Chapter 13 periodic motion

Collapse of the Tacoma Narrowssuspension bridge in America in 1940(p 415)

SHMSHM

dynamicsdynamics kinematicskinematics

Dynamic equation

Dynamic equation

Circle of reference

Circle of reference

Kinematicsequation

Kinematicsequation

oscillationoscillation

EnergyEnergy Superposition of shm

Superposition of shm

Damped oscillation

Damped oscillation

Forcedoscillation

Forcedoscillation

periodic motion / oscillationrestoring forceamplitude

cycleperiod

frequencyangular frequencysimple harmonic motionharmonic oscillator

circle of referencephasor

phase anglesimple pendulum

Key terms:

physical pendulumDampingDamped oscillationCritical dampingoverdampingunderdampingdriving forceforced oscillationnatural angular frequencyresonancechaotic motionchaos

Ideal model:

A) spring mass system

1)　 dynamic equation

xkdt

xdm

xkamF

2

2

02

2

xm

k

dt

xd

)tcos(Ax

022

2

xdt

xd

m

k

2

T

§1 Dynamic equation§1 Dynamic equation

Small angle approximation sin

02

2

L

g

dt

d

B) The Simple Pendulum

2

2

2

2

sin

sin

dt

dLg

Lsdt

sdmmg

maF tt

)cos(0 t

l

g2

mg

d

O

I

mgdI

mgd

dt

d

dt

dImgd

I

2

2

2

2

2

sin

C) physical pendulum

Example: Tyrannosaurus rex and physical pendulum

the walking speed of tyrannosaurus rex can be estimated from its leg length L and its stride length s

Conclusion:Equation of SHM

0xdt

xd 22

2

Solution: )tcos(Ax

Solution:

2

'

g r

GmMF ,Rr M

R

rr

3

4

R34

MM

3

33

3

'

RO

gF r

M

rR

GmMF

3g

2

2

3 dt

rdmr

R

GmM 0r

R

GM

dt

rd32

2

2

Example:A particle dropped down a hole that extends from one side of the earth, through its center, to the other side. Prove that the motion is SHM and find the period.

mTkM 22 )

4(

Example:An astronaut on a body mass measuring device (BMMD),designed for use on orbiting space vehicles,its purpose is to allow astronauts to measure their mass in the ‘weight-less’ condition in earth orbit.

The BMMD is a spring mounted chair,if M is mass of astronaut and m effective mass of the BMMD,which also oscillate, show that

02

2

2

RIm

kydt

ydWe have

Example:the system is as follow,prove the block

will oscillate in SHM

)4(

)3(

)2()(

)1(

2

21

1

Ra

kxT

IRTT

maTmg

Solution:

o

y

Alternative solution

222

2

1

2

1

2

1mvIkymgy

Rv

(1)

(2)

Take a derivative of y with respect to x

0xdt

xd 22

2

Solution: )tcos(Ax

)tsin(Adt

dxv

)tcos(Adt

xda 2

2

2

2.1 Equation

§ 2 kinematic equation§ 2 kinematic equation

)tcos(Ax )tsin(Av

xtAa 22 )cos(

1) Amplitude (A): the maximum magnitude of displacement from equilibrium.

2) Angular frequency(): T

2

Spring-mass:m

k

Simple pendulum:l

g

is not angular frequency rather than velocity .it depends on the system

Caution:

A) Basic quantity:

2.2) 　 the basic quantity——amplitude、 period,phase

In phase

2,1,0k

k212

2,1,0k

)1k2(12

Out of phase0

012 012

Lag in phaseAhead in phase

3) Phase angle ( = t+ ): the status of the object.

)tcos(Ax

sinAv,cosAx,then 00

2020 )

v(xA

0

0

x

vtg

0xdt

xd 22

2

Caution:Caution:

B) The formula to solve: A, ,

1) is predetermined by the system.

2) A and are determined by initial condition:

if t=0, x=x0, v=v0 ,

Is fixed by initial condition

Solution:

m

k ,5

4

100 2

202

0

vxA

m4.1

10

0 x

vtg

00 vand

4

)4

5cos(4.1 tx

An object of mass 4kg is attached to a spring of k=100N.m-1. The object is given an initial velocity of v0=-5m.s-1 and an initial displacement of x0=1. Find the kinematics equation

)cos( 0 tAx

Circle of reference method

)tcos(Ax

)tsin(Av

)tcos(Aa 2

Compare SHM with UCM

x(+), v(-), a(-)In first quadrant

Angle between OQ and axis-x

Phase

Angular VelocityAngular Frequency

ProjectionDisplacement

x

Radius

UCM

Amplitude

SHM

A

Ox

AA

Q

P

20

30

o xo x /3

6

3

1

x(m)

o t(s)

0.8

1

x(cm)

o t(s)

63

Example:Find the initial phase of the two oscillation

1

2

3

4

1

2

a b

10

-2

x (m)

t (s)

