Problem 5-10, A.P French, Vibrations and Waves · 2015-04-05 · Problem 5-10, A.P French,...
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Problem 5-10, A.P French, Vibrations and Waves Calculating ω 2 Equations of Motion ¨ x 2 + ω 2 0 (2x 2 - x 3 )=0 (1) ¨ x 3 + ω 2 0 (x 3 - x 2 )=0 (2) Substituting in Trial Solution and Using Matrix Form Trial solution: x i = A i cos(ωt)= ⇒ -ω 2 A 2 + ω 2 0 (2A 2 - A 3 )=0 (3) -ω 2 A 3 + ω 2 0 (A 3 - A 2 )=0 (4) M A 2 A 3 = 2ω 2 0 - ω 2 -ω 2 0 -ω 2 0 ω 2 0 - ω 2 A 2 A 3 =0 (5) For a non trivial solution to the above equations we need det(M) = 0: ∴ (2ω 2 0 - ω 2 )(ω 2 0 - ω 2 ) - ω 4 0 =0 = ⇒ ω 4 - 3ω 2 ω 2 0 + ω 4 0 =0 (6) ω 2 = 3ω 2 0 ± p (9 - 4)ω 4 0 2 = 3 ± √ 5 2 ω 2 0 (7) Ratio of Frequencies f 2 α f 2 β = ω 2 α ω 2 β = 3+ √ 5 3 - √ 5 (8) Using the hint given in the question, we see that: 3+ √ 5= (1 + √ 5) 2 2 3 - √ 5= ( √ 5 - 1) 2 2 (9) ∴ f α f β = √ 5+1 √ 5 - 1 (10) Ratio of Amplitudes ω 2 α = 3 - √ 5 2 ω 2 0 (11) ω 2 β = 3+ √ 5 2 ω 2 0 (12) (13) By using the first row of matrix M with ω = ω α : 1+ √ 5 2 A 2 = A 3 = ⇒ A 2 A 3 = 1 - √ 5 2 (14) 1
Transcript of Problem 5-10, A.P French, Vibrations and Waves · 2015-04-05 · Problem 5-10, A.P French,...
Problem 5-10, A.P French, Vibrations and Waves
Calculating ω2
Equations of Motion
x2 + ω20(2x2 − x3) = 0 (1)
x3 + ω20(x3 − x2) = 0 (2)
Substituting in Trial Solution and Using Matrix Form
Trial solution: xi = Aicos(ωt) =⇒
−ω2A2 + ω20(2A2 −A3) = 0 (3)
−ω2A3 + ω20(A3 −A2) = 0 (4)
M
(A2
A3
)=
(2ω2
0 − ω2 −ω20
−ω20 ω2
0 − ω2
)(A2
A3
)= 0 (5)
For a non trivial solution to the above equations we need det(M) = 0:
∴ (2ω20 − ω2)(ω2
0 − ω2)− ω40 = 0 =⇒ ω4 − 3ω2ω2
0 + ω40 = 0 (6)
ω2 =3ω2
0 ±√
(9− 4)ω40
2=
3±√
5
2ω20 (7)
Ratio of Frequencies
f2α
f2β
=ω2α
ω2β
=3 +√
5
3−√
5(8)
Using the hint given in the question, we see that:
3 +√
5 =(1 +
√5)2
23−√
5 =(√
5− 1)2
2(9)
∴fαfβ
=
(√5 + 1√5− 1
)(10)
Ratio of Amplitudes
ω2α =
3−√
5
2ω20 (11)
ω2β =
3 +√
5
2ω20 (12)
(13)
By using the first row of matrix M with ω = ωα:
1 +√
5
2A2 = A3 =⇒ A2
A3=
1−√
5
2(14)
1
Following the same method with ω = ωβ :
1−√
5
2A2 = A3 =⇒ A2
A3=
1 +√
5
2(15)
c©Ben Williams, 2015 2
