Analytical mechanics solutions manual fowles03

CHAPTER OSCILLATIONS 3.1 x = 0.002sin  2π ( 512 s −1 ) t  [ m ] m m xmax = ( 0.002 )( 2π )( 512 )   = 6.43   s s 2 m m xmax = ( 0.002 )( 2π ) ( 512 )   = 2.07 × 104   s  s  3.2 x = 0.1sin ω t [m] When t = 0, x = and ω = s −1 3.3 m x = 0.1ω cos ω t   s m x = 0.5   = 0.1ω s 2π T= = 1.26 s ω x ( t ) = x cos ω t + x ω sin ω t and ω = 2π f x = 0.25cos ( 20π t ) + 0.00159sin ( 20π t ) [ m ] 3.4 cos (α − β ) = cos α cos β + sin α sin β x = A cos (ω t − φ ) = A cos φ cos ω t + A sin φ sin ω t x = Α cos ω t + Β sin ω t , Α = A cos φ , Β = A sin φ 3.5 2 2 mx1 + kx1 = mx2 + kx2 2 2 2 2 k ( x1 − x2 ) = m ( x2 − x1 ) k  x22 − x12  = ω =  m  x12 − x22  2 kA = mx1 + kx1 2 m x2 x2 − x2 x2 A2 = x12 + x12 = 12 22 + x12 k x2 − x1 x x −x x  A=   x −x  2 2 2 2 1 3.6 3.7 1 l =π T =π s ≈ 2.5 s 9.8 g For springs tied in parallel: Fs ( x ) = − k1 x − k2 x = − ( k1 + k2 ) x (k + k ) 2 ω=   m  For springs tied in series: The upward force m is keq x Therefore, the downward force on spring k2 is keq x The upward force on the spring k2 is k1 x′ where x′ is the displacement of P, the point at which the springs are tied Since the spring k2 is in equilibrium, k1 x′ = keq x Meanwhile, The upward force at P is k1 x′ The downward force at P is k2 ( x − x′ ) Therefore, k1 x′ = k2 ( x − x′ ) x′ = k2 x k1 + k2  k x  And keq x = k1    k1 + k2  ω= 3.8  k1k2 2 =  m  ( k1 + k2 ) m  keq For the system ( M + m ) , − kX = ( M + m ) X The position and acceleration of m are the same as for ( M + m ) : k xm M +m   k k xm = A cos  t + δ  = d cos t + + M m M m   The total force on m, Fm = mxm = mg − Fr xm = − Fr = mg + mk mkd k cos xm = mg + t M +m M +m M +m For the block to just begin to leave the bottom of the box at the top of the vertical oscillations, Fr = at xm = −d : mkd = mg − M +m g ( M + m) d= k 3.9 x = e −γ t A cos (ω d t − φ ) dx = −e −γ t Aω d sin (ω d t − φ ) − γ e−γ t A cos (ω d t − φ ) dt dx maxima at = = ω d sin (ω d t − φ ) + γ cos (ω d t − φ ) dt tan (ω d t − φ ) = − γ ωd thus the condition of relative maximum occurs every time that t increases by ti +1 = ti + 2π 2π ωd : ωd For the i th maximum: xi = e −γ ti A cos (ω d ti − φ ) xi +1 = e −γ ti +1 A cos (ω d ti +1 − φ ) = e −γ 2π ωd xi 2π −γ xi = e ωd = eγ Td xi +1 3.10 (a) (b) (c) c = s −1 2m ω d = ω − γ = 16 s −2 γ= k = 25 s −2 m ω r = ω d − γ = s −2 ω2 = ∴ω r = s −1 F 48 = Amax = m = 0.2 m Cωd 60.4 tan φ = 2γω r 2γω r ω r = = = 2 (ω − ω r ) 2γ γ ∴φ ≈ 41.4 3.11 17 β mx = 17 γ = β and ω = β 2 2 2 ω r = ω − 2γ = β mx + 3β mx + (a) Amax = (b) = ω d2 = ω − γ = 25 β ∴ω d = β 2A 15β e −γ Td = 3.12 F 2mγω d ∴ω r = β γ= ln = f d ln Td ω d = (ω − γ ) (a) 2 So, ω = (ω + γ ) d 1 1   γ 2    ln   f =  fd +    = f d 1 +     2π     2π    f = 100.6 Hz (b) ω r = (ω d2 − γ )   γ 2    ln 2  fr =  fd −    = f d 1 −     2π     2π    f r = 