l 20 course review w= mg, where g=9.8 m/s 2 in previous slide w (=f g ) = f n

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L 20 Course Review

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Page 1: L 20 Course Review W= mg, where g=9.8 m/s 2 In Previous slide W (=F G ) = F N

L 20Course Review

Page 2: L 20 Course Review W= mg, where g=9.8 m/s 2 In Previous slide W (=F G ) = F N
Page 3: L 20 Course Review W= mg, where g=9.8 m/s 2 In Previous slide W (=F G ) = F N

W= mg, where g=9.8 m/s2

In Previous slide W (=FG) = FN

Page 4: L 20 Course Review W= mg, where g=9.8 m/s 2 In Previous slide W (=F G ) = F N

Simple Harmonic Motion• Position x vs. time t• Definition of period T• Definition of amplitude A

Page 5: L 20 Course Review W= mg, where g=9.8 m/s 2 In Previous slide W (=F G ) = F N

Frequency and Periodf = 1/T or T = 1/f or f T =1

T period, in seconds (s)f = frequency in Hertz (Hz)

Metric prefixes:centi- (c), milli- (m), micro- ( )m

kilo- (k), mega- (M)

Page 6: L 20 Course Review W= mg, where g=9.8 m/s 2 In Previous slide W (=F G ) = F N
Page 7: L 20 Course Review W= mg, where g=9.8 m/s 2 In Previous slide W (=F G ) = F N

Wave velocity for a periodic vibration

Let the wavelength be λand the frequency of the

vibration be f.The wave velocity v is just

V=λ/T, or

V= λf

Page 8: L 20 Course Review W= mg, where g=9.8 m/s 2 In Previous slide W (=F G ) = F N

/Tv

Page 9: L 20 Course Review W= mg, where g=9.8 m/s 2 In Previous slide W (=F G ) = F N

More specifically,

we consider a force acting through a distance.Work = Force x distance or W = F.dUnits - newtons x meters = joules (J), or pounds x feet (foot pounds, ft.lbs)BTU = 778 ft.lbs (energy of one wooden kitchen

match)Pushing on a wall and wall doesn’t move

(no work done on the wall)Conversion: 1J= 0.738 ft.lb

Page 10: L 20 Course Review W= mg, where g=9.8 m/s 2 In Previous slide W (=F G ) = F N

Potential Energy

Energy of position or configuration

Other examples - Springs, bow, sling shot, chemical energy, and gravitational potential energy

The latter is GPE = mgh (the force required to lift at constant speed times the distance )

Page 11: L 20 Course Review W= mg, where g=9.8 m/s 2 In Previous slide W (=F G ) = F N

W

Power = Work/time or P = W/t

Units - J/s =

Watt

2. POWER

550 ft.lb/s = 1

hp

1 hp = 746 J/s = 746 W

1 BTU/hr = 0.293 W

100 W bulb = 0.1341 hp

250 hp engine = 186,450

W

Page 12: L 20 Course Review W= mg, where g=9.8 m/s 2 In Previous slide W (=F G ) = F N
Page 13: L 20 Course Review W= mg, where g=9.8 m/s 2 In Previous slide W (=F G ) = F N

Conditions for standing waves

Page 14: L 20 Course Review W= mg, where g=9.8 m/s 2 In Previous slide W (=F G ) = F N
Page 15: L 20 Course Review W= mg, where g=9.8 m/s 2 In Previous slide W (=F G ) = F N

overpressure

L

Page 16: L 20 Course Review W= mg, where g=9.8 m/s 2 In Previous slide W (=F G ) = F N
Page 17: L 20 Course Review W= mg, where g=9.8 m/s 2 In Previous slide W (=F G ) = F N

Closed tubes(closed on one end)

overpressure

Closed end: antinode

open end:node

L

Page 18: L 20 Course Review W= mg, where g=9.8 m/s 2 In Previous slide W (=F G ) = F N
Page 19: L 20 Course Review W= mg, where g=9.8 m/s 2 In Previous slide W (=F G ) = F N
Page 20: L 20 Course Review W= mg, where g=9.8 m/s 2 In Previous slide W (=F G ) = F N

We define the Sound Intensity I as the Audio Power crossing a unit

area,or I = P/A

Units- W/m2

Page 21: L 20 Course Review W= mg, where g=9.8 m/s 2 In Previous slide W (=F G ) = F N

12-2 Intensity of Sound: Decibels

An increase in sound level of 3 dB, which is a doubling in intensity, is a very small change in loudness.

