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# Sect5 6

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### Sect5 6

1. 1. SECTION 5.6 FORCED OSCILLATIONS AND RESONANCE 1. Trial of x = A cos 2t yields the particular solution xp = 2 cos 2t. (Can you see that — because the differential equation contains no first-derivative term — there is no need to include a sin 2t term in the trial solution?) Hence the general solution is x(t) = c1 cos 3t + c2 sin 3t + 2 cos 2t. The initial conditions imply that c1 = -2 and c2 = 0, so x(t) = 2 cos 2t - 2 cos 3t. 2. Trial of x = A sin 3t yields the particular solution xp = -sin 3t. Then we impose the initial conditions x(0) = x′(0) = 0 on the general solution x(t) = c1 cos 2t + c2 sin 2t - sin 3t, and find that x(t) = (3/2)sin 2t - sin 3t. 3. First we apply the method of undetermined coefficients — with trial solution x = A cos 5t + B sin 5t — to find the particular solution xp = (3/15)cos 5t + (4/15)sin 5t 1 3 4  1 = 3 5 cos 5t + 5 sin 5t  = 3 cos (5t − β )  where β = tan-1(4/3) ≈ 0.9273. Hence the general solution is x(t) = c1 cos 10t + c2 sin 10t + (3/15)cos 5t + (4/15)sin 5t. The initial conditions x(0) = 25, x′(0) = 0 now yield c1 = 372/15 and c2 = -2/15, so the part of the solution with frequency ω = 10 is xc = (1/15)(372 cos 10t - 2 sin 10t) = 138388  372 2  =  cos10t − sin10t  15  138388 138388  138388 = cos (10t − α ) 15 where α = 2π - tan-1(1/186) ≈ 6.2778 is a fourth-quadrant angle.
2. 2. 4. Noting that there is no first-derivative term, we try x = A cos 4t and find the particular solution xp = (10 / 9) cos 4t. Then imposition of the initial conditions on the general solution x(t ) = c1 cos 5t + c2 sin 5t + (10 / 9) cos 4t yields x(t) = [(-10/9)cos 5t + 2 sin 5t] + (10/9)cos 4t 2 10 = ( −5cos 5t + 9 sin 5t ) + cos 4t 9 9 2 106  5 9  10 = − cos 5t + sin 5t  + cos 4t 9  106 106  9 where α = π - tan-1(9/5) ≈ 2.07789 is a second-quadrant angle. 5. Substitution of the trial solution x = C cos ω t gives C = F0 /(k - mω2). Then imposition of the initial conditions x(0) = x0 , x′(0) = 0 on the general solution x(t ) = c1 cos ω 0t + c2 sin ω 0t + C cos ω t (where ω 0 = k / m ) gives the particular solution x(t) = (x0 - C)cos ω0t + C cos ωt. 6. First, let's write the differential equation in the form x′′ + ω 0 x = ( F0 / m ) cos ω 0t , which 2 is the same as Eq. (13) in the text, and therefore has the particular solution xp = ( F0 / 2mω 0 ) t sin ω 0t given in Eq. (14). When we impose the initial conditions x(0) = 0, x′(0) = v0 on the general solution x(t ) = c1 cos ω 0t + c2 sin ω 0t + ( F0 / 2mω 0 ) t sin ω 0t we find that c1 = 0, c2 = v0 / ω 0 . The resulting resonance solution of our initial value problem is 2mv0 + F0 t x(t ) = sin ω 0t. 2mω 0 In Problems 7-10 we give first the trial solution xp involving undetermined coefficients A and B, then the equations that determine these coefficients, and finally the resulting steady periodic solution xsp. 7. xp = A cos 3t + B sin 3t ; − 5 A + 12 B = 0, 12 A + 5B = 0 50 120 10  5 12  10 xsp (t ) = − cos 3t + sin 3t =  − cos 3t + sin 3t  = cos(3t − α ) 169 169 13  13 13  13
