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Chapter 13

Periodic Motionor

Oscillation

Examples:

swinging chandeliersboats bobbing at anchorguitar stringsquartz crystals in wristwatchesatoms in solidsair molecules that transmit sound…

Oscillations in real word are usually damped; that is, the motion dies out gradually, transferring mechanical energy to thermal energy

Oscillations – motions that repeat themselves

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Periodic motionAny motion that repeat itself at regular intervals is called periodic motion or harmonic motion.

Key characteristicsPeriod (T) – time required for one cycle of periodic

motion. Units: s (seconds/cycle)Frequency (f) – number of oscillations that are

completed each second. Units: s-1 (cycle/second) that is 1 hertz = 1 Hz = 1 oscillation per second = 1 s-1

Tf 1=

Periodic motion – Simple harmonic motionSimple harmonic motion (SHM) – a special case of a periodic motion when the displacement x of the particle (object) from the origin is given as a sinusoidal function of time

)cos(2cos)( φωφπ+=⎟

⎠⎞

⎜⎝⎛ += txt

Txtx mm

SI unit for ω is the radian per second

fT

ππω 22==

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ExamplesExamples

)cos(2cos)( φωφπ+=⎟

⎠⎞

⎜⎝⎛ += txt

Txtx mm

The velocity and acceleration of SHM

The velocity of SHM

[ ])cos()()( φω +== txdtd

dttdxtv m

)sin()( φωω +−= txtv m

The acceleration of SHM

[ ])sin()()( φωω +−== txdtd

dttdvta m

)()cos()( 22 txtxta m ωφωω −=+−=

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ExamplesExamples

)cos()( φω += txtx m

)sin()( φωω +−= txtv m

)cos()( 2 φωω +−= txta m

CheckpointCheckpoint

)cos()( 2 φωω +−= txta m

In simple harmonic motion, the magnitude of the acceleration is: A) constant B) proportional to the displacement C) inversely proportional to the displacement D) greatest when the velocity is greatest E) never greater than g

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The force law for simple harmonic motion

From Newton’s second law

)( 2xmmaF ω−==

For a SHM – a restoring force that is proportional to the displacement but opposite in sign

Example: Hook’s law for a spring

kxF −= then 2ωmk =

mk

=ωkmT π2=

Car springs

When a family of four people with a total mass of 200 kg step into their 1200-kg car, the car springs compress 3.0 cm

(a) What is the spring constant for the car springs, assuming that they act as a single spring?

(b) How far will the car lower if loaded with 300 kg?(c) What are the period and frequency of the car after

hitting a bump? Assume the shock absorbers are poor, so the car really oscillates up and down.

example

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Car springs (cont.)example

HzT

f

skmT

kFx

xFkkxF

09.11

92.0105.6

)0.3000.1200(22

105.4

105.603.0

)8.9()0.200(

4

2

4

==

=⋅+

==

⋅==

⋅=⋅

===

−

frequency The

N/mkg period The

m

kg 300.0 with loaded car the if

N/mm

m/s kg then 2

ππ

Two springs

Suppose that two springs in figure have different spring constant k1 and k2. What is the frequency of oscillations of the block?

example

kxF −= mkf

π21

=

22

21

21

2121

22

11

21

)(21

21

fffm

kkf

xkkxkxkFmkf

mkf

net

+=+

=

+−=−−=

==

and

π

ππ

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Energy in simple harmonic motion

The potential energy

)(cos21

21)( 222 φω +== tkxkxtU m

The kinetic energy

)(sin21

21)( 222 φω +== tkxmvtK m

The total mechanical energy

2

21

mkxKUE =+=

ExamplesExamples

)(cos21)( 22 φω += tkxtU m

)(sin21)( 22 φω += tkxtK m

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Problem: energy in SHMAn object of mass m on a a horizontal frictionless surface is attached to a spring with spring constant k. The object is displaced from equilibrium of x0 horizontally and given an initial velocity of v0 back toward the equilibrium position.(a) What is the frequency of the motion?(b) What is the initial potential energy?(c) What is the initial kinetic energy?(d) What is the amplitude of the oscillation?

