Lecture 3

Normal Modes

♦ Spring-mass systems

♦ Coupled oscillators with damping and forcing terms

SPRING-MASS SYSTEMS

• Two masses m moving on a straight line without friction

under the action of three springs:

kk kα α

u1 2u

mu1 = −αku1 − k(u1 − u2)

mu2 = −αku2 + k(u1 − u2)

• Determine normal modes of the system.

• Express the energy in terms of normal modes.

• Find solution for initial conditions u1 = u0, u2 = 0, u1 = 0, u2 = 0.

mu1 = −αku1 − k(u1 − u2)

mu2 = −αku2 + k(u1 − u2)

Setting

S =u1 + u2√

2, D =

u1 − u2√2

normal

coordinates

gives

mS + αkS = 0

mD + (α+ 2)kD = 0

⇒ ωS =

√

αk

m, ωD =

√

(α+ 2)k

mnormal frequencies

Kinetic energy : K =1

2m (u2

1+ u2

2) =

1

2m (S2 + D2)

Potential energy : V =1

2αku2

1+

1

2k (u2 − u1)

2 +1

2αku2

2

=1

2αkS2 +

1

2(α+ 2)kD2 =

1

2mω2

SS2 +

1

2mω2

DD2

⇒ E =1

2mS2 +

1

2mω2

SS2

︸ ︷︷ ︸

E1

+1

2mD2 +

1

2mω2

DD2

︸ ︷︷ ︸

E2

= E1 + E2 sum of the energies of each normal mode

General solution of the equations of motion:

S(t) = CS sin(ωSt+ ϕS) , D(t) = CD sin(ωDt+ ϕD)

that is, u1(t) =1√2

[CS sin(ωSt+ ϕS) + CD sin(ωDt+ ϕD)]

u2(t) =1√2

[CS sin(ωSt+ ϕS)− CD sin(ωDt+ ϕD)]

Given the initial conditions at time t = 0

u1 = u0 , u2 = 0 , u1 = 0 , u2 = 0

the solution satisfying these conditions is given by

u1(t) =1

2u0 (cosωSt+ cosωDt)

u2(t) =1

2u0 (cosωSt− cosωDt)

Homework

Three equal masses m are constrained to move on a circle without friction, subject to

the action of three springs of elastic constant k connecting the three masses pairwise

to each other. The equations of motion are given by

mu1 = −k(u1 − u2) + k(u3 − u1) ,

mu2 = −k(u2 − u3) + k(u1 − u2) ,

mu3 = −k(u3 − u1) + k(u2 − u3) .

Determine normal frequencies and normal coordinates of the system.

Answ.: Normal frequencies ω1 = 0, ω2 = ω3 =√

3k

m.

Normal coordinates q1 = 1√

3(u1 + u2 + u3), q2 = 1

√

6(u1 − 2u2 + u3), q3 = 1

√

2(u1 − u2).

Example

[CP4 Paper June 2008]

8. Two massless springs each have spring constant k. Masses 2m and m are attachedas shown in the figure.

2m

m

The masses make small vertical oscillations about their equilibirium positions.Show that the respective displacements x and y of the masses 2m and m satisfy thecoupled differential equations

d2x

dt2=

k

2m(y − 2x)

d2y

dt2=

k

m(x − y)

and explain why there is no term involving the acceleration due to gravity. [7]

Find expressions for the normal frequencies for small oscillations of the masses. [7]

Find the ratio of the amplitudes for each normal mode. [6]

9. a) Prove the identity

k

k

x

y m

2 m

m y = − k ( y − x )

2 m x = − k x − k ( x − y )

. .

. .

g does not appear because x and y are

(gravity will determine shift mg/k of the zero)displacements from equilibrium

y = (k/m)(x− y)

x = [k/(2m)](y − 2x)

Matrix method

Ansatz

(x

y

)

=

(X

Y

)

eiωt −→(−ω2 + k/m −k/2m

−k/m −ω2 + k/m

) (X

Y

)

= 0

∣∣∣∣

−ω2 + k/m −k/2m

−k/m −ω2 + k/m

∣∣∣∣= (−ω2 + k/m)2 − (k/m)2/2 = 0

⇒ ω2 − k

m= ± 1√

2

k

m

i.e.,

ω2 =k

m

(

1± 1√2

)

normal frequencies

• Normal mode 1: ω2 = ω2

1= (k/m)

(1 + 1/

√2)

