Chapter (a) 2. At equilibrium, V = 0. x = (c) V( x) ke bkxe = 0. At equilibrium. b bx bx 2 bx. V = bke bkxe + b kxe = + = < 0.

Transcription

1 Chpter. () V( x) = x + x V = x x At equiibriu, V = x = V = + x V = + = > x= Stbe (b) V( x) = xe bx bx bx V = e bxe At equiibriu bx bx e bxe = x = b bx bx bx V = be bxe + b xe V be be = + = < x= b Unstbe (c) V( x) = ( x 4 b x ) V = ( 4x b x) At equiibriu ( 4 ) x=, ± b ( ) V = x b V x b x = = b < Unstbe x= = ( 6 ) = 4 > Stbe x=± b V b b b (d) for cse () = T π π = = π = s for cse (c) t x=± b 4b π = T = = π = π s 4b. V ( x, y) = x + y bx 4by

2 V V = x ( b) = y ( b) x y t equiibriu x = b nd y = b V V V = = = x x y y = > = = = = 4 > The equiibriu is stbe.. V( x) = x dv ( x) dx F( x) = = x= x = x dx dx x dx = x dx x v v xdx = x dx ( x x ) = x dx v= = ( x x ) dt x t dx = dt ; α x x x where α = n ( ) x + x x n x = α t x + x x = x e α t x x = x e xx e + x α t αt αt αt e + e x = x cosh = x αt.4 Let the ength of the unstretched, estic cord be d. Then d = + y y V = ( d ) gy

3 ( ) V = + y + y + gy V = + y + y gy The first ter, 4, is n dditive constnt to the potenti energy, so with pproprite djustent of the zero reference point V( y) = y + y gy dv ( y) = y y( + y ) g dy At equiibriu, the bove expression is zero, so 4y 4y g = + y etting 4y g = 4y + y 6y 6 y 8gy+ g = + y 4 6 y 8gy g 6 y 8 gy + g = y g g g g 6 6 y g u = nd = 4 4 u u + u u+ = 4 y + y y+ = h h θ φ θ b d V = g h + h h = bcosθ h = dsin ( θ + ϕ ) h = d( θ ϕ+ θ ϕ ) sin cos cos sin bθ cosϕ= sinϕ= d d h = bθ sinθ + cosϕ V = g ( + b)cosθ + bθsin θ V = g ( + b)( sinθ ) + bsinθ + bθcosθ

4 4 [ sinθ θcosθ] [ cosθ θsinθ cosθ] V = g + b V = g b b V = g b Equiibriu Stbe Unstbe < b > b.6 4 θ θ cosθ = +...! 4! 5 θ θ sin θ = θ +...! 5! 4 5 θ θ θ θ V = g ( + b) bθ θ +...! 4!! 5! b b 4 V = g + b θ + θ For = b, V = g θ +... g V = θ +... ters in higher order of θ V = gθ +... V = gθ +... V = g< Equiibriu is unstbe.7 h h θ b φ θ The center of ss (CM) of the heisphere is fro the ft side (see Eqution 8..8). 8 The height of CM) bove the point of contct between the two heispheres is designted by h in the figure. h is the height of the point of contct bove the ground. V = g( h + h ) = g bcosθ + cosθ cos( θ + ) 8 ϕ nd ϕ = b θ

5 5 bθ V = g( h+ h) = g ( + b) cosθ cosθ + 8 sin + b V g b θ sin + b = + + θ 8 Equiibriu occurs t θ = o cos + b V g b cos + θ b = + + θ 8 + b g V = g ( + b) + = ( + b)( b V > for b> 5 b Therefore, the equiibriu is stbe for < 5.8 Fro Probe.4, we hve V ( y) ( y y ) gy y y = + = + gy Expnding the squre root for s y y 4 4 V( y) y y + gy V y g t equiibriu, V = g y = 6 V = y 6 g g V = 6 y= g = ) 4 y y y +... = V 6 g 6 g = = = 6

6 6.9 Fro Probe.5, V = g b V = = g b π π T = = g b. ( ) g = + b 5) 8 T = π = 4π V g( + b)( b 5) Fro Probe.7, V ( b)(. b θ φ T = vc + Iϕ v = b θ c I = 5 The retionship between the nges, θ nd ϕ (see Figure), cn be deterined fro the condition tht there is no sipping s the b ros in the heisphere, so the ength of ro esured ong the b, ( θ + ϕ ), ust equ the ength of ro esured ong the heisphere, bθ so we hve bθ = θ + ϕ nd b θ = θ + ϕ b nd ϕ = θ ( b ) T = ( b ) + = b V = g( b )cosθ θ θ θ V = g ( b )sin θ θ = o