2

2

,40

22 2

att

btt

2

3

2

ba

t

tfor

0

4

3

1

)4(20

tt

a 0t

SHM: x-t graph,find 0 , a , b , and the angular frequency

Solution:

From circle of reference

§ 3 Energy in SHM§ 3 Energy in SHM

)(sin2

1)(sin

2

1

2

1 222222 tkAtAmmvKE

)(cos2

1

2

1 222 tkAkxPE

2222

2

1)(cos)(sin

2

1kAttkAPEKE

Kinetic energy:

Potential energy:

Total energy of the system:

Mk

X

O

Solution:

1v

,kA2

1Mv

2

1 21

21 ,v)mM(Mv 21

,kA2

1v)mM(

2

1 22

22

mM

MAA 12

Mm

k

Example:Spring mass system.particle move from left to right, amplitude A1. when the block passes through its equilibrium position, a lump of putty dropped vertically on to the block and stick to it. Find the kinetic equation suppose t=0 when putty dropped on to the block

)cos( 02 tAx

20

Example:A wheel is free to rotate about its fixed axle,a spring is attached to one of its spokes a distance r from axle.assuming that the wheel is a hoop of mass m and radius R,spring constant k. a) obtain the angular frequency of small oscillations of this system b) find angular frequency and how about r=R and r=0

)cos( 111 tAx )cos( 222 tAx

)cos(21 tAxxx

)cos(2 1221

2

2

2

1 AAAAA2211

2211

coscossinsin

AAAA

arctg

k212 max2121

2

2

2

1 2 AAAAAAAA

)12(12 kmin2121

2

2

2

1 2 AAAAAAAA

4.1 mathematics method

§ 4. Superposition of SHM§ 4. Superposition of SHM

2211

2211

coscos

sinsintg

AA

AA

(2-1)M1

A1

1

M

A2

xo

xx1 x2

ω

A

A2

2

M2

)cos(2 12212

22

1 AAAAA

B) circle of reference

x= x1+x2= Acos( t+ )

x1 =A1cos( t+1 )

x2 =A2cos( t+2 )

Solution:

x3 3O1A

2AA

Draw a circle of reference,

)tcos(Axxx 21

cmt )4

32cos(23

Example:x1=3cos(2t+)cm, x2=3cos(2t+/2)cm, find the superposition displacement of x1 and x2.

dt

dxbkx

dt

xdm

bvkxmaF

2

2

5.1 Phenomena

)cos()( 2

tAetx mbt

5.2 equation

If damping force is relative small

2

2

4m

b

m

k

2

2

4m

b

m

k

)(0

0

0

22

11

tata ececxgoverdampin

dampingcritical

ngunderdampi

§ 5 Damped Oscillations§ 5 Damped Oscillations

0.0 0.5 1.0 1.5 2.0

-0.8

-0.6

-0.4

-0.2

0.0

0.2

0.4

0.6

0.8

1.0

underdamping

overdamping

No oscillation

Amplitude decrease

0

)cos()( 2

tAetx mbt

Critical damping

tFdt

dxbkx

dt

xdm sin02

2

The steady-state solution is

2

220

2

0

)cos()(

mb

mF

A

tAtx

where 0=(k/m)½ is the natural frequency of the system.

The amplitude has a large increase near 0 - resonance

:§6 Forced Oscillations:§6 Forced Oscillations

drive an oscillator with a sinusoidally varying force:

Projectile motion with air resistance

(case study:p147)

1. Projectile motion with air resistance

A plane moves in constant velocity due eastward,a missile trace it,suppose at anytime the missile direct to plane,speed is u,u>v,draw the path of missile

h

x

(x,y)

(X,Y)(X0,h) v

u

2. Tracing problem

vy

O

u

y(0)=0, x(0)=0

Y=h,X(0)=0

h

x

(x,y)

(X,Y)

22 )()( xXyY

yY

udt

dy

22 )()( xXyY

xX

udt

dx

tnvnyny

tnvnxnx

y

x

)()()1(

)()()1(

22 )()( xXyY

xXu

dt

dxvx

22 )()( xXyY

yYu

dt

dyv y

tVnXnX )()1(

0YY

Solution:

Example:the orbits of satellites in the gravitational field

d

dr

r

v

d

dr

dt

drv

Er

GmM

mr

Lmv

vvv

Er

GmMmv

mrLvLmrv

Lvmr

r

er

r

e

2

22

222

2

22

1

2

1

v

rvv

r

me

ms

3. Planets trajectory

We get:222 22 LrmGmEmrr

Ldrd

e

cos1 e

Rr

reference : 《大学物理》吴锡珑 p 149

2r/r̂GMmUF

Solution1: Newton’s laws

)rr(mma)r(F 2r

0)r2r(mma)(F

GM/hp,

hyperbola1e

parabola1e

ellipse1e

cose1

pr 2

3. Planets trajectory

r/GMmU

Solution2: conservation of mechanical energy

E)r(Umv2

1 2 0)r2r(m

GM/hp

hyperbola0E

parabola0E

ellipse0E

cos))GM(

h(

mE2

11

pr

2

2

2