In open areas, the

intensity of sound diminishes with distance:

However, in enclosed spaces this is complicated by reflections, and if sound travels through air the higher frequencies get preferentially absorbed.

Page 22: L 20 Course Review W= mg, where g=9.8 m/s 2 In Previous slide W (=F G ) = F N

12-2 Intensity of Sound: Decibels

The loudness of a sound is much more closely related to the logarithm of the intensity.

Sound level is measured in decibels (dB) and is defined:

(12-1)

I0 is taken to be the threshold of hearing:

Page 23: L 20 Course Review W= mg, where g=9.8 m/s 2 In Previous slide W (=F G ) = F N
Page 24: L 20 Course Review W= mg, where g=9.8 m/s 2 In Previous slide W (=F G ) = F N

12-2 Intensity of Sound: Decibels

The intensity of a wave is the energy transported per unit time across a unit area.

The human ear can detect sounds with an intensity as low as 10-12 W/m2 and as high as 1 W/m2.

Perceived loudness, however, is not proportional to the intensity.

Page 25: L 20 Course Review W= mg, where g=9.8 m/s 2 In Previous slide W (=F G ) = F N
Page 26: L 20 Course Review W= mg, where g=9.8 m/s 2 In Previous slide W (=F G ) = F N

12-3 The Ear and its Response; LoudnessThe ear’s sensitivity varies with frequency. These curves translate the intensity into sound level at different frequencies.

Page 27: L 20 Course Review W= mg, where g=9.8 m/s 2 In Previous slide W (=F G ) = F N
Page 28: L 20 Course Review W= mg, where g=9.8 m/s 2 In Previous slide W (=F G ) = F N

Note span Interval Frequency ratioC - C unison 1/1C - C# semitone 16/15C - D whole tone (major second) 9/8C - D# minor third 6/5C - E major third 5/4C - F perfect fourth 4/3C - F# augmented fourth 45/32C - G perfect fifth 3/2C - G# minor sixth 8/5C - A major sixth 5/3C - A# minor seventh 16/9 (or 7/4)C - B major seventh 15/8C3 - C4 octave 2/1C3 - E4 octave+major third 5/2

Intervals12-tone scale (chromatic) 8-tone scale (diatonic)

Page 29: L 20 Course Review W= mg, where g=9.8 m/s 2 In Previous slide W (=F G ) = F N

Pythagorean ScaleBuilt on 5ths

Page 30: L 20 Course Review W= mg, where g=9.8 m/s 2 In Previous slide W (=F G ) = F N

A pleasant consonance was observed playing strings whose lengths were

related by the ratio of 3/2 to 1 (demo).Let’s call the longer string C, and the

shorter G, and the interval between G and C a 5th

Denote the frequency of C simply by the name C, etc.

Page 31: L 20 Course Review W= mg, where g=9.8 m/s 2 In Previous slide W (=F G ) = F N

The major triad is the basis for the just scale, which we now develop

in a way similar to that of the Pythagorean scale.

Page 32: L 20 Course Review W= mg, where g=9.8 m/s 2 In Previous slide W (=F G ) = F N
Page 33: L 20 Course Review W= mg, where g=9.8 m/s 2 In Previous slide W (=F G ) = F N

We wish to make a chromatic scale- 12 tones including both octaves- and we want all the

intervals (ratios of adjacent notes to all be the same).