(−ω2

1+ k/m)X = [k/(2m)]Y =⇒ −X/

√2 = Y/2 i.e., X/Y = −1/

√2

• Normal mode 2: ω2 = ω2

2= (k/m)

(1− 1/

√2)

(−ω2

2+ k/m)X = [k/(2m)]Y =⇒ X/

√2 = Y/2 i.e., X/Y = 1/

√2

X

Y

X

Y

move against move together

NM 1 NM 2

The damped driven pendulum

m1!!x = !" !x ! m

1gx / l

1+ k y ! x( ) + F cos#t

m2!!y = !" !y ! m

2gy / l

2! k y ! x( )

d2

dt2+

!m1

d

dt+

g

l1

+k

m1

"

#$%

&'(k

m1

(k

m2

d2

dt2+

!m2

d

dt+

g

l2

+k

m2

"

#$%

&'

"

#

$$$$

%

&

''''

x

y

"#$

%&'=F

m1

1

0

"#$

%&'Re e

i) t( )

x

y

!"#

$%&= Re

X

Y

!"#

$%&ei' t!

"#$

%&CF

d2

dt2+

!m1

d

dt+

g

l1

+k

m1

"

#$%

&'(k

m1

(k

m2

d2

dt2+

!m2

d

dt+

g

l2

+k

m2

"

#$%

&'

"

#

$$$$

%

&

''''

x

y

"#$

%&'=F

m1

1

0

"#$

%&'Re e

i) t( )

Substitute (complex) eigenvalues in (1) to obtain eigenvectors

!" 2+ i

#m1

" +g

l1

+k

m1

$

%&'

()!k

m1

!k

m2

!" 2+ i

#m2

" +g

l2

+k

m2

$

%&'

()

$

%

&&&&

'

(

))))

X

Y

$%&

'()=0

0

$%&

'()

!" 2+ i

#m1

" +g

l1

+k

m1

$

%&'

()!k

m1

!k

m2

!" 2+ i

#m2

" +g

l2

+k

m2

$

%&'

()

= 0Eigenvalue eq.

(1)

!"#$%&'(#%$%)*+#,$(#,%-.-'$/$0"#$1&)23-.&)$4,0#5)&.$

m1!!x = !" !x ! m

1gx / l

1+ k y ! x( ) + F cos#t

m2!!y = !" !y ! m

2gy / l

2! k y ! x( )

d2

dt2+

!m1

d

dt+

g

l1

+k

m1

"

#$%

&'(k

m1

(k

m2

d2

dt2+

!m2

d

dt+

g

l2

+k

m2

"

#$%

&'

"

#

$$$$

%

&

''''

x

y

"#$

%&'=F

m1

1

0

"#$

%&'Re e

i) t( )

d2

dt2+

!m1

d

dt+

g

l1

+k

m1

"

#$%

&'(k

m1

(k

m2

d2

dt2+

!m2

d

dt+

g

l2

+k

m2

"

#$%

&'

"

#

$$$$

%

&

''''

x

y

"#$

%&'=F

m1

1

0

"#$

%&'Re e

i) t( )

x

y

!"#

$%&= Re

P

Q

!"#

$%&ei' t!

"#$

%&!"#

!" 2+ i

#m1

" +g

l1

+k

m1

$

%&'

()!k

m1

!k

m2

!" 2+ i

#m2

" +g

l2

+k

m2

$

%&'

()

$

%

&&&&

'

(

))))

P

Q

$%&

'()*MP =

F

m1

1

0

$%&

'()

P !P

Q

"#$

%&'=M

(1 F

m1

1

0

"#$

%&'

M =

!" 2+ i

#m1

" +g

l1

+k

m1

$

%&'

()!k

m1

!k

m2

!" 2+ i

#m2

" +g

l2

+k

m2

$

%&'

()

$

%

&&&&

'

(

))))

x

y

!"#

$%&= Re M

'1 F

m1

1

0

!"#

$%&ei( t!

"#$

%&

!" 2 ! i#m" +

g

l+k

m

$%&

'()

!k

m

!k

m!" 2 ! i

#m" +

g

l+k

m

$%&

'()

= 0

Simple case m1= m

2= m l

1= l

2= l

!" 2 ! i#m" +

g

l

$%&

'()= 0 or !" 2 ! i

#m" +

g

l+

2k

m

$%&

'()= 0

!1,2 = i"2m

± !1,2

2 #"2m

$%&

'()2

Eigenvalues

!1

2=g

lor !

2

2=g

l+ 2

k

m! = 0 eigenvalues (c.f. previous result)