7 7 cosθ V = g b V = g b ( ) 5g ( ) ( ) V g b M 7 b 7 b 5 = = = π 7 T = = π. ( b ) 5g The potenti energy of the steite shped ie thin rod is (See Expe.., Figure..) M e d V = G but d = dx where is the ength of the rod. r GM e GM e dx V = dx= r r cos r= r + x + r x φ xr r= r + x + ε cosφ where ε = r + x GM e V = φ ( + ε cos ) ( r + x ) dx x r >> x, r + x r, ε nd r r + εcosφ For r Thus, for s ε, the expression for the potenti energy, V, cn be pproxited GM e x x V cos cos d r φ+ φ x r 8 r GM e cos φ V + + r r GM e cos φ V + r r GM e GM e V cos sin sin φ φ= φ r r r

8 8 φ = GM V cos φ nd r e M = I = T V M = = π = = π GMe r r GM e GM V e r. The pitude of the syetric coponent is A nd the pitude of the nti-syetric coponent is A (See Equtions..9 through..b). A [ ] = x() + x() 4 A A= [ x() + x() ] = A [ ] = x() x() 4 A A= [ x() x() ] = A= A Fro Eqution..8 the soution for x is A x() t = ( cost+ cost) (The phse δ is 8 o, which insures tht x() = A) A ( + ) t ( ) t x () t = cos cos + Letting = nd = x () t = A cost cos t Fro Eqution..8, the soution for x is A x() t = ( cost cost) A ( + ) t ( ) t x () t = sin sin x () t = A sintsin t

9 9.4 At tie t = nd short ties therefter cos t nd sin t. Thus, x A cost nd x. This sitution occurs gin when t = π. + = = = + for <<, = = π π T = = π = = T T.5 θ θ r T = θ + θ ( ) ( cosθ ) ( cosθ ) V = g + r= r + sinθ sinθ r V cosθ = g sinθ θ r + θ θ ( sin sin ) V sinθ cos θ g r + r + θ = cosθ θ ( sinθ sinθ) ( sinθ sin ) V = = g θ r θ = θ = V θ = + θ cos cos ( r sinθ sinθ ) θ θ

10 V = = V = θ θ θ θ r θ = θ = θ = θ = V cosθ = g sinθ + θ r + θ θ ( sin sin ) V sinθ cos θ g r + r + θ = cosθ θ ( sinθ sinθ) ( sinθ sin ) V = = g θ r θ = θ = Thus, fro Eqution..7 or b, we hve V = θ + θθ + θ But fro Eqution..9, for the couped oscitor, we hve V = x + ( x xx + x ) + x The fors of the potenti energy function re siir with = + = nd = In the cse here = g nd = r

11 Chpter (continued).6 T = x + x V = x + x x + x L= T V d L L = x = x + dt x x x x + ( ) = x x x x d L dt x = x x + x + x x L x = x x x = ( ) ( ) ( )( ) ( ) ( ) = = ( + ) + ( + ) ± ( + ) + ( + ) 4 + ( + ) = = =, =, =, =, = For + ± ( + ) ( )( + ) ( ) = 6 = ± 4 4 = nd where =

12 .7 x x Note: As discussed in Section., the effect of ny constnt extern force on hronic oscitor is to shift the equiibriu position. x nd x re the positions of the hronic oscitor sses wy fro their respective shifted equiibriu positions. T = x + ( ) x V = x + x x L= T V d L L = x, = x+ ( x x) dt x x x + x x = d L L = x, = ( x x) dt x x x + x x= The secur eqution (.4.) is thus + = = The eigenfrequencies re thus 5± 7 = 4 The hoogeneous equtions (Equtions.4.) for the two coponents of the j th eigenvector re + j = + j For the first eigenvector (the nti-syetric ode, j = ) 5+ 7 Inserting = into the first of the two hoogeneous equtions yieds = 4 7 = 4 Letting =, then = -.8 (Thus, in the nti-syetric nor ode, the pitude of the vibrtion of the second ss is.8 tht of the first ss nd 8 o out of phse with it.)