Page 34: L 20 Course Review W= mg, where g=9.8 m/s 2 In Previous slide W (=F G ) = F N
Page 35: L 20 Course Review W= mg, where g=9.8 m/s 2 In Previous slide W (=F G ) = F N
Page 36: L 20 Course Review W= mg, where g=9.8 m/s 2 In Previous slide W (=F G ) = F N
Page 37: L 20 Course Review W= mg, where g=9.8 m/s 2 In Previous slide W (=F G ) = F N

Beats

f1-f2 = beat frequency

Average frequency “heard” = (f1+f2)/2

Page 38: L 20 Course Review W= mg, where g=9.8 m/s 2 In Previous slide W (=F G ) = F N

Modes

• Ionian – Major Scale• Dorian – 2nd of Major Scale• Phrygian – 3rd of Major Scale• Lydian – 4th of Major Scale• Mixolydian – 5th of Major Scale• Aolian – 6th of Major Scale (Minor)• Locrian – 7th of Major Scale

Page 39: L 20 Course Review W= mg, where g=9.8 m/s 2 In Previous slide W (=F G ) = F N

Non-Western Scales

Page 40: L 20 Course Review W= mg, where g=9.8 m/s 2 In Previous slide W (=F G ) = F N
Page 41: L 20 Course Review W= mg, where g=9.8 m/s 2 In Previous slide W (=F G ) = F N
Page 42: L 20 Course Review W= mg, where g=9.8 m/s 2 In Previous slide W (=F G ) = F N

Resonance

Page 43: L 20 Course Review W= mg, where g=9.8 m/s 2 In Previous slide W (=F G ) = F N
Page 44: L 20 Course Review W= mg, where g=9.8 m/s 2 In Previous slide W (=F G ) = F N

Fourier SynthesisDemo- PhET (Physics,Fourier)

Page 45: L 20 Course Review W= mg, where g=9.8 m/s 2 In Previous slide W (=F G ) = F N
Page 46: L 20 Course Review W= mg, where g=9.8 m/s 2 In Previous slide W (=F G ) = F N

String Instruments

Page 47: L 20 Course Review W= mg, where g=9.8 m/s 2 In Previous slide W (=F G ) = F N
Page 48: L 20 Course Review W= mg, where g=9.8 m/s 2 In Previous slide W (=F G ) = F N
Page 49: L 20 Course Review W= mg, where g=9.8 m/s 2 In Previous slide W (=F G ) = F N
Page 50: L 20 Course Review W= mg, where g=9.8 m/s 2 In Previous slide W (=F G ) = F N
Page 51: L 20 Course Review W= mg, where g=9.8 m/s 2 In Previous slide W (=F G ) = F N
Page 52: L 20 Course Review W= mg, where g=9.8 m/s 2 In Previous slide W (=F G ) = F N
Page 53: L 20 Course Review W= mg, where g=9.8 m/s 2 In Previous slide W (=F G ) = F N
Page 54: L 20 Course Review W= mg, where g=9.8 m/s 2 In Previous slide W (=F G ) = F N
Page 55: L 20 Course Review W= mg, where g=9.8 m/s 2 In Previous slide W (=F G ) = F N
Page 56: L 20 Course Review W= mg, where g=9.8 m/s 2 In Previous slide W (=F G ) = F N
Page 57: L 20 Course Review W= mg, where g=9.8 m/s 2 In Previous slide W (=F G ) = F N
Page 58: L 20 Course Review W= mg, where g=9.8 m/s 2 In Previous slide W (=F G ) = F N
Page 59: L 20 Course Review W= mg, where g=9.8 m/s 2 In Previous slide W (=F G ) = F N
Page 60: L 20 Course Review W= mg, where g=9.8 m/s 2 In Previous slide W (=F G ) = F N
Page 61: L 20 Course Review W= mg, where g=9.8 m/s 2 In Previous slide W (=F G ) = F N
Page 62: L 20 Course Review W= mg, where g=9.8 m/s 2 In Previous slide W (=F G ) = F N
Page 63: L 20 Course Review W= mg, where g=9.8 m/s 2 In Previous slide W (=F G ) = F N
Page 64: L 20 Course Review W= mg, where g=9.8 m/s 2 In Previous slide W (=F G ) = F N
Page 65: L 20 Course Review W= mg, where g=9.8 m/s 2 In Previous slide W (=F G ) = F N
Page 66: L 20 Course Review W= mg, where g=9.8 m/s 2 In Previous slide W (=F G ) = F N

The Vocal Tract

epiglottis

Page 67: L 20 Course Review W= mg, where g=9.8 m/s 2 In Previous slide W (=F G ) = F N

Vocal Formants

“had”

Page 68: L 20 Course Review W= mg, where g=9.8 m/s 2 In Previous slide W (=F G ) = F N

To calculate T, consider a room with a hole in one wall of area A.

