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Consider a driven damped oscillator with k = 32.0 N/m, m = 0.5 kg and b = 1Ns/m. The driving force is F(t)=F0 cos ωt with F0 = 10N and ω= 2ω0 where ω0 is the natural frequency of the oscillator. Show that the solution with initial conditions x(t=0) = 2m and v(t=0) = 0 is: x(t) = 0.103 cos(16t − 2.98) + 2.11e^(−t)cos(7.94t − 0.109)

Consider a driven damped oscillator with k = 32.0 N/m, m = 0.5 kg and b = 1Ns/m. The driving force is F(t)=F0 cos ωt with F0 = 10N and ω= 2ω0 where ω0 is the natural frequency of the oscillator. Show that the solution with initial conditions x(t=0) = 2m and v(t=0) = 0 is: x(t) = 0.103 cos(16t − 2.98) + 2.11e^(−t)cos(7.94t − 0.109)

Elements Of Electromagnetics
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
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Consider a driven damped oscillator with k = 32.0 N/m, m = 0.5 kg and b = 1Ns/m. The driving force is F(t)=F0 cos ωt with F0 = 10N and ω= 2ω0 where ω0 is the natural frequency of the oscillator. Show that the solution with initial conditions x(t=0) = 2m and v(t=0) = 0 is:
x(t) = 0.103 cos(16t − 2.98) + 2.11e^(−t)cos(7.94t − 0.109)

(Use the symbols to solve the problem and plug in the numbers at the end of your calculation. You need to solve a differential equation instead of using the result for the driven damped harmonic oscillator.

Use picture attached as a hint. 

12. Consider a damped ancillator with = 32, m= 5,6= 1 in MKS () Find the solution with (0)=2,0)=0. It wing symbole till the every end. (4)Now add on a driving force Fcont with F₂ = 10N and 24. Find the solution with z(0) = 2, v(0) = 0. (3) From the begiven, we can clearly see that ²-4k will be low than , so this oscillator will have an underdamped solution. From class, you know that the solution for an underdamped clator takes the pen Yale courses werversity 2012. sto er Creative Com C (0)=-4 cum - z(t) = | = √3 = 7.94. Plugging in initial conditions we find VG) z(0) A(-) = 2 => A= -4G Therefore, A=2.02 and the full solution for the damped oscillator without a driving force is z(t) = Acom/t - da) = 2(Qx* com(7.94-0.125). (4) The full solution with a driving force is the solution we solved for above where the right hand side of the differential equation is to (kon as the complimentary solution) plus the solution that gives the right hand side equal to Foconut (the particule solution). I will write this as z(t)=(t) + 2y(1) where 2+(1) take the form from part (1) From this, we that The particular solutions is the same solutions we used in problem 10, ¹/²-) 7000 + (-₂) 1₂(t) = ₂ ( + 6) = where =√√8¹ and = 20 = 16 se"". We can find the value of ở by plugging back into the differential equation given by Equation 1. Doing this we find that F₁ con=(-²³) con(ut - 0) — 7 wake(ust – p). This equation has to hold at all times, so choosing t=0 to simply things we have in Open a course a 3.0 Un teret under the Creative C rations and further Falm ماه + ف م (ی - ا = ZOM Vậy là người luy ban thư kh the √(- w²1² +²² VI₂²-²³P+²² 2.98. We can also plug in numbers to calculate a value for zo and we Plugging in numbers we find, find 0.103m. Now, we need to use the initial conditions to solve for the fall solution where I have suppressed all the units. z(t) = 1₂(t) + 1₂(t) = ±¸¢xx(ust − 6) + A−79/7 con(²t - (a) where we have already solved for zo. 6, and u/ above. At t=0 we have z(0) = 2 = + Ac r(0) 6 =imo-A Tando-u'a -). Solving these two equations for A and we find that A-211 m and 0.100. Therefore, our full solution is z(t)=0.103(16-298) +2.11ec(7.94-0.109)
Transcribed Image Text:12. Consider a damped ancillator with = 32, m= 5,6= 1 in MKS () Find the solution with (0)=2,0)=0. It wing symbole till the every end. (4)Now add on a driving force Fcont with F₂ = 10N and 24. Find the solution with z(0) = 2, v(0) = 0. (3) From the begiven, we can clearly see that ²-4k will be low than , so this oscillator will have an underdamped solution. From class, you know that the solution for an underdamped clator takes the pen Yale courses werversity 2012. sto er Creative Com C (0)=-4 cum - z(t) = | = √3 = 7.94. Plugging in initial conditions we find VG) z(0) A(-) = 2 => A= -4G Therefore, A=2.02 and the full solution for the damped oscillator without a driving force is z(t) = Acom/t - da) = 2(Qx* com(7.94-0.125). (4) The full solution with a driving force is the solution we solved for above where the right hand side of the differential equation is to (kon as the complimentary solution) plus the solution that gives the right hand side equal to Foconut (the particule solution). I will write this as z(t)=(t) + 2y(1) where 2+(1) take the form from part (1) From this, we that The particular solutions is the same solutions we used in problem 10, ¹/²-) 7000 + (-₂) 1₂(t) = ₂ ( + 6) = where =√√8¹ and = 20 = 16 se"". We can find the value of ở by plugging back into the differential equation given by Equation 1. Doing this we find that F₁ con=(-²³) con(ut - 0) — 7 wake(ust – p). This equation has to hold at all times, so choosing t=0 to simply things we have in Open a course a 3.0 Un teret under the Creative C rations and further Falm ماه + ف م (ی - ا = ZOM Vậy là người luy ban thư kh the √(- w²1² +²² VI₂²-²³P+²² 2.98. We can also plug in numbers to calculate a value for zo and we Plugging in numbers we find, find 0.103m. Now, we need to use the initial conditions to solve for the fall solution where I have suppressed all the units. z(t) = 1₂(t) + 1₂(t) = ±¸¢xx(ust − 6) + A−79/7 con(²t - (a) where we have already solved for zo. 6, and u/ above. At t=0 we have z(0) = 2 = + Ac r(0) 6 =imo-A Tando-u'a -). Solving these two equations for A and we find that A-211 m and 0.100. Therefore, our full solution is z(t)=0.103(16-298) +2.11ec(7.94-0.109)
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