# Driven linear oscillations

by Benzoate
Tags: driven, linear, oscillations
HW Helper
P: 5,004
 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?
P: 569
 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
 HW Helper P: 5,004 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
P: 569
 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)
 HW Helper P: 5,004 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)?
P: 569
 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?
 HW Helper P: 5,004 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?
P: 569
 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?
 HW Helper P: 5,004 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.
P: 569
 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
 P: 569 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

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