3. 3. α = π − tan −1 (12 / 5) ≈ 1.9656 (2nd quadrant angle) 8. xp = A cos 5t + B sin 5t ; − 20 A + 15 B = − 4, 14 A + 20 B = 0 16 12 4 4 3  4 xsp (t ) = cos 5t − sin 5t =  cos 5t − sin 5t  = cos(5t − α ) 125 125 25  5 5  25 α = 2π − tan −1 (3 / 4) ≈ 5.6397 (4th quadrant angle) 9. xp = A cos10t + 10sin 5t ; − 199 A + 20 B = 0, − 20 A − 199 B = 3 60 597 xsp (t ) = − cos10t − sin10t 40001 40001 3  20 199  3 = − cos10t − sin10t  = cos(10t − α ) 40001  40001 40001  40001 α = π + tan −1 (199 / 20) ≈ 4.6122 (3rd quadrant angle) 10. xp = A cos10t + 10sin 5t ; − 97 A + 30 B = 8, − 30 A − 97 B = 6 956 342 xsp (t ) = − cos10t − sin10t 10309 10309 2 257725  478 171  = − cos10t − sin10t  ≈ 0.09849 cos(10t − α ) 10309  257725 257725  α = π + tan −1 (171/ 478) ≈ 3.4851 (3rd quadrant angle) Each solution in Problems 11-14 has two parts. For the first part, we give first the trial solution xp involving undetermined coefficients A and B, then the equations that determine these coefficients, and finally the resulting steady periodic solution xsp. For the second part, we give first the general solution x(t) involving the coefficients c1 and c2 in the transient solution, then the equations that determine these coefficients, and finally the resulting transient solution xtr. 11. xp = A cos 3t + B sin 3t ; − 4 A + 12 B = 10, 12 A + 4 B = 0 1 3 10  1 3  10 xsp (t ) = − cos 3t + sin 3t = − cos 3t + sin 3t  = cos (3t − α ) 4 4 4  10 10  4 α = π − tan −1 (3) ≈ 1.8925 (2nd quadrant angle) x(t ) = e−2 t ( c1 cos t + c2 sin t ) + xsp (t ); c1 − 1/ 4 = 0, − 2c1 + c2 + 9 / 4 = 0
4. 4. 1 7  50  1 7  50 xtr (t ) = e−2 t  cos t − sin t  =  cos t − sin t  = cos (t − β ) 4 4  4  50 50  4 β = 2π − tan −1 (7) ≈ 4.8543 (4th quadrant angle) 12. xp = A cos 5t + B sin 5t ; − 12 A − 30 B = 0, − 30 A − 12 B = 10 25 10 xsp (t ) = − cos 5t − sin 5t 87 87 5 29  5 2  5 = − cos 3t − sin 3t  = cos (3t − α ) 87  29 29  3 29 α = π + tan −1 (2 / 5) ≈ 3.5221 (3rd quadrant angle) x(t ) = e −3 t ( c1 cos 2t + c2 sin 2t ) + xsp (t ); c1 − 25 / 87 = 0, − 3c1 + 2c2 − 50 / 87 = 0  50 125  xtr (t ) = e−3t  cos 2t + sin 2t   174 174  25 29  2 5  25 =  cos 2t + sin 2t  = cos ( 2t − β ) 174  29 29  6 29 α = tan −1 (5 / 2) ≈ 1.1903 (1st quadrant angle) 13. xp = A cos10t + B sin10t ; − 94 A + 20 B = 3, 20 A + 94 B = 0 141 30 xsp (t ) = − cos10t + sin10t 4618 4618 3 2309  47 10  3 = − cos10t + sin10t  = cos (10t − α ) 4618  2309 2309  2 2309 α = π − tan −1 (10 / 47) ≈ 2.9320 (2nd quadrant angle) ( ) x(t ) = e − t c1 cos t 5 + c2 sin t 5 + xsp (t ); c1 − 141/ 4618 = 10, − c1 + 5 c2 + 150 / 2309 = 0 e−t xtr (t ) = 4618 5 (46321 5 cos t 5 + 46021sin t 5 ) = 3 309083 − t 23090 ( e cos t 5 − β ) ≈ 10.9761 e −t ( cos t 5 − β ) β = tan −1 (46321 5 / 46021) ≈ 1.1527 (1st quadrant angle)
5. 5. 