13 For the second eigenvector (the syetric ode, j = ) 5 7 Inserting = into the first of the two hoogeneous equtions yieds = = 4 Letting =, then =.78 (Thus, in the syetric nor ode, the pitude of the vibrtion of the second ss is.78 tht of the first ss nd in phse with it.) The two eigenvectors (Eqution.4. nd see ccopnying Tbe) re Q= cos( t δ) = cos( t ) δ.8 Q= cos( t δ) = cos( t ) δ.78.8 T = θ + θ + φ V = g cosθ g cosθ + cosφ For s ngur dispceents θ φ L= T V θ + ( θ + φ) + g + g d L = θ + ( θ + φ), L = gθ dt θ θ θ + φ + gθ = d L L = ( θ + φ), = gφ dt φ φ θ + φ + gφ = The secur eqution (Eqution.4.) is + g = + g 4 g + + g = 4 Soving for the eigenfrequencies ( ) + ± g g g g = = + + +

14 4 The hoogeneous equtions (Equtions.4.) for the two coponents of the j th eigenvector re + g j = + g j Inserting the rger eigenfrequency (the (+) soution for bove) into the upper hoogeneous eqution yieds the soution for the coponents of the st eigenvector + g = g = g + = for the higher frequency, nti-syetric ode. Inserting the ser eigenfrequency (the (-) soution for ) into the upper hoogeneous eqution yieds the soution for the coponents of the nd eigenvector + g = + + = for the ower frequency, syetric ode. Agin, we et = nd =, since ony rtios of the coponents of given eigen vector cn be deterined. The two eigenvectors re thus (Eqution.4. nd ccopnying tbe) Q = cos( + t δ ) nti-syetric Q = cos( + + t δ ) syetric As chec, set = = nd copre with the soution for Expe....9 T = ( x + x + x ) V = x + x x + x x + x L= T V d L L = dt x x x x x ( ) ( ) i i

15 5 x x x x + = x + x x = + ( ) ( ) = x x x x x x + x + x x = + ( ) + = x x x x x + x + x = The secur eqution (Eqution.4.) is + + = = + =, or + = = = + =± = ( ± ) = ( ± ) Fro Eqution.5.7 (N = nd n = ) π = sin 8 cos Becuse sin θ = θ cosπ 4 = = = π = sin = = 8 π cosπ 4 = sin = 8 = + = +

16 6. X M Generized coordintes: X, s( = θ ) nd x= X + sin θ y = ( cos θ ) x = X + θ cosθ y = θsinθ y θ x (x,y) T = MX + ( x + y ) = MX + ( X + θcosθ) + ( θsinθ) V = gy = g cosθ For s oscitions, in ters of the generized coordintes X nd s T MX + X + s V gs L= T V MX + X + s gs Lgrnge s equtions of otion yied g X + s+ s= ( M + ) X + s = Assuing i t i t X Ae = s = Be we obtin the trix eqution g A = ( M ) + B Setting the deterinnt of the bove trix equ to zero yieds g( + M) =, = M The ode corresponding to is g A = ipies tht A= B= B Thus, ode exhibits no oscition! It is pure trnstion with θ = nd X = At + A The ode corresponding to is ( ) A+ g B=

17 7 B + M or = = A g Setting A =, we hve for the nd ode i t X = e + M i t nd s = θ = e This ode corresponds to n oscition bout the CM where ( + M) X = s( = θ ) The nor ode vectors re Q At+ A = nd i t Q = + e. X M M () We cn sove for the nor odes using Eqution.4.9 K M = K nd M re the potenti energy nd inetic energy trices respectivey. is twocoponent vector whose eeents re the r pitudes of the coordintes q. The inetic θ energy in Expe.., ssuing s dispceents fro equiibriu, re T X + ( X + ( r θ ) + Xr θ ) T X + ( Xr θ ) + ( r θ ) or, in trix for X T = qmq where q = rθ Thus, M = The potenti energy is g V X + ( rθ ) r g V X + ( rθ) = X + ( rθ) r where = = M +

18 8 Thus K = The bove trix eqution is thus = ( ) = Fro the top eqution in the trix eqution bove, we get = ( ) = The pitudes for the two nor odes cn be found by setting = or. In ech cse we set be deterined. = For ( ) ( ) ( ) = = ( ) = = we obtin, which we re free to do since ony rtios of the pitudes cn = Thus, for ode, the ower frequency, syetric ode i t q = e For ( ) ( ) ( + ) = = ( + ) = = + we obtin = Thus, for ode, the higher frequency, nti-syetric ode i t q = e (b) In this cse, << or M >>. We so ssue tht the spring is sc, + M g i.e., <<, n ssuption not stted in the probe (which needs to be rectified in M + r the next edition, I suppose) g Aso, g r, so we et = nd Ω = M + M r The inetic energy is

