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Fundamentals Of Differential Equations And Boundary Value Problems
Found in: Page 288
Fundamentals Of Differential Equations And Boundary Value Problems

Fundamentals Of Differential Equations And Boundary Value Problems

Book edition 9th
Author(s) R. Kent Nagle, Edward B. Saff, Arthur David Snider
Pages 616 pages
ISBN 9780321977069

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Short Answer

A double pendulum swinging in a vertical plane under the influence of gravity (see Figure 5.35) satisfies the system


When θ1 and θ2 are small angles. Solve the system when


The solutions for θ1 and θ2 are:


See the step by step solution

Step by Step Solution

Substituting the values of m1,m2,l1,l2

Substituting the given values for m1,m2,l1,l2 into the given system and taking g=9.8m/s2 one will get:


Dividing the first equation by 25 and the second by 10 one will get:


One will rewrite this system in operator form:


Finding  θ1

One will eliminate θ2 from the system. To do so one will multiply the first equation by D2+9.8 and the second by -2D2 and then add those two equations together.


Now, one can find a general solution for θ1. The auxiliary equation is:

15r4+98r2+96.04=0 , and its roots are:

r2=-98±9604-5762.430=-98±61.9830r1,22=-1.2, r3,42=-5.33r1,2=±1.2i, r3,4=±5.33ir1,2=±1.1i, r3,4=±2.31i

So, the general solution for θ1 is θ1(t)=c1cos1.1t+c2sin1.1t+c3cos2.31t+c4sin2.31t


Now, one can find a solution for θ2 from the first equation of the system



Integrating the previous equation twice one will get:


But this equation must satisfy the second equation of the system, which is:


Finding the solution for θ1 and  θ2

Since one does not have any constant nor a term multiplying t in D2θ1 and D2θ2, one cannot have it in θ1, and therefore A=B=0, so the solution for θ2is θ2(t)=1.57c1cos1.1t+1.57c2sin1.1t-1.57c3cos2.31t-1.57sin2.31t

The initial conditions give us:

θ1(0)=c1+c3=π6, θ2(0)=1.57c1-1.57c3=0c1=c3,2c1=π6,c1=c3=π12θ1'(0)=1.1c2+2.31c4=0, θ2'(0)=1.73c2-3.65c4=0c2=2.11c4,4.62c4=0,c2=c4=0

Finally, the solutions for θ1 and θ2 are


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