example

Problem: energy in SHMexample

2

2

2

12

,,,,,,

2

00

20

0

20

0

00

00

m

m

kxKUE

mvK

kxU

Tf

kmT

xKUfvxkm

=+=

=

=

== and

:Find :Given

π

0x

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The simple pendulumThe restoring force

sL

mgLsmgmgF ==≈ θ

Comparing to kxF −=

Lmgk =

gL

kmT ππ 22 ==

)sin(θmgF =For small angles

An experiment to measure g

22

22

/5.390.1

)2(

smT

g

LTLg

=

=

=

m for

π

gLT π2=

example

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Problem: Pendulum

A 2500 kg demolition ball sings from the end of the crane. The length of the swinging segment of cable is 17 m. (a) Find the period of the swinging, assuming that the system can be treated as a simple pendulum(b) Does the period depends on the ball’s mass?

ssm

mgLT 3.8

/8.90.1722 2 === ππ

Example

Two blocks (m=1.0 kg and M=10 kg) and a spring (k=200 N/m) are arranged on a horizontal, frictionless surface. The coefficient of static friction between the two blocks is 0.40. What amplitude of simple harmonic motion of the spring-block’s system puts the smaller block on the verge of slipping over the larger block?

mk

Mmggx

gx

xa

mgmaF

ss

s

s

59.0)(2max

max2

max2

max

maxmax

=+

==

=

=

==

μωμ

μω

ω

μ

and

then

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Equations

)()(

)sin()()cos()(

2 txta

txtvtxtx

m

m

ω

φωωφω

−=

+−=+=

Periodic Motion

Any motion that repeat itself at regular intervals

Simple harmonic motion)cos()( φω += txtx m

fT

ππω 22==

Restoration force

cmT

mc

cxmaF

πω 2==

−==

Spring

kmTkxF π2=−=

Pendilum

gLTx

LmgF π2=−=

Energy

22

21

21 cxmvE x +=

Damped OscillationsReal-world always have some dissipative (frictional) forces.The decrease in amplitude caused by dissipative forces is called by damping.Quite often damping forces are proportional to the velocity of the oscillating object.

02

2

=++

−−=

kxdtdxb

dtxdm

bvkxF xx Second-order differential equation (we need two initial conditions)

2

2

2/

4'

)'cos()(

mb

mk

textx mbtm

−=

+= −

ω

φω

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A pendulum + periodic external force

Model: 3 forces

• gravitational force

• frictional force is proportional to velocity

• periodic external force

)cos(,.),sin(

2

2

tFdtdmgL

dtdI

extfg

extfg

ωτθβτθτ

τττθ

=−=−=

++=

example 1

2220

202

2

,,

)cos()sin(

mLFf

mLLg

ImgL

tfdtd

dtd

====

+−−=

βαω

ωθαθωθ

)sin(202

2

θωθ−=

dtd

0 10 20 30 40 50 60-1.2

-1.0

-0.8

-0.6

-0.4

-0.2

0.0

0.2

0.4

0.6

0.8

1.0

1.2 θ(0)=1.0

θ(t)

time

dtd

dtd θαθωθ

−−= )sin(202

2

0 10 20 30 40 50-1.2

-1.0

-0.8

-0.6

-0.4

-0.2

0.0

0.2

0.4

0.6

0.8

1.0

1.2

θ(0)=1.0, α=0.1

θ(t)

time

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example 2: beats )cos()sin(202

2

tfdtd ωθωθ

+−=

When the magnitude of the force is very large – the system is overwhelmed by the driven force (mode locking) and the are no beats

When the magnitude of the force is comparable with the magnitude of the natural restoring force the beats may occur

0 20 40 60 80 100 120-2.5

-2.0

-1.5

-1.0

-0.5

0.0

0.5

1.0

1.5

2.0

2.5

θ(0)=0.2, α=0.0, f=0.2, ω=1.1

θ(

t)

time

example 3: resonance

)cos()sin(202

2

tfdtd ωθωθ

+−=

When the frequency of an external force is close to a natural frequency

0 20 40 60 80 100 120-2.5

-2.0

-1.5

-1.0

-0.5

0.0

0.5

1.0

1.5

2.0

2.5

θ(0)=0.2, α=0.0, f=0.1, ω=1.0

θ(t)

time

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Oscillations: The Tacoma bridge (1940)