19 9 ( ( θ ) θ ) T MX + X + r + Xr T ( M + ) X + ( r θ ) + Xr θ MX + Xr θ + r θ nd the M-trix is M M The potenti energy is g V X + rθ = M + Ω r The K-trix is thus M K Ω To sove for, we set K M = ( ) M ( ) Ω = which yieds 4 4 Ω + Ω M = 4 M +Ω M + Ω M = Negecting with respect to M nd sipifying yieds 4 +Ω + Ω = Soving for ( ) ( ) +Ω +Ω 4 Ω = ± ( +Ω ) ( Ω ) = ± Thus, we get = nd = Ω Now, we sove for the pitudes of the nor ode vectors K M = M = ( Ω ) Using the first of the bove trix equtions for = = gives = nd = Using the second eqution for = =Ω gives

20 = nd = Thus, the nor odes re pproxitey i t iωt Q = e nd Q = e Note tht we coud hve guessed this ost ieditey. The bove ssuption is tntount to oitting the cross eeents. This copetey eiintes the s couping between the two oscitors, which reduces the trix K M to the purey digon ters ( ) M, which eds directy to the bove soution. ( Ω ). θ We sce the force constnts nd sses to unit, ney, = nd =. Let = K nd = θ K θ such tht = circuference So: T = ( θ + θ + θ ) V = θ Coecting ters V = 4 θ + ( + K) θ + ( + K) θ 4 θθ 4 θθ 4 Kθθ K - Mtrix M - Mtrix K = - +K -K M = - -K + K The ttnsfortion trix A tht digonizes these trices is de up of the three eigenvectors Q i whose pitudes re i we guess tht ( θ θ ) ( θ θ ) ( θ θ ) ( θ θ ) K( θ θ ) K( θ ). Unifor rottion θ = θ = θ or =

21 . Anti-syetric oscition of &, whie reins fixed θ = ; θ = θ or =. Anti-syetric oscition of & together with respect to θ = ; θ = θ or = Thus, A nd A= = We cn now digonize Knd M K -K AKA = - - = K - -K + K Kdig = 4+ 8K Liewise M dig = 6 4 The eigenfrequencies re i = K dig M dig = ; = + 4 K ; = 4 The gener soution cn be generted fro the foowing tbe Q Q Q θ θ θ where Q = cos( t δ), Q = cos( t δ), Q = cos( t δ) i Thus, θ = cos( t δ) + cos ( t δ) θ = cos( t δ) + cos( t δ) cos( t δ) θ = cos( t δ) cos( t δ) cos( t δ) dig

22 o Initi conditions re: θ = θ = ; θ = θ = ; θ = θ = θ = The conditions generte 6 equtions with 6 unnowns nd soving gives θ θ θ = cost+ cost θ θ θ = cost cost θ θ θ = cost cost. See Ex..4., pge The pitudes of the eigenvectors re = = = M K - Mtrix M - Mtrix K -K K = -K K -K M = M -K K K K Kdig = AKA = K K K M M K K K dig = K K + 8K M 8K M Mdig = AMA = M M M + M M dig = + 4 M So, we hve =, = K K =, nd K + 8K M + 8K M K( + 4 M + 4 M ) = = + 4 M + M

23 ( + M) ( + ) K K = = + M.4 ( M) y M x y α α y x x Seect coordintes (x i, y i ) s shown, then T = x + y + x + y + M x + y The potenti energy depends ony on the copression (or stretching) of the two springs connecting ech to M (hydrogen to sufur). Let δ nd δ be increent chnges in the distnces or HS( ) nd or HS( ). We hve δ = ( x x) sinα + ( y y) cosα = ( x x + y y ) δ = ( x x) sinα + ( y y) cosα = ( x x+ y y ) V = ( δ) + ( δ) We cn reduce the degrees of freedo fro 6 to by ignoring the two trnstion odes nd the rottion ode. Thus we consider ony vibrtion odes. The coordintes ust obey the foowing constrints No center of ss otion: y + y + My = nd x + x + Mx = No ngur oentu bout ny point. We choose tht point to be the sufur to (M) y sinα y sinα x sinα x sinα =