14. xp = A cos t + B sin t ; 24 A + 8B = 5, − 8 A + 24 B = 13 1 22 485  1 22  485 xsp (t ) = cos t + sin t =  cos t + sin t  = cos (t − α ) 40 40 40  485 485  40 α = tan −1 (22) ≈ 1.5254 (1st quadrant angle) x(t ) = e− 4 t ( c1 cos 3t + c2 sin 3t ) + xsp (t ); c1 + 1/ 40 = 5, − 4c1 + 3 c2 + 11/ 20 = 0  199 258  xtr (t ) = e−4 t  cos 3t + sin 3t   40 40  106165 −4 t  199 258  106165 −4 t = e  cos 3t + sin 3t  = e cos (3t − β ) 40  106165 106165  40 β = tan −1 (288 /199) ≈ 0.9138 (1st quadrant angle) In Problems 15-18 we substitute x(t ) = A(ω ) cos ω t + B(ω ) sin ω t into the differential equation mx′′ + cx′ + kx = F0 cos ω t with the given numerical values of m, c, k , and F0 . We give first the equations in A and B that result upon collection of coefficients of cos ω t and sin ω t , and then the values of A(ω ) and B(ω ) that we get by solving these equations. Finally, C = A2 + B 2 gives the amplitude of the resulting forced oscillations as a function of the forcing frequency ω . 15. (2 − ω 2 ) A + 2ω B = 2, − 2ω A + (2 − ω 2 ) B = 0 A = 2 (2 − ω 2 ) / (4 + ω 4 ) , B = 4ω / ( 4 + ω 4 ) C(ω ) = 2 / 4 + ω 4 begins with C(0) = 1 and steadily decreases as ω increases. Hence there is no practical resonance frequency. 16. (5 − ω 2 ) A + 4ω B = 10, − 4ω A + (5 − ω 2 ) B = 0 A = 10 (5 − ω 2 ) / ( 25 + 6ω 2 + ω 4 ) , B = 40ω / ( 25 + 6ω 2 + ω 4 ) C(ω ) = 10 / 25 + 6ω 2 + ω 4 begins with C(0) = 2 and steadily decreases as ω increases. Hence there is no practical resonance frequency. 17. (45 − ω 2 ) A + 6ω B = 50, − 6ω A + (45 − ω 2 ) B = 0 A = 50 ( 45 − ω 2 ) / ( 2025 − 54ω 2 + ω 4 ) , B = 300ω / ( 2025 − 54ω 2 + ω 4 )
6. 6. C(ω ) = 50 / 2025 − 54ω 2 + ω 4 so, to find its maximum value, we calculate the derivative −100 ω ( −27 + ω 2 ) C ′(ω ) = . (2025 − 54ω 2 + ω 4 )3 / 2 Hence the practical resonance frequency (where the derivative vanishes) is ω = 27 = 3 3. 18. (650 − ω 2 ) A + 10ω B = 100, − 10ω A + (650 − ω 2 ) B = 0 A = 100 ( 650 − ω 2 ) / ( 422500 − 1200ω 2 + ω 4 ) , B = 1000ω / ( 422500 − 1200ω 2 + ω 4 ) C(ω ) = 100 / 422500 − 1200ω 2 + ω 4 so, to find its maximum value, we calculate the derivative −200 ω ( −600 + ω 2 ) C ′(ω ) = . ( 422500 − 1200ω 2 + ω 4 )3 / 2 Hence the practical resonance frequency (where the derivative vanishes) is ω = 600 = 10 6. 19. m = 100/32 slugs and k = 1200 lb/ft, so the critical frequency is ω 0 = k /m = 384 rad/sec = 384 / 2π ≈ 3.12 Hz. 20. Let the machine have mass m. Then the force F = mg = 9.8m (the machine's weight) causes a displacement of x = 0.5 cm = 1/200 meters, so Hooke's law F = kx, that is, mg = k (1/ 200) gives the spring constant is k = 200mg (N/m). Hence the resonance frequency is ω = k /m = 200 g ≈ 200 × 9.8 ≈ 44.27 rad / sec ≈ 7.05 Hz , which is about 423 rpm (revolutions per minute). 21. If θ is the angular displacement from the vertical, then the (essentially horizontal) displacement of the mass is x = Lθ, so twice its total energy (KE + PE) is m(x')2 + kx2 + 2mgh = mL2(θ')2 + kL2θ2 + 2mgL(1 - cos θ) = C. Differentiation, substitution of θ ≈ sin θ, and simplification yields θ''+ (k/m + g/L)θ = 0 so
7. 7. ω0 = k / m + g / L. 22. Let x denote the displacement of the mass from its equilibrium position, v = x′ its velocity, and ω = v/a the angular velocity of the pulley. Then conservation of energy yields mv2 /2 + Iω2 /2 + kx2 /2 - mgx = C. When we differentiate both sides with respect to t and simplify the result, we get the differential equation (m + I/a2)x″ + kx = mg. Hence ω = k / ( m + I / a 2 ). 