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