1LECTURE 4 CP Ch 14

Damped Oscillations

Forced Oscillations and Resonance

2

max ma

2

2 2 2max

2 2 2 2 2 2max

x

a

max

2m x

cos sin

cos

1 1cos

2 21 1 1

sin sin2 2 2

x t v t

a t a x

PE k x k x t

KE m v m x t

x x

x

k x t

2 2max max

1 1= constant

2 2totalE KE PE k x m v

2 2 2 2 2max max

1 1 1

2 2 2m v k x k x v x x

2

22 f

Tk

m

CP 445

SHM

3

0 10 20 30 40 50 60 70 80 90 100-10

0

10SHM

posi

tion

x

0 10 20 30 40 50 60 70 80 90 100-5

0

5

velo

city

v

0 10 20 30 40 50 60 70 80 90 100-1

0

1

acce

lera

tion

a

time t

CP445

4

0 2 4 6 80

0.02

0.04

0.06

0.08

0.1

0.12b = 0

ener

gy

K U

E (

J)

time t (s)

KE PE

E

CP 455

5

Mathematical modelling for harmonic motion

Newton’s Second Law can be applied to the oscillating system

F = restoring force + damping force + driving force

F(t) = - k x(t) - b v(t) + Fd(t)

For a harmonic driving force at a single frequency

Fd(t) = Fmaxcos(t + ).

This differential equation can be solved to give x(t), v(t) and a(t).

CP 463

6

Damped oscillations

Oscillations in real systems die away (the amplitude steadily decreases) over time - the oscillations are said to be damped

For example:The amplitude of a pendulum will decrease over time due to air resistance

If the oscillating object was in water, the greater resistance would mean the oscillations damp much quicker.

CP 463

7

Damped oscillations

CP 463

8

0 2 4 6 80

0.02

0.04

0.06

0.08

0.1

0.12b = 6

ener

gy

K U

E (

J)

time t (s)

KE

PE

E

CP 463

9

Forced oscillations Driven Oscillations & Resonance

If we displace a mass suspended by a spring from equilibrium and let it go it oscillates at its natural frequency

f 1

2k

m

If a periodic force at another frequency is applied, the oscillation will be forced to occur at the applied frequency - forced oscillations

CP 465

10

Resonance

Forced oscillations are small unless the driving frequency is close to the natural frequency

When the driving frequency is equal to the natural frequency the oscillations can be large - this is called resonance

Away from resonance, energy transfer to the oscillations is inefficient. At resonance there is efficient transfer which can cause the oscillating system to fail - see wine glass experiment.

Famous example of resonance: soldiers marching on bridge

CP 465

11

Resonance phenomena occur widely in natural and in technological applications:

Emission & absorption of lightLasersTuning of radio and television setsMobile phonesMicrowave communicationsMachine, building and bridge designMusical instrumentsMedicine – nuclear magnetic resonance magnetic resonance imaging – x-rays – hearing

Nuclear magnetic resonance scanCP 465

12

0 0.5 1 1.5 2 2.5 30

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4am

plitu

de A

(m

)

d /o

b = 2

b = 8

b = 10

CP 465

Response Curve

fd / fO

13Sinusoidal driving force fd / fo = 0.1

0 20 40 60 80 100-1

-0.5

0

0.5

1b = 2

posi

tion

x (

m)

time t (s)

Sinusoidal driving force fd / fo = 1

0 20 40 60 80 100-1

-0.5

0

0.5

1b = 2

pos

ition

x

(m

)time t (s)

CP 465

14

Sinusoidal driving force fd / fo = 2

0 20 40 60 80 100-1

-0.5

0

0.5

1b = 2

pos

ition

x

(m

)

time t (s)

Impulsive force – constant force applied for a short time interval.

0 20 40 60 80 100-1

-0.5

0

0.5

1b = 2

pos

ition

x

(m

)time t (s)

CP 465

15

http://www.acoustics.salford.ac.uk/feschools/waves/wine3video.htm

http://www.acoustics.salford.ac.uk/feschools/waves/shm3.htm

An optical technique called interferometry reveals the oscillations of a wine glass

Great Links to visit

16

Eardrum

Auditory canal

Cochlea

Basilar membrane

vibrations of small bones of the middle ear

vibration of eardrumdue to sound waves

Inner air – basilar membrane – as the distance increases from the staples, membrane becomes wider and less stiff – resonance frequency of sensitive hair cells on membrane decreases

staples

1

2o

kf

m

3000 Hz 30 Hz

Resonance and Hearing

CP467

staples

17

Self excited oscillationsSometimes apparently steady forces can cause large oscillations at the natural frequency

Examples

singing wine glasses (stick-slip friction)

Tacoma Narrows bridge (wind eddies)

CP 466

18

CP 466

19

What is a good strategy for answering

examination questions ???

201 Read and answer the question

2 Type of problem

Identify the physics – what model can be used? Use exam formula sheet

3 Answer in point form

Break the question into small parts, do step by step showing all working and calculations (if can’t get a number in early part of a question, use a “dummy” number. Explicit physics principles (justification, explanation)

Annotated diagrams (collect and information & data – implicit + explicit

Equations Identify Setup Execute Evaluate

21Problem 4.1

Why do some tall building collapse during an earthquake ?

22I S E E

Vibration motion can be resolved into vertical and horizontal motions

23Vertical motion

m

k

1

2o

kf

m

natural frequency of vibration

driving frequency fd

0 0.5 1 1.5 2 2.5 30

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

ampl

itude

A (

m)

d /o

b = 2

b = 8

b = 10

Resonance fd fo large amplitude oscillations – building collapses

24

Driver frequency fd

nodes

antinodes

2nd floor disappeared – driving frequency matches natural frequency (3rd harmonic)

Horizontal Motion

ResonanceStanding Waves setup in building

2nd floor

25

Problem 4.2

Consider a tractor driving across a field that has undulations at

regular intervals. The distance between the bumps is about 4.2

m. Because of safety reasons, the tractor does not have a

suspension system but the driver’s seat is attached to a spring to

absorb some of the shock as the tractor moves over rough

ground. Assume the spring constant to be 2.0104 N.m-1 and

the mass of the seat to be 50 kg and the mass of the driver, 70

kg. The tractor is driven at 30 km.h-1 over the undulations.

Will an accident occur?

26

x = 4.2 m

k = 2x104 N.m-1

v = 30 km.h-1

m = (50 + 70) kg = 120 kg

Solution 4.2 I S E E

27Tractor speed v = x / t = 30 km.h-1 = (30)(1000) / (3600) m.s-1 = 8.3 m.s-1

The time interval between hitting the bumps (x = 4.2 m)

t = x / v = (4.2 / 8.3) s = 0.51 s

Therefore, the frequency at which the tractor hits the bumps and energy is

supplied to the oscillating system of spring-seat-person

f = 1 / t = 1 / 0.51 = 2.0 Hz.

The natural frequency of vibration of the spring-seat-person is

42.

1 1 2 10

2 2 1201 Hz

kf

m

This is an example of forced harmonic motion. Since the driving frequency (due to hitting the bumps) is very close to the natural frequency of the spring-seat-person the result will be large amplitude oscillations of the person and which may lead to an unfortunate accident. If the speed of the tractor is reduced, the driving frequency will not match the natural frequency and the amplitude of the vibration will be much reduced.