23. (a) In ft-lb-sec units we have m = 1000 and k = 10000, so ω 0 = 10 rad/sec ≈ 0.50 Hz. (b) We are given that ω = 2π / 2.25 ≈ 2.79 rad/sec, and the equation mx'' + kx = F(t) simplifies to x′′ + 10 x = (1/ 4) ω 2 sin ω t. When we substitute x(t) = A sin ωt we find that the amplitude is A = ω 2 / 4 (10 − ω 2 ) ≈ 0.8854 ft ≈ 10.63 in. 24. By the identity of Problem 43 in Section 5.5, the differential equation is mx′′ + kx = F0 (3cos ω t + cos 3ω t ) / 4. Hence resonance occurs when either ω or 3ω equals ω 0 = k / m , that is, when either ω = ω 0 or ω = ω 0 / 3. 25. Substitution of the trial solution x = A cos ω t + B sin ω t in the differential equation, and then collection of coefficients as usual yields the equations (k − mω ) A + (cω ) B 2 = 0, − (cω ) A + ( k − mω 2 ) B = F0 with coefficient determinant ∆ = ( k − mω 2 ) + ( cω ) and solution A = − ( cω ) F0 / ∆, 2 2 B = ( k − mω 2 ) F0 / ∆. Hence
8. 8. F0  k − mω 2 cω  x(t ) =  sin ω t − cos ω t  = C sin (ω t − α ) , ∆ ∆ ∆  where C = F0 / ∆ and sin α = cω / ∆ , cos α = ( k − mω 2 ) / ∆ . 26. Let G0 = E02 + F02 and ρ = 1/ (k − mω 2 ) + (cω ) 2 . Then x sp (t ) = ρE0 cos(ω t − α ) + ρF0 sin(ω t − α ) = ρG0 E cos(ω t − α ) + F sin(ω t − α )# !G \$ 0 0 0 G 0 = ρG0 cos β cos(ω t − α ) + sin β sin(ω t − α ) x sp (t ) = ρG0 cos(ω t − α − β ) where tan β = F0 / E0 . The desired formula now results when we substitute the value of ρ defined above. (k − mω ) + (cω ) 2 The derivative of C (ω ) = F0 / 2 2 27. is given by ω F0 (c 2 − 2km) + 2(mω ) 2 C ′(ω ) = − . ( k − mω ) + ( cω ) 3/ 2 2  2 2 2    (a) Therefore, if c 2 ≥ 2km, it is clear from the numerator that C ′(ω ) 0 for all ω, so C(ω) steadily decreases as ω increases. (b) But if c 2 2km, then the numerator (and hence C ′(ω ) ) vanishes when ω = ω m = k / m − c 2 / 2m 2 k / m = ω 0 . Calculation then shows that 16 F0 m3 (c 2 − 2km) C ′′(ω m ) = 0, c3 ( 4km − c 2 ) 3/ 2 so it follows from the second-derivative test that C (ω m ) is a local maximum value. 28. (a) The given differential equation corresponds to Equation (17) with F0 = mAω 2 . It therefore follows from Equation (21) that the amplitude of the steady periodic vibrations at frequency ω is
9. 9. F0 mAω 2 C(ω ) = = . ( k − mω 2 )2 + (cω )2 (k − mω 2 ) 2 + (cω )2 (b) Now we calculate mAω  2k 2 − (2 mk − c 2 )ω 2    C ′(ω ) = , ( k − mω 2 )2 + ( cω )2  3/ 2     and we see that the numerator vanishes when 2k 2 k  2mk  k ω = = = ω0 . 2mk − c 2 m  2mk − c 2    m 29. We need only substitute E0 = acω and F0 = ak in the result of Problem 26. 30. When we substitute the values ω = 2π v / L, m = 800, k = 7 × 104 , c = 3000 and L = 10, a = 0.05 in the formula of Problem 29, simplify, and square, we get the function 25 (9π 2v 2 + 122500 ) Csq (v) = 16 (16π 4 v 4 − 64375π 2 v 2 + 76562500 ) 2 giving the square of the amplitude C (in meters) as a function of the velocity v (in meters per second). Differentiation gives 50π 2 v(9π 4 v 4 + 245000π 2 v 2 − 535937500) Csq′(v) = − . (16π 4 v 4 − 64375π 2 v 2 − 76562500) 2 Because the principal factor in the numerator is a quadratic in v2, it is easy to solve the equation Csq′(v) = 0 to find where the maximum amplitude occurs; we find that the only positive solution is v ≈ 14.36 m/sec ≈ 32.12 mi/hr. The corresponding amplitude of the car's vibrations is Csq(14.36) ≈ 0.1364 m = 13.64 cm.