OscillationsMechanical• Mass-spring system• 2nd order differential eq.• Energy tossing between mass (kinetic energy) and

spring (potential energy)• Effect of friction, critical damping (shock absorber)• Simple pendulum, physical pendulum (sweet points)

• Tacoma Narrow Bridge: Example of Torsional Oscillation http://www.youtube.com/watch?v=3mclp9QmCGs

Mass-Spring System

( ) ( )

( )

2

2

2

2

by the spring: (N): spring force constant in N/m

Equation of motion for the mas

Restoring force

mass accelerations:

0

Try = co

=

F kxk

dv d dx d xM M M kxdt dt dt dt

d x t k x tdt M

x

F

t A

= −

= = = −

+ =

×

22

2

2 2

s .

cos sin , cos cos

Subsitution to the diff. eq. yields

cos cos 0

(rad/sec) oscillation angular frequency

t

d dA t A t A t A tdt dt

kMA t kA tM

kM

ω

ω ω ω ω ω ω

ω ω ω ω

ω

= − = −

− + = → =

=

Example: A spring 2 m longhanging from the ceiling elongates by 30 cm when a mass of 1.5 kg is attached to the end. If the mass is pulled down another5 cm and released, what is the oscillation frequen

2

cy?

The spring constant is from1.5 9.8 N 49 N/m

0.3 m49 N/m= 5.7 rad/s, 0.91 Hz1.5 kg

9.8 m/sAlternativley, 5.7 rad/s0.3 m

Example: Discuss energy exchange in the abov

F MgF k l kl l

k fM

Mg gM l l

ω

ω

×= ∆ → = = = =

∆ ∆

= = =

= = = =∆ ∆

( ) ( )

( )

2 2 2 20 0

22 2 2 2 2

0 0

2 2 20

e Example.Initial potential energy is1 1 149 0.05 0.061 J. At time t, cos2 2 2Kinetic energy of the mass is

1 1 1sin sin2 2 2

1 1The total enengy: cos sin2 2

k x kx t

dxM M x t kx tdt

kx t t k

ω

ω ω ω

ω ω

= × × =

= =

+ = 20 =const.x

Energy Tossing

( )

( )

0

22 2 2 2 2 2

0 0

2 2 2 20 0

Potential energy (red) + kinetic energy (green) = constant (black)If cos ,

1 1 1 1cos sin2 2 2 2

1 1cos sin2 2

x t x t

dxkx M kx t M x tdt

kx t t kx

ω

ω ω ω

ω ω

=

+ = +

= + =

0

0.1

0.2

0.3

0.4

0.5

0.5 1 1.5 2 2.5 3x

2 2

2 2

2

2

2 2

2 2

2

2

Energy conservation oscillation equation 1 1 const.2 2

1 1 0 02 2

,

Then 0 0

Effect of Friction

+ + 0 damping (dissipa

Mv kx

d dv dxMv kx Mv kxdt dt dt

dx dv d xvdt dt dt

d x d x kM kx xdt dt M

d x dxM f kxdt dt

→

+ =

+ = → + =

= =

+ = → + =

= →

2

2

tion of energy)

Compare with LCR circuit1+ + 0 d q R dq q

dt L dt LC=

2 4 6 8 10 12 14 16 18 20

-1.0

-0.5

0.0

0.5

1.0

t

x(t)

1 2 3 4 5 6 7 8 9 10

-1.0

-0.5

0.0

0.5

1.0

t

x(t)

0 1 2 3 4 5 6 7 8 9 100.0

0.5

1.0

t

x(t)( )

2 20 0

20

0

0

0

0

2 20

Damped oscillation

'' ' 0

" 2 ' 0

, 2

From top( ) 0.1( ) =0.5( )

Solution

cos sint

f kx x xM M

x x xf kM M

abc

x t x e t tγ

γ ω ω

γ ω

γ ωγ ωγ ω

γω ωω

ω ω γ

−

+ + =

+ + =

= =

=

=

= +

= −

Two springs

( )

1 1 2 22

1 2 1 12

21

1 1 122

21 2

1 1 121 2 1 2

Common force

=

1

1 1 1, effeff

F k x k xdM x x k xdtd kM x k xdt k

d k kM x x k xdt k k k k k

= − = −

+ −

+ = −

= − = − = ++

( )1 1 2 2 1 2 1

1 2

In this case, the force is additive,

eff

F k x k x k k xk k k= − − = − +

= +

Pendulum

2 2

2 2

Pendulm of length and mass Torque about the pivot sin (small )

0 0

Oscilaltion frequency independent of mass

Example: If 1 m,

9.8 rad/s =3.13 rad/s

0

L MMgL MgL

d d gI MgLdt dt L

gL

L

gL

f

θ θ θ

θ θθ θ

ω

ω

− −

+ = → + =

=

=

= =

=

.498 Hz

The oscillation frequency can be used for measuring .g

Nonlinear Oscillation

2

2

3 2

22

2

The pendulum equation

sin 0

is . The sine function can be expanded for small as1 1sin 16 6

1Then 1 0 and the oscillation frequency is modified a6

nonlinear

d gdt L

d gdt L

θ θ

θ

θ θ θ θ θ

θ θ θ

+ =

− + ⋅ ⋅ ⋅ = −

+ − =

( )

( )

20

0

20 0

s

1 / 6

where is the amplitude which reduces the oscillation frequency.