Call the reverberation time T.T ˜ volume V, 1/A

T= K V/AIt has been worked out that, for V in

m3 , A in m2

T= 0.16 V/A

Page 69: L 20 Course Review W= mg, where g=9.8 m/s 2 In Previous slide W (=F G ) = F N

Let us now replace the open window area with an absorbing

material of area S and absorption coefficient a.

Then A= Sa. If there is more than one type of absorbing material, the

A= S1 a1+s2a2 +S3a3+…

Page 70: L 20 Course Review W= mg, where g=9.8 m/s 2 In Previous slide W (=F G ) = F N
Page 71: L 20 Course Review W= mg, where g=9.8 m/s 2 In Previous slide W (=F G ) = F N

Basic Analog Electronics

Ohm’s Law Links: Bob Holtzworth part 1 slides 1-

11,12,16

Page 72: L 20 Course Review W= mg, where g=9.8 m/s 2 In Previous slide W (=F G ) = F N

Ohm’s LawThe current (charge per unit time)

flowing through a circuit element is equal to the potential drop across

this element divided by the resistance of the element.

I= V/R

Page 73: L 20 Course Review W= mg, where g=9.8 m/s 2 In Previous slide W (=F G ) = F N

Digital Electronics

Introduction to Binary Numbers

Page 74: L 20 Course Review W= mg, where g=9.8 m/s 2 In Previous slide W (=F G ) = F N

We can write the number 752 as2x100 + 5x101 + 7x102

SimilarlyWe could use the base 2, e.g.

3 = 1x20 + 1x21, which we represent as 11.

Hence 01 is 2

These are 2-binary digit (bit) numbers.

Page 75: L 20 Course Review W= mg, where g=9.8 m/s 2 In Previous slide W (=F G ) = F N

Digital Sampling

Page 76: L 20 Course Review W= mg, where g=9.8 m/s 2 In Previous slide W (=F G ) = F N
Page 77: L 20 Course Review W= mg, where g=9.8 m/s 2 In Previous slide W (=F G ) = F N

Calculating Bit-rates (CD quality)

Sampling Rate

x Resolution x# of

Channels= Bit-rate

44,100 x 16 x 2 = 1,411,200

Calculating File Sizes (one minute of CD audio)

Sampling Rate

x Resolution x Number of Channels x Time in

Seconds /

Bits /

Byte = File Size(in Bytes)

44,100 x 16 x 2 x 60 / 8 = 10,584,000

MP3 compression at 128 kbps compresses this by a factor of 11

Page 78: L 20 Course Review W= mg, where g=9.8 m/s 2 In Previous slide W (=F G ) = F N

MP 3 Compression

Page 79: L 20 Course Review W= mg, where g=9.8 m/s 2 In Previous slide W (=F G ) = F N

The most important principle in MP3 compression is the psychoacustic selection of sound signals to cut away. Those signals, we are unable to hear are removed. These include weaker sounds that are present but are not heard because

they are drowned out (masked) by louder instruments/sounds.

Many encoders use the fact that the human ear is most sensitive to midrange sound frequencies (1 to 4 KHz). Hence

sound data within this range is left unchanged. An other compression used is to reduce the stereo signal into

mono, when the sound waves are so deep, that the human ear cannot register the direction. Also the contents of common

information in the two stereo channels is compressed. The Huffman algorithm reduces the file size by optimizing the

data code for the most often used signals. This is a lossless compression working within the MP3 system.

Page 80: L 20 Course Review W= mg, where g=9.8 m/s 2 In Previous slide W (=F G ) = F N

More on CDs

750 Mbytes

75 minutes of audioLink: “how Edison got his groove back”

Page 81: L 20 Course Review W= mg, where g=9.8 m/s 2 In Previous slide W (=F G ) = F N

The elongated bumps that make up the track are each 0.5 microns wide, a minimum of 0.83 microns, they look something like this: