Chapter 10

Sinusoidally Driven Oscillations

Question of Chapter 10

How do the characteristic frequencies generated in one object (say a piano string) excite vibrations in another object (say a sounding board)?

A Simple Driving System

Natural Frequency (o)

If the board is a door, then the natural frequency is around 0.4 Hz.

If the system is driven at 0.4 Hz, large amplitudes result.

Smaller amplitudes result for driver frequency different from 0.4 Hz.

Actual motions

The door starts with complex motions (transient) that settle down to sinusoidal, no matter the motor rate.

The final frequency is always the driving frequency of the motor ().

The amplitude of the oscillations depends on how far from the natural frequency the motor is.

Driving at Various Rates

Amplitude vs. Frequency

Natural Frequency

<< o

Motor frequency is far below the natural frequency ( << o) door moves almost in

step with motor.

Door moves toward motor when bands are stretched most.

< o

Door lags behind the motor.

= o

Door lags by one quarter cycle.

>> o

Door lags by one-half cycle.

Summarizing

/o Door Lags

<< 1 0

< 1 Small

= 1 ¼-cycle

>> 1 ½-cycle

Computer Model

Click on the link and experiment

Nature of the Transient

Transients are reproducible If crank starts in the same position, we get the

same transient

Damped Harmonic Oscillations Shown by changing the damping Imagine the bottom of the door immersed in an

oil bath The amount of immersion gives the damping

Small DampingTransient Part

-0.3

-0.2

-0.1

0.0

0.1

0.2

0.3

0.00 0.25 0.50 0.75 1.00

Time

Am

plit

ud

e

Steady State Part

-0.3

-0.2

-0.1

0.0

0.1

0.2

0.3

0 1 2 3 4 5

Time

Am

plit

ud

e

Driven Oscillator - Lightly Damped

-0.3

-0.2

-0.1

0

0.1

0.2

0.3

0.4

0 1 2 3 4 5

Time

Am

plit

ud

e

Heavier DampingTransient Part

-0.3

-0.2

-0.1

0.0

0.1

0.2

0.00 0.25 0.50 0.75 1.00

Time

Am

plit

ud

e

Steady State Part

-0.2

-0.1

0.0

0.1

0.2

0.3

0 1 2 3 4 5

Time

Am

plit

ud

e

Driven Oscillator - Heavier Damping

-0.2

-0.1

0

0.1

0.2

0.3

0 1 2 3 4 5

TimeA

mp

litu

de

Two Part Motion

Damped harmonic oscillation (transient) is at the natural frequency

Driven (steady state) oscillation is at the driver frequency

Driver Frequency = Natural Frequency

Transient Part

-0.2

-0.1

0.0

0.1

0.2

0.3

0.00 0.25 0.50 0.75 1.00

Time

Am

plit

ud

e

Steady State Part

-0.4

-0.3

-0.2

-0.1

0.0

0.1

0.2

0.3

0 0.5 1 1.5 2 2.5 3

Time

Am

plit

ud

e

Driving Force = Natural Frequency

-0.4

-0.3

-0.2

-0.1

0

0.1

0.2

0.3

0 0.5 1 1.5 2 2.5 3

TimeA

mp

litu

de

Damping and the Steady State

As long as we are far from natural frequency, damping doesn’t affect the steady state.

Near the natural frequency, damping does have an effect.

Damping and the Steady State

As damping is increased the height of the peak decreases

Small damping

Large damping

Am

plit

ude

Frequency

W½

Trends with Damping

As damping increases we expect the halving time to decrease ( ) Oscillations die out quicker for larger damping.

12

1T

D

max

1A

D

As damping increases the maximum amplitude decreases ( )

Also notice W½ D. Larger damping

means a broader curve.

Percentage Bandwidth (PBW)

Range of frequencies for which the response is it least half the maximum amplitude.

Let N be the number of oscillations that the pendulum makes in T½.

Direct measurement yields

PBW = 38.2/N measured in %

Example of PBW

Imagine tuning an instrument by using a tuning fork (A 440) while playing A.

If you are not matching pitch, the tuning fork is not being driven at its natural frequency and the amplitude will be small.

Only at a frequency of 440 Hz will the amplitude of the tuning fork be large

Example of PBW - continued

T½ = 5 sec (it takes about 5 seconds for the tuning fork to decay to half amplitude)N = (440 Hz) (5 sec) = 2200 cyclesSo when you get a good response from the tuning fork, you have found pitch to better than

PBW = 38.2/2200 = 0.017%or 0.076 Hz!

Caution!

You must play long, sustained tones

Short “toots” will stimulate the transient which recall is at the natural frequency of the tuning fork (440 Hz) Without the sustained driving force of the instrument,

we will never get to the steady state and the tuning fork will ring due to the transient.

You will think the instrument is in pitch when it is not.

Systems with Two Natural Modes

Each mode has its own frequency, decay time, and shape.

The modes are always damped sinusoidal.

Superposition applies.

Simple Two Mass Model

Normal Modes of Two Mass Model (Chapter 6)

Let Mode 1 have a natural frequency of 10 Hz and Mode 2 a natural frequency of 17.32 Hz.

Driving Point Response Function or Resonance Curve

Am

plit

ude

Frequency

10 Hz 17.32 Hz

Frequencies Between Peaks

Mass one has a mode one component and should lag a half-cycle behind the driver( > o1)

Mass one also has a mode two component to its motion, and here the driving frequency is less than the natural frequency ( << o2) Mass one keeps in step with the driver

These conflicting tendencies account for the small amplitude here

New Terms

Driving Point Response Curve – measure the response at the mass being driven

Transfer Point Response Curve – measure the response at another mass in the system (not a driven mass)

Properties of a Sinusoidally Driven System

At startup there is a transient that is made up of the damped sinusoids of all of the natural frequencies.

Once the transient is gone the steady state is at the driving frequency. When the driving frequency is close to one of the natural frequencies, the amplitude is a maximum and resembles that natural mode.

A Tin Tray

The tray is clamped at three places. Sensors ( )and drivers ( )are used as pairs in the locations indicated.

Response Curves

General Principles

Sensor cannot pickup any mode whose nodal line runs through it. Notice that Sensor 2 is on the centerline It cannot pick up modes with nodal lines

through the center, such as…

Sensor 2 is blind to…

Sensor 2 is very sensitive to…

Drivers Ability to Excite Modes

If a driver falls on the nodal line of a mode, that mode will not be excited

If a driver falls between nodal lines of a mode, that mode will be excited

Steady State Response

Superposition of all the modes excited and their amplitudes at the detector positions. Some modes may reinforce or cancel other modes.

Example – consider the modes on the next screen Colored sections are deflected up at this time and the

uncolored sections are deflected down The vertical lines show where in the pattern of each we

are for a particular position on the plate

Superposing Two Modes

Summary

Altering the location of either the driver or the detector will greatly alter what the transfer response curve will be.

Altering the driver frequency will also change the response.

Three Cases Presented

Deflections of the same sign (giving a larger deflection) Add

Deflections of opposite sign (canceling each other out) Subtract

Deflection of one mode lined up with the node of the other (deflection due to one mode only) Single

The G4 Phantom at G3

392, 784, 1176, 1568, 1960, 2352, …

196, 392, 588, 784, 980, 1176, …

Depress G3 slowly

Press & release G4