More advanced calsulation yields 1 / 8

gL

gL

ω θ θ

θ

ω θ θ

= −

= −

http://physics.usask.ca/~hirose/ep225/animation/pendulum/anim-pendulum.htm

2 4 6 8 10 12 14 16 18 20 22 24 26 28 30

-0.10-0.050.000.050.10

t

theta

2 4 6 8 10 12 14 16 18 20 22 24 26 28 30

-1.4-1.2-1.0-0.8-0.6-0.4-0.20.00.20.40.60.81.01.21.4

x

y

Numerical solutions small and large amplitude

Oscillation involving moment of inertia

2 2 2 2

2 2

A cylinder attached to a spring rolls on floor. Energy conservation is1 1 1 1const. where (moment of inertia of cylinder)2 2 2 2

3 1and . Then const.4 2

Oscillation frequency is =

Mv I kx I Ma

v a Mv kx

ω

ω

ω

+ + = =

= + =

23

kM

Stick pivoted at its end

2

2

2

2

2

A stick of length is pivoted at its end. The moment of inertia1about the end is . The restoring torque for small is3

1 1sin2 2

12

1 02

3 =2 2

If 1 m, =

L

I ML

LMg LMg

dI LMgdtdI LMgdt

LMg gI L

L

θ

θ θ

θ θ

θ θ

ω

ω

=

− −

= −

+ =

=

=

3.83 rad/sec, 0.61 Hzf =

Mg

-MgθL

θ

Circular loop

2 2 2 2

The moment of inertial of a circular loop about a point on itself is

+ =2Then the oascilaltion frequency

= 2

If fact, an incomplete loop also has the same oscillation frequency. (Ho

CMI I Ma Ma Ma Ma

ga

ω

= + =

mework)

pivot

Mg

θ

θ

a

a

CM

( )

( )

2/ 2

0

22 / 2

0

2 2/ 2

0

22

Half CircleThe moment of inertia about the pivot

2 2sin2

8 sin2

4 1 cos 4 12

2CM is at 1 from the pivot. Then

1 2 /

4 12

MI a ada

Ma d

Ma Mad

a

agMa

π

π

π

θ θπ

θ θπ

πθ θπ π

ππ

ωπ

π

=

=

= − = −

− −

=−

∫

∫

∫

2ga

=

θ/2θ/2

θ/2sin2a

a

θ

θ0

a(cosθ−cosθ0)

Sloshing Oscillation of Water in Pan, Lake, Bay (Seiche)

22

2

The center of mass follows the trajectory

6

where is the depth of water and is the channel length. The oscillation frequency is

= 12

Hy xaH a

H ga

ω

=

Sloshing Oscillation (cont.)

( )Let the water displacement at the edge be . The CM of the rectangle BDEC is at

10, ( )2

The CM of the triangle ABC is at2 1,

6 3 3Then the CM of the entire water is at

1 1 ,6 6

h t

x y H h

ax y h H h H h

a ax ah h yaH H

= = −

= = + − = −

= × = ( )2

2

22

2

2

1 1 12 3 2 6

Eliminating , we find 6

12

A ball rolling along a parabolic curve oscillates at

2

a H hH h ah H haH HHh y xa

Hga

y Ax

Ag

ω

ω

= − + − = +

=

=

=

= y = Ax2

Sweet Point

2

Sweet point of baseball bat, tennis racket, etc. canbe determined as follows. The grip point can be regardedas a pivot for physical pendulum. Its oscillation frequency is

=

Simple pendulum of

CM

MgdI Md

ω+

2

. Force actin

length : '=

g at the gri

. From '

p (pivot) becomes minimum if the ball is hit at thesweet point. (H

,

This determines the position of s

omework

weet poi

)

nt

CM CM

gLL

I Md IL dMd Md

ω ω ω=

+= = +

Electromagnetic Oscillations

2

2

2

2

Charged capacitor is suddenly connected to aninductor. Voltage balance is

10 But 0

The charge oscillates at1

If a resistance is inlcuded, 10

q di dq d qL i qC dt dt dt LC

LC

q di d q R dqL Ri qC dt dt L dt LC

ω

+ = = → + =

=

+ + = → + +

( ) -10

0

Weakly damped solution is

cos , (s )2

t Rq t q e tL

γ ω γ−

=

= =

( )

( )

22

2

2220

0

2 22 2 2 2 20

0

1 1The capacitor stores energy of 2 2

1The inductor stores energy of 2

1 1For cos , cos2 2

1 1 1 1sin sin2 2 2 2Total energy is constant as in spring-ma

qCVC

Li

qqq t q t tC C

dq t qLi L L q t tdt C

ω ω

ω ω ω

=

= =

= = =

ss mechanical oscillation.

The capacitor and inductor exchange energy periodically.

0

0.5

1

1.5

2

y

5e-05 0.0001 0.00015 0.0002 0.00025 0.0003t

33 -1

6

Example: 5 F capacitor charged to 1 kV is suddenlyconnected to a 2 H inductor. Assuming a 5 m circuit resistance, describe how the initial energy decays.

5 10 1.25 10 s is smaller t2 2 2 10RL

µµ

−

−

Ω

×= = ×

× ×

( ) ( )

( ) ( )

5 -1

3 1250 50

20

2500 2 5

han

1 =3.16 10 s . Weakly damped.

cos 10 cos 3.16 10

1Intial energy 2.5 J. Then2

2.5 cos 3.16 10

t t

t

LCV t V e t e t

CV

E t e t

γ

ω

ω− −

−

= ×

= = ×

=

= ×

Forced Oscillation

( )

0

02

2

0

If an circuit is driven by an oscillator with voltage cos , the amplitude of the current is

which exhibits a resonance1

1when .

Example: 1 mF, 1 mH, 0.1 1 V

1

10

LCRV t

Vi

L RC

LCC L R

V

i

ω

ωω

ω

ω

= − +

=

= = = Ω=

=2

33

1 0.0110

ωω

−−

− +

0

2

4

6

8

10

y

1000 2000 3000 4000 5000x