99.4 Hz 3.13 Since the amplitude diminishes by e −γ Td in each complete period, n e −γ Td = = e −1 e γ Td n = ω γ= = d Td n 2π n ( Now So ) ( ωd = ω − γ ω = (ω d2 + γ ) 2 ) 1 2  = ω d 1 + 2   4π n  2π Td ω d ω  2 = = = 1 + 2  2π ω d  4π n  T ω For large n, Td ≈ 1+ 2 8π n T 3.14 ( (a) ω r = ω − 2γ 2 ) 2  ω   = ω −    = 0.707ω      2 ω 1 −  ω −γ ωd 4 (b) Q = = =  = 0.866 2γ 2γ ω  2    ω    ( 2ω ) 2γω 2 (c) tan φ = =   =− ω −ω ω − 4ω 2 ( )  2   φ = tan −1  −  = 146.3  2 ω  2 (d) D (ω ) = (ω − 4ω ) +   ( 4ω )  = 3.606 ω     F F A (ω ) = m = 0.277 D (ω ) mω 3.15 A (ω ) ≈ Amaxγ  (ω − ω ) + γ    1 γ for A (ω ) = Amax , = 2  (ω − ω ) + γ    (ω − ω ) + γ = 4γ ω − ω = ±γ ω =ω ±γ 3.16 (b) Q = ω2 −γ ωd = 2γ 2γ ω2 = , LC γ= R 2L    R   −   LC   L   L  1 Q= =  −   R   R C  4 2   2L  L ω  R (c) Q =   = C = R R  2γ  3.17 Fext = F sin ω t = Im  F eiω t  and x ( t ) is the imaginary part of the solution to: mx + cx + kx = F eiω t i.e i ω t −φ x ( t ) = Im  Ae ( )  = A sin (ω t − φ ) where, as derived in the text, F A= ( k − mω )2 + c 2ω    and 2γω tan φ = ω −ω2 3.18 Using the hint, Fext = Re ( F e β t ) , where β = −α + iω , and x(t) is the real part of the solution to: mx + cx + kx = F e β t x = Ae β t −iφ Assuming a solution of the form: ( mβ + cβ + k ) x =  FA  xeiφ F mα − 2imαω − mω − cα + icω + k = ( cos φ + i sin φ ) A F m (α − ω ) − cα + k = cos φ A F ω ( −2mα + c ) = sin φ A φ = tan −1 ω ( c − 2mα ) m (α − ω ) − cα + k Using sin φ + cos φ = , F2 =  m (α − ω ) − cα + k  + ω ( c − 2mα ) A F A= {m (α − ω ) − cα + k  + ω ( c − 2mα ) } 2 2 2 and x ( t ) = Ae −α t cos (ω t − φ ) + the transient term 3.19 l  A2  T ≈ 2π 1 −  g  (a) for A = (b) π − l ×1.041 g , T ≈ 2π 4π 2l × 1.084 T2 l 4π 2l Using T = 2π gives g = , approximately 8% too small g T (c) g= λ A3 ω2 B=− and λ = 32ω B A2 = A 192 for A = 3.20 π , f ( t ) = ∑ cn einω t B = 0.0032 A n = 0, ±1, ±2, n f ( t ) = ∑ cn cos nω t + ∑ cn i sin nω t , n = 0, ±1, ±2, n T T − T T − n and cn = T∫ f ( t ) e − inωt dt , cn = T∫ f ( t ) cos ( nω t ) dt − n = 0, ±1, ±2, i T T T − ∫ f ( t ) sin ( nω t ) dt The first term on cn is the same for n and − n ; the second term changes sign for n vs − n The same holds true for the trigonometric terms in f ( t ) Therefore, when terms that cancel in the summations are discarded: 1 T  f ( t ) = c + ∑  ∫ 2T f ( t ) cos ( nω t ) dt  cos nω t − n T  T 1  + ∑  ∫ 2T f ( t ) sin ( nω t ) dt  sin nω t , − n T  T2 n = ±1, ±2, , and c = ∫ T f ( t ) dt T −2 Now, due to the equality of terms in ± n :  T2  f ( t ) = c + ∑  ∫ T f ( t ) cos ( nω t ) dt  cos nω t − n T  T 2  + ∑  ∫ 2T f ( t ) sin ( nω t ) dt  sin nω t , − n T  n = 1, 2,3, Equations 3.9.9 and 3.9.10 follow directly 3.21 f ( t ) = ∑ cn e inω t n T= 2π ω so , T2 cn = ∫ T f ( t ) e− inωt dt , T −2 ω cn = 2π = π ω π − ω ∫ and n = 0, ±1, ±2, … f ( t ) e− inω t dt π  ω  − inω t ω − inω t − + e dt e dt  ( ) π  ∫− ∫ 2π  ω   ω  −inω t = e 2π  inω  − π ω π  − inω t ω  − e  inω  1 − e + inπ − e− inπ + 1 2π in For n even, einπ = e − inπ = and the term in brackets is zero For n odd, einπ = e − inπ = −1 cn = , n = ±1, ±3, 2π in inω t f (t ) = ∑ e , n = ±1, ±3, n 2π in 1 inω t − inω t =∑ ( e − e ) , n = 1,3,5, n π n 2i 41 =∑ sin ( nω t ) , n = 1,3,5, n π n 4 1  f ( t ) = sin ω t + sin 3ω t + sin 5ω t + … π  = 3.22 In steady state, x ( t ) = ∑ An e ( i nω t −φn ) n An = Fn m (ω − n 2ω )2 + 4γ n 2ω    4F , n = 1,3,5, and Now Fn = nπ ω 9ω Q = 100 ≈ so γ ≈ 40, 000 2γ 4F A1 = ⋅ mπ 2 2  9ω ⋅ ω  2 ( 9ω − ω ) + 40000    F A1 ≈ 2mπω 4F A3 = ⋅ 3mπ 2 2   9ω  ( 9ω − 9ω ) +      200   400 F A3 ≈ 27mπω 4F ⋅ A5 = 5mπ 2   2  3ω  2 − + ω ω ω 25 )  200  ( )  (   F A5 ≈ 20mπω i.e., A1 : A3 : A5 = : 29.6 : 0.1 3.23 (a) x +ω2x = y=x ω = 3ω Thus y = −ω x x= y y dy ω x = =− x dx y divide these two equations: (b) Solving ydy ω2 + xdx = and Integrating Let 2C = A2 y2 x2 + =1 ω A2 A → y2 x2 + =C 2ω 2 an ellipse 3.24 The equation of motion is F ( x ) = x − x = mx For simplicity, let m=1 Then (a) (b) x = x − x This is equivalent to the two first order equations … x = y and y = x − x3 The equilibrium points are defined by x − x = x (1 − x )(1 + x ) = Thus, the points are: (-1,0), (0,0) and (+1,0) We can tell whether or not the points represent stable or unstable points of equilibrium by examining the phase space plots in the neighborhood of the equilibrium points We’ll this in part (c) dy y x − x = = or The energy can be found by integrating dx x y ∫ y dy = ∫ ( x − x ) dx + C or y x2 x4 = − +C 2 In other words … E = T + V = ` (c) y  x4 x2  +  −  = C The total energy C is 4 2 constant The phase space trajectories are given by solutions to the above equation  2 x4 y = ±  x − + 2C    The upper right quadrant of the trajectories is shown in the figure below The trajectories are symmetrically disposed about the x and y axes They form closed paths for energies C0 Thus, (0,0) is a point of unstable equilibrium 0.5 0 0.5 1.5 10 3.25 θ + sin θ =  d θ  − cos θ  = dt   ⇒ θ2 Integrating: θ θ = cos θ θ ) θ ∴T = ∫ θ = ( cosθ − cosθ or dθ  ( cos θ − cosθ )  Time for pendulum to swing from θ = to θ = θ is Now—substitute sin φ = θ sin sin so φ = θ and after some algebra … at θ = θ cos θ = − 2sin and use the identity π T θ dφ  2θ  1 − sin  = θ dθ   2θ θ  sin sin −    2     ∴T = ∫ dθ   2θ θ    sin − sin      1 or π dφ T = 4∫ (a) 1 − α sin φ  (1 − α sin φ ) (b) − where α = sin 2 θ ≈ + α sin φ + α sin φ + … π   T = ∫ dφ 1 + α sin φ + α sin φ + …   θ θ  θ2 (c) α = sin ≈ − + … ∼  48   θ2  T = 2π 1 + + …  16  -2 θ  α  T = 2π 1 + + α + …  64  11 ... x2  +  −  = C The total energy C is 4 2 constant The phase space trajectories are given by solutions to the above equation  2 x4 y = ±  x − + 2C    The upper right quadrant of the