## Driven linear oscillations

 Quote by gabbagabbahey But p doesn't need to be one of those two values (p1,p2). p1 and p2 are the values needed to obey the homogeneous part of the ODE, not the inhomogeneous part. The general form of the particular solution should be Ae^ipt +Be^-ipt +C not just e^ipt. The first two terms are necessary to account for the 36cos(pt) term on the RHS, while the constant term is needed to account for the -mg constant term. What do you get when you plug this particular solution into the ODE?
So I wouldn't write out use Euler term because of the extra constant term?

x= Ae^ipt +Be^-ipt +C

x'= ipAe^ipt -ipBe^-ipt + 0

x''=-p^2Ae^ipt -B^2e^-ipt+0

x''-1000x= 36eipt -10 , but F(t)=36 cos(t), so trig terms will not completely go away .
 Blog Entries: 1 Recognitions: Gold Member Science Advisor Staff Emeritus I'm going to jump out here, two people helping in a thread isn't constructive. Let me know if you need me to jump back in.

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 Quote by Benzoate So I wouldn't write out use Euler term because of the extra constant term? x= -p^2Ae^ipt -B^2e^-ipt x'= ipAe^ipt -ipBe^-ipt + 0 x''=-p^2Ae^ipt -B^2e^-ipt+0 x''-1000x= 36eipt -10 , but F(t)=36 cos(t), so trig terms will not completely go away .
You're missing a 1/m and cos(pt)≠e^ipt:

x''-1000x= (36/m)cos(pt) -10=18cos(pt)-10

=>-p^2Ae^ipt -Bp^2e^-ipt-1000(Ae^ipt -B^2e^-ipt)=18cos(pt)-10

But what is cos(pt) in terms of complex exponentials?

 Quote by gabbagabbahey You're missing a 1/m and cos(pt)≠e^ipt: x''-1000x= (36/m)cos(pt) -10=18cos(pt)-10 =>-p^2Ae^ipt -Bp^2e^-ipt-1000(Ae^ipt -B^2e^-ipt)=18cos(pt)-10 But what is cos(pt) in terms of complex exponentials?
Euler Formula:

ei$$\theta$$= cos($$\theta$$)+sin($$\theta$$)
therefore

18*eipt-10= 18*(cos(pt) + sin (pt))-10
 Recognitions: Homework Help But you don't have 18 e^ipt! You have 18 cos(pt). What is cos(pt) in term of complex exponentials? Hint: look under the section "Relationship to Trignometry" here

 Quote by gabbagabbahey But you don't have 18 e^ipt! You have 18 cos(pt). What is cos(pt) in term of complex exponentials? Hint: look under the section "Relationship to Trignometry" here

sorry if you are becoming frustrated. I have a copy of Gregory Douglass's Classical mechanics books and there is an example like this in that book on p. 10 cos(t) and they say the comple counter part is 10e^it

18 cos(pt)= 18e^ipt - 10(e^0)
 Recognitions: Homework Help cos(pt)≠e^ipt Euler's formula gives cos(pt)=Real[e^ipt]=(e^ipt+e^-ipt)/2 (Real[z] is the real part of the complex number z) so, 18cos(pt)-10=9e^ipt+9e^-ipt-10 =>-p^2Ae^ipt -Bp^2e^-ipt+1000(Ae^ipt +Be^-ipt+C)=9e^ipt+9e^-ipt-10 Compare the coefficients in front of each of the e^ipt,e^-ipt, and constant terms...what must A,B and C be (in terms of p)?

 Quote by gabbagabbahey cos(pt)≠e^ipt Euler's formula gives cos(pt)=Real[e^ipt]=(e^ipt+e^-ipt)/2 (Real[z] is the real part of the complex number z) so, 18cos(pt)-10=9e^ipt+9e^-ipt-10 =>-p^2Ae^ipt -Bp^2e^-ipt-1000(Ae^ipt -B^2e^-ipt+C)=9e^ipt+9e^-ipt-10 Compare the coefficients in front of each of the e^ipt,e^-ipt, and constant terms...what must A,B and C be (in terms of p)?
A(p^2-1000)=9 ==> A=9/(p^2-1000)
B(1000-p^2)=9 ==>B=9/(1000-p^2)
C=-10/1000?
 Recognitions: Homework Help Close, there were some typos in my last equation; you should get A=9/(1000-p^2), your B and C are correct though. Now, since A and B are equal you have: x(t)=A(e^ipt+e^-ipt)+C=2Acos(pt)+C Now, if the maximum extension of the spring is 4cm, what must the Value of A be?

 Quote by gabbagabbahey Close, there were some typos in my last equation; you should get A=9/(1000-p^2), your B and C are correct though. Now, since A and B are equal you have: x(t)=A(e^ipt+e^-ipt)+C=2Acos(pt)+C Now, if the maximum extension of the spring is 4cm, what must the Value of A be?
4=9/(1000-p^2)*cos(pt)-1/100 am I trying to find p?
 Recognitions: Homework Help Yes, since x(t)=Acos(pt)+C, it should be clear that the angular frequency of oscillation is p. So you want to find p. You need to be careful of your units though; 4cm =0.04m so you should have: 0.04=9/(1000-p^2)*cos(pt)-1/100 since the rest of the quantities in the equation are in meters.

 Quote by gabbagabbahey Yes, since x(t)=Acos(pt)+C, it should be clear that the angular frequency of oscillation is p. So you want to find p. You need to be careful of your units though; 4cm =0.04m so you should have: 0.04=9/(1000-p^2)*cos(pt)-1/100 since the rest of the quantities in the equation are in meters.
won't there be two solutions to p? The p's I finding for the xD are entirely different from the p's in the complementary function
 Sorry to bumped this thread again, even though its been two days since its been active. for my final solution I get c=36/(1000-p^2) and c=.04 ==> p1=10 and p2=-10; I'm not sure what my text means when it says 'spring is safe if p=<20 rad/s and p>=40. rad/s