lecture 17 october 29, 2004. on wednesday - thumping

Lecture 17October 29, 2004

On Wednesday - Thumping

The Drum

Each of these modes are usually excited.The tension of the drum determines the frequency of each mode.The modes may NOT be harmonicEach mode dies out at a different rate.The player can change the basic “tone” of the drum by changing the tension of the drum head.

Kettle Drum INITIAL Spectrum

Drum Frequencies

The Sonogram

Modes

All modes are excited at first strike.These vibrations may excite others … resonance. Each mode decays in a different time. Amplitude

time

So … back to the tuning forks

Objects will resonate when

They are in contact with something that vibrates at its resonant frequency. Buzz in cars is a good example

Sound can cause resonance if it is at a frequency that is the resonant frequency of another object nearby. It must have enough energy.

Back to the trip from the instrument into our heads

The Trip

CREATIONOf Sound

TRAVELAND

ROOM ACOUSTICS

Sound Travels

ENERGY

Something is Missing

As time progresses, the amount of energy received by the ear increases.We need a measure of energy per unit time.

ENERGY

ENERGY PER UNIT TIME

sec

Joule1 watt 1

Second

JoulesPOWER

TimeUnit

Energy

Example = The Light Bulb

Consider a 60 Watt light bulb.It requires 60 Joules of energy each second.One Joule = 1 Newton Meter Joule=1( N-m)x (1 lb/4.45N) x

(3.28ft/m) Joule=0.738 ft-lbs

Thinking about light bulbs

60 Joules = 44.2 ft-lbsLift a ~4 pound weight 10 ft. or about one story of a building.Do this every second for 60 watts.

Joule=0.738 ft-lbs

A 100 lb woman would have to run up about 2 floorsof a building per second to generate this much power!

About Spheres ….

2

3

4

3

4

rAreaSurface

rVolume

r

Energy Spreads Out

These AREAS increase with r2.

Power per unit area therefore DECREASES with r2.

Let’s go to a concert.50 Meters

30 watts Ear Canal ~ 0.5 cm = 0.005 m

Area = (0.005)2=0.000025 m2

Houston we have another problem

Ear Area = (0.005)2=0.000025 m2

30 watts, 50 meters

To the ear ….

50m

30 watt

Area of Sphere =r2

=3.14 x 50 x 50 = 7850 m2

Ear Area = 0.000025 m2

In the ear…

wattsmm

watt

Ear

m

watt

m

watt

areaunit

power

000000095.0000025.00038.0

:

0038.07850

30

22

22

How do we deal with all of these zeros???

Answer: Scientific NotationChapter 1 in Bolemon, Appendix 2 in Johnston

0.000000095 watts = 9.5 x 10-8 watts

NOTE

10a/10b=10a-b

Example 1000/10=103/101=10(3-1)=102=100 10000/0.005=104/5 x 10-3=(1/5)x10(4-(-3))

=(1/5)x 107 =(10/5) x 106 = 2,000,000

You can actually get used to doing it this way! But you probably won’t!

Q:Can we hear 9.5 x 10-8 watts?

?

Acoustic Power (Watts)

INSTRUMENT Acoustic Power

Clarinet 0.05

Double Bass 0.16

Trumpet 0.31

Cymbals 9.5

Bass Drum 25

Entire Orchestra 67

Decibels - dB

The decibel (dB) is used to measure sound level, but it is also widely used in electronics, signals and communication.

Decibel continued (dB)Suppose we have two loudspeakers, the first playing a sound with power P1, and another playing a louder

version of the same sound with power P2, but

everything else (how far away, frequency) kept the same.

The difference in decibels between the two is defined to be

10 log (P2/P1) dB       

where the log is to base 10.

?

What the **#& is a logarithm?

Bindell’s definition:

Take a big number … like 23094800394

Round it to one digit: 20000000000Count the number of zeros … 10The log of this number is about equal to the number of zeros … 10.Actual answer is 10.3Good enough for us!

Back to the definition of dB:

The dB is proportional to the LOG10 of a ratio of intensities.Let’s take P1=Threshold Level of Hearing which is 10-12 watts/m2

Take P2=P=The power level we are interested in.

10 log (P2/P1)

An example:

The threshold of pain is 1 w/m2

1201210)10log(1010

1log 10

:PAIN of thresholdfor the rating dB

1212-

Take another look.

The sensitivity range for human hearing depends on the loudness and pitch. Noises along each black line would be heard with the same volume.