24 4 y y x x = We introduce three generized coordintes Q, q nd q tht shoud be cose to wht we guess woud be nor odes Q= x+ x; q = x x; q = y+ y Sove for x i, y i in ters of Q, q nd q using the bove equtions of constrint x = ( Q+ q) ; x = Q; = x x; x = ( Q q ) M y = ( q Q) ; y = q; y = ( Q+ q ) M Thus, the inetic energy, in ters of these generized coordintes, is µ T = + Q + q + q M 4 4 M where µ = M + is the ss of the H S oecue The potenti energy is µ µ V = + Q + q + q qq M 8 8 M 4 M Note, tht in the Lgrngin L= T V, the ony cross ter is one invoving qq. therefore, Q is nor ode with eigenfrequency given by the rtio Q = KQ MQ = + M Constructing the residu x K nd M trices invoving ony q nd q ters, which we wi c Kq nd Mq gives µ M Kq = nd Mq = 4 µ M µ M µ M We hve oitted the fctors nd, which we repce in the fin soution, reebering tht the eigenfrequencies tht we find s soution to Kq - Mq = wi be utipes of /. µ M Kq - Mq = = 4 µ M µ M µ M which reduces to ( + µ M ) = which hs the non-trivi soution, = ( + µ M ) in which, we hve put bc the fctor /. These two odes re degenerte.

25 Pugging bc into the trix eqution ( ) Kq - Mq q = gives q = q or x+ x = y+ y which cn be stisfied in vriety of wys, for expe, with x = x nd y = y, etc Pictoriy, the three nor odes re 5 Q-ode: nt-syetric bout y-xis: x = x nd y = y Q = + M Brething ode: syetric bout y-xis: x = x nd y = y = + µ ( M ) Stretching ode: syetric bout y-xis: x = x nd y = y = ( + µ M ) () (b) (c) Pug ech of these functions into the wve eqution nd it is stisfied! ( ) Q= q + q = e e + e e i t ix i + t i + x i( + ) t i( + ) x i ( ) t ( ) x + i ( ) t ( ) x e e = e + e The re prt of the bove is ( ) t ( ) x Q= cos cos + t x + ( ) t ( ) x cos cos( t x) The group speed is dx ug = (the phse of the pitude reins the se) dt

26 6 u g =.6 Fro Eqution Nπ π π 4π sin sin sin sin N 4 =, = =.8, = =.6, = =.8 π sin π π π sin sin sin

27 7.7 F = is the tension in the cord + d = is its stretched ength n + Fro the eqution foowing Eqution.5.7 F K = d Fro Eqution.6.4c v = + vtrns = + for trnsverse wves For ongitudin vibrtions, we use Y, the tension in the cord per unit stretched ength Y = = ( + ) + Y K = (Eqution.6.8) d d ( + ) nd v = = (Eqution.6.9) n nd n + d = + v ong.8 = + ( ) Fro Eqution.6.7b F v = µ As in probe.7, F = vtrns = µ Fro Eqution.6.9b Y v = µ = + so v nd fro probe.7 Y ( ) ong + = µ

28 8.9 The gener soution to the wve eqution (Eqution.6.) tht yieds stnding wve of ny rbitrry shpe cn be obtined s iner cobintion of stnding sine wves of for given by Eqution.6.4, i.e., π x yxt (, ) = ( Ansinnt+ Bncosnt) sin n= λn π where = n T n since the speed of wve is λ n nλn F v = = = Tn π µ we hve F π n = µ λn But the wveength of the stnding wve is constrined by the fixed endpoints of the string, i.e. λ n =, so n F n = µ nπ Now, t t =, the wve strts fro rest in the configurtion specified, so nπ x yx (,) = Bn sin Fro the discussion of Fourier nysis in Appendix G or in Section.9, the Fourier coefficients re given by nπ x Bn = yx (,)sin dx Since the string strts fro rest, we hve yxt (, ) nπ x = nansin t t= Therefore, A n = The initi configurtion is shown in the Figure P.9. Thus yx (,) = x < x<

29 9 yx (,) = ( x) < x< We cn now deterine B n nπx nπx Bn = xsin dx+ ( x) sin Using integrtion tbes, we obtin the resut n 8 nπ B n = sin n=,,5,... n π = n=,4,6,... Bn dx Thus, the gener soution is 8 πvt πx πvt πx 5πvt 5πx yxt (, ) = cos sin cos sin cos sin... + π 9 5 None of the hronics which hve node t the idpoint hve been stiuted ony the odd hronics hve been excited.. By nogy with the genertion of the trveing sine wve of Eqution.6.4 fro Eqution.6., we get 4 π ( x + vt) π ( x vt ) π ( x + vt ) π ( x vt ) yxt (, ) = sin sin sin sin + + π 9 5π ( x+ vt) 5π sin sin ( x vt)