Chapter 4 Part I

Chapter 4 Part I - CHAPTER 4 Second-Order Linear...

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CHAPTER 4 Second-Order Linear Differential Equations 4.1 The Harmonic Oscillator ! The Undamped Oscillator 1. !! xx += 0, x 01 () = , ! x 00 The general solution of the harmonic oscillator equation 0 is given by x t ct xt c t c t () =+ ()=− + 12 cos sin ! sin cos . Substituting this expression into the initial conditions x , ! x , gives xc 1 2 == ! so c 1 1 = , c 2 0 = . Hence, the IVP has the solution t cos . 2. 0, x , ! x The general solution of the harmonic oscillator equation 0 is given by x t c t + cos sin ! sin cos . Substituting this expression into the initial conditions x , ! x , gives 1 2 ! or cc 1 . Hence, the IVP has the solution t t cos sin . 253
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254 CHAPTER 4 Second-Order Linear Differential Equations In polar form, this would be xt t bg =− F H G I K J 2 4 cos π . 3. !! xx += 90 , x 01 () = , ! x The general solution of the harmonic oscillator equation is given by x t ct c t c t () =+ ()=− + 12 33 3 3 cos sin ! sin cos . Substituting this expression into the initial conditions x , ! x , gives xc 03 1 1 2 == ! so c 1 1 = , c 2 1 3 = . Hence, the IVP has the solution t t cos sin 3 1 3 3. In polar form, this would be t b g 10 3 3 δ where = tan 1 1 3 . This would be in the first quadrant. 4. 40 , x , ! x 02 The general solution of the harmonic oscillator equation is given by c t c t c t c t + 22 2 2 cos sin ! sin cos . Substituting this expression into the initial conditions x , ! x , gives 2 1 2 ! so c 1 1 = , c 2 1 . Hence, the IVP has the solution t t cos sin . In polar form, this would be t F H G I K J 4 .
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SECTION 4.1 The Harmonic Oscillator 255 5. !! xx += 16 0, x 01 ()=− , ! x 00 () = The general solution of the harmonic oscillator equation 16 0 is given by xt c t c t c t c t () =+ + 12 44 4 4 cos sin ! sin cos . Substituting this expression into the initial conditions x , ! x , gives xc 04 0 1 2 == ! so c 1 1 =− , c 2 0 = . Hence, the IVP has the solution t cos4 . 6. 16 0, x , ! x 04 The general solution of the harmonic oscillator equation 16 0 is given by c t c t c t + 4 4 cos sin ! sin cos . Substituting this expression into the initial conditions x , ! x , we get 04 4 1 2 ! so c 1 0 = , c 2 1 = . The IVP has the solution t sin4 . ! Computer Lab 7. ytt cos sin The equation tells us T = 2 π and because T = 2 ω , 0 1 = . We then measure the delay δ 0 08 . which we can compute as the phase angle 081 . . . The amplitude A can be measured directly giving A 14 . . Hence, cos sin . cos . tt t +≈ 08 . Compare with the algebraically exact form in Problem 13. –1.5 1.5 t T = 2 A 1.4 0.8 y
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256 CHAPTER 4 Second-Order Linear Differential Equations 8. yt t =+ 2cos sin The equation tells us T = 2 π and because T = 2 ω , 0 1 = . We then measure the delay δ 0 05 . , which we can compute as the phase angle () = 051 . . . The amplitude A can be measured directly giving A 22 . . Hence, 2 0 5 cos sin . cos . tt t +≈ () . 8 –4 –2.5 2.5 4 T = 2 A 2.2 0.5 y t 9. t 53 3 cos sin The equation tells us that period is T = 2 3 and because T = 2 , 0 3 = . We then measure the delay 0 005 . , which we can compute as the phase angle = 3005 015 ..
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Chapter 4 Part I - CHAPTER 4 Second-Order Linear...

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