m1 + m2 m2 (m1 + m2 ) m1 . 3. [20 pts.] If the three particles, which have masses Mi and positions Ri , i = 1, 2, 3, are in an equilateral triangle of side L, then |Ri Rj | = L for i = j. Then the force on particle i is Fi = GMi j=i Mj (Rj Ri ) |Rj Ri |3 = GMi j=i Mj (Rj Ri ) L3 = GMi all Mj (Rj Ri ) GMi MT = (R Ri ), L3 L3 i where MT = Mi is the total mass and R is the position of the center of mass. If the particles are rotating about R with angular velocity , the centripetal force required maintain to this motion is Mi 2 (R Ri ), so if 2 = GMT L3 , this is exactly what the gravitational forces do, and the rotation satises the equations of motion. (b) If the third mass M3 = m is x much smaller than the other two, y so that its eect on the motion of R the others is negligible, M1 and 3 M2 will circle their common center of mass at distances M2 L/(M1 + M2 ) and M1 L/(M1 + M2 ) respecL L tively. We choose rotating cartesian coordinates with the origin and z-axis at the center of mass M2 and along respectively. From M 0 1 the general equation for the acM 1L M 2L celeration in a rotating reference M1 + M M1 + M 2 2 frame, we have mr = F 2m v m ( r), If r diers from the equilibrium position R3 by a vector , the force on m is F = Gm M 1 R1 R3 |R1 R3 |3 + M2 R2 R3 |R2 R3 |3 Gm +3 R1 R3 R2 R3 MT + M1 + M2 3 3 L L L3 L5 R1 R3 + 3 M2 R2 R3 L5 (1) R2 R3 , M1 R1 R3 where we have expanded to rst order in , using |R1 R3 |3 R1 R3 2 2 R1 R3 . 3/2 L3 1 + 3 R 1 R3 L2 The second and third terms of (1), which are independent of , add as at the equilibrium point to GmMT R3 /L3 = m 2 R3 , where we have made use of R = 0. Let = (x, y, z). Then 2 R2 R3 /L = x 3y, 2 R1 R3 /L = x 3y, 2 R1 R3 /L = (1, 3, 0), 2 R2 R3 /L = (1, 3, 0), so the last two terms in (1) are 3 GmMT + 3y x 3x + 3y 4 L3 where = (M1 M2 )/(M1 + M2 ). Thus 3 2 + 3y x = , 3x + 3y 4 x+ 3y 3 = 2 2 R3 + 2 3x+3y 2v (+R3 ) 4 0 0 y x + 3y 3 2 = 2 0 + 2 x + 3x + 3y . 4 z 0 0 The z motion decouples simply: z = 2 z, which is stable oscillatory mo tion. The x and y coordinates have coupled second order linear dierential equations. A trial solution x = aeit , y = beit will satisfy the equations if 3 a2 = 2ib + (a + 3b), 4 3 2 b = 2ia + ( 3a + 3b). 4 A nonzero solution for (a, b) requires a vanishing determinant: 2 + 3/4 2i + 3 3/4 0 = det 2i + 3 3/4 2 + 9/4 27 2 = 2 + 3/4 2 ...

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