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Physics 17 Part N Dr. Alward

Waves

Video Lecture 1: String Waves Intro

Video Lecture 2: Standing String Waves

Video Lecture 3: Intro to Sound Waves

Video Lecture 4: Doppler Effect for Sound

Video Lecture 5: Open-Closed Tube Resonances

Video Lecture 6: Open-Open Tube Resonances

Waves on Strings

Shake once: A traveling “pulse” is

created.

Pulse Speed

L = Length of String

m = Mass

μ = Linear Mass Density

= m/L

T = Tension

v = Pulse Speed

= √(T/μ)

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Example:

T = 4.9 N

μ = 0.10 kg/m

v = (4.9/0.10)1/2

= 7.0 m/s

Frequency

The frequency of a wave on a string is the number

of pulses delivered to the string each second.

The units of frequency are “hertz” (Hz).

1.0 Hz = 1.0 s-1

Wavelength

The wavelength of a wave is the distance

between consecutive maxima, or two

consecutive minima.

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Nodes

Nodes are places where there is zero

deviation from normal.

The distance between consecutive nodes

is a half-wavelength, /2.

Amplitude

The amplitude of a wave is the maximum deviation from

the “relaxed” (nodal) condition.

The symbol for a wave’s amplitude is A.

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The Wave Equation

f = v

Example A:

The speed of waves

on a string is 12.0 m/s.

At what frequency will

the wavelength of a

wave be 4.0 m?

f = v/λ

= (12.0 m/s )/(4.0 m)

= 3.0 s-1

= 3.0 Hz

Example B:

T = 100 N

m = 0.4 kg

L = 1.6 m

f = 10 Hz

λ = ?

µ = 0.4/1.6

= 0.25 kg/m

v = (T/µ)1/2

= (100/0.25)1/2

= 20 m/s

λ = v/f

= 20/10

= 2 m

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Resonances

At certain frequencies, a “resonance,” called a “standing wave,”

occurs in which comparatively small vibrations at the left end of the

string below leads to comparatively large vibrations called

“antinodes” (A) at certain points along the string, and zero vibrations

at other points called “nodes” (N).

They’re called “standing waves” because they appear to be

stationary.

Examples of resonances are shown below.

The standing waves above are in the shape of a

series of “loops.” At the ends of each loop are

nodes. The A and N labels alternate.

Recall that the distance between consecutive

nodes in a wave is a half-wavelength. Thus,

each loop above has a width equal to half a

wave-length: λ/2.

Note that there is an antinode at the center of a

loop, so the number of loops in a standing wave

is the same as the number of antinodes.

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VIDEO DEMONSTRATION OF STANDING WAVE

Click Here

Example A:

The speed of pulses on a string of

length 1.40 m is 2.60 m/s.

What frequency of oscillation will

create a standing wave with four

antinodes?

Each of the four loops has a width

of λ/2. The sum of these widths

equals the length of the string.

4(λ/2) = 1.40

λ = 0.70 m

f = v/λ

= 2.60/0.70

= 3.71 Hz

Example B:

Resonance on 1.60 m string

occurs with six antinodes

when the string is oscillated

at 6.0 Hz. The tension in the

string is 14.0 N.

What is the linear mass

density of the string?

Solution:

6(λ/2) = 1.60

λ = 0.53 m

v = λf

= 0.53 (6.0)

= 3.18 m/s

(T/µ)1/2 = v

(14.0/µ)1/2 = 3.18

Solve for µ:

µ = 1.38 kg/m

The assumption that the sum of the loop widths equals the length of

the string requires that the amplitude of the wave be negligibly small

compared to the length of the string.

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Resonances on Hanging Ropes

If one end of a handing rope is free, then

resonances on the rope have an anti-node at the

free end.

Example:

A rope 0.90 meters long is hanging vertically;

the bottom of the rope is free.

An oscillator attached to the top of the rope is

vibrating at a frequency of 4.0 Hz, which

causes a resonance with four antinodes. What

is the speed of waves on this rope?

Solution:

Note that there are 3.5 loops.

3.5 (λ/2) = 0.90

λ = 0.51 m

v = λf

= 0.51 (4.0)

= 2.1 m/s

For later use in the discussion of sound wave

resonances, note that the distance between

consecutive labels (N-A or A-N) is a quarter-

wavelength:

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Sound Waves

Sound waves are propagated in the manner of compressions set in

motion on a “slinky.” Sound waves are “longitudinal” in nature in

the sense that the medium (air) oscillates back and forth along the

direction as the propagation of the wave. The type of wave we

with strings was “transferse”: the medium (string particles) moved

perpendicular to the direction of propagation of the wave.

Click the link below to gain a sense of how sound waves are

created and propagate.

Video of Slinky Motion as Sound Wave Analogy

Recall facts about atmospheric pressure at sea-level:

1.0 millibar (mb) = 100 Pa

Po = 1.01 x 105 Pa

= 101 kPa

= 14.7 psi

=1010 millibar (mb)

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The maxima above are 5 mb above normal air pressure.

Atmospheric Pressure: 1010 millibars (mb)

Amplitude: 5 millibars

Node:

N = Normal Pressure: 1010 mb

Antinodes:

H = Higher Pressure: 1015 mb

L = Lower Pressure: 1005 mb

Compression: H

Rarefaction: L

Speed of Sound in Air: 340 m/s

Speed of Sound in Water: 1500 m/s

Most Sound: 1000-4000 Hz.

Range of Human hearing: 20-20,000 Hz

Ultrasound: Greater than 20,000 Hz

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The indicated portion of the pressure graph above is one wavelength

wide. The AN and NA distances are the same width. Note that

NA = λ/4.

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Waves Interfering with Matter

Waves whose wavelength are

comparable to the length of objects

over which the wave is traveling may

shake apart the object. Longer

wavelengths may have no effect at all.

Consider the example of a ship at sea

under which waves travel.

In Figure 1, the wave rises and falls

gently. The distance between

consecutive wave crests (the

“wavelength”) is many times the

width of the ship: The ship is scarcely

aware of the presence of the wave.

However, the wave traveling under

the ship in Figure 2 has a wavelength

comparable to the length of the ship.

The ship rapidly rises and falls and is

at risk of breaking apart.

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Example:

“Lithotripsy” is a medical

procedure in which ultrasound is

used to break up kidney stones.

What is the optimum ultrasound

frequency for breaking up a kidney

stone whose diameter is one

centimeter (0.01 m)?

In kidney tissue, the speed of

sound is about 1500 m/s.

f = v/λ

= 1500/0.01

= 150,000 Hz

= 150 kilohertz

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The Doppler Effect for Sound Click here to view doppler effect video.

The listener hears a higher frequency as the car

approaches. As it is moving away, the listener

hears a lower frequency.

If either one (source or observer) is moving

toward the other, choose the sign for the

speed term that makes the ratio larger.

If either one is moving away from the other,

choose the sign for the speed term that

makes the ratio smaller.

Note: vo and vs are speeds, not velocities.

Example:

Suppose an observer is at rest, and the source

is moving away from the observer at 30 m/s.

If the siren’s frequency is 5,000 Hz, what

does the observer hear?

vo = 0

Source is moving away, so the observed

(heard) frequency will be lower, so we need

to choose the sign for the source speed that

will make the ratio smaller: We add 30 to the

340 in the denominator.

fo = 5000 (340 + 0) / (340 + 30)

= 4595 Hz

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Example:

The police car below is chasing a

speeder. What frequency does the

speeder hear?

The speeder is moving away from the

police car, so we choose the sign for

the observer speed that makes the ratio

smaller: We subtract 50.

The police car is moving toward the

speeder, so we choose the sign for the

source speed term that makes the ratio

larger: We subtract 30.

fo = 4000 (340 - 50) / (340 - 30)

= 3742 Hz

Example:

A fire truck emitting 2500 Hz and traveling

east at 40 m/s is racing toward an automobile

traveling to the west. The automobile driver

hears 3000 Hz.

What is the automobile’s speed?

The observer and the source are each traveling

toward each other, so we choose the sign for

each speed term that makes the ratio larger.

3000 = 2500 (340 + vo) / (340 – 40)

vo = 20 m/s

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Sound Wave Resonances

Facts:

Closed ends are always nodes (N).

Open ends are always antinodes (A).

A and N labels alternate.

Recall our earlier observation:

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Open-Closed Tubes

Example:

What are the four lowest frequencies of sound that will resonate

in an open-closed tube 0.85 meter long?

1(λ/4) = 0.85

λ = 3.40 m

f = 340/λ

= 100 Hz

3(λ/4) = 0.85

λ = (3.40/3) m

f = 340/λ

= 300 Hz

5(λ/4) = 0.85

λ = (3.40/5) m

f = 340/λ

= 500 Hz

7(λ/4) = 0.85

λ = (3.40/7) m

f = 340/λ

= 700 Hz

Note: The resonant frequencies (“harmonics”) are odd-integer

multiples of 100 Hz. Other harmonics not shown above include

900, 1100, 1300, 1500, and 1700 Hz, and so on.

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Open-Open Tubes

Example:

What are the four lowest frequencies of sound that will

resonate in the 0.85 meter tube in the previous

example if both ends are open?

2(λ/4) = 0.85

λ = (3.40/2) m

f = 340/λ

= 200 Hz

4(λ/4) = 0.85

λ = (3.40/4) m

f = 340/λ

= 400 Hz

6(λ/4) = 0.85

λ = (3.40/6) m

f = 340/λ

= 600 Hz

8(λ/4) = 0.85

λ = (3.40/8) m

f = 340/λ

= 800 Hz

Note: The resonant frequencies (“harmonics”) are

even-integer multiples of 200 Hz. Other harmonics

include 1000, 1200, 1400, 1600, and 1800 Hz, and so

on.

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If we include all of the frequencies that

resonate in a tube of length 0.85 meters,

irrespective of whether one of the tube’s

ends is closed, or not, we have the

following list:

Frequency

(Hz)

Harmonic

f1 = 100 1st Open-Closed

f2 = 200 2nd Open-Open

f3 = 300 3rd Open-Closed

f4 = 400 4th Open-Open

f5 = 500 5th Open-Closed

f6 = 600 6th Open-Open

f6 = 700 7th Open-Closed

f8 = 800 8th Open-Open

The frequency of the nth harmonic in any

tube is fn = nf1, where f1 is the lowest

frequency of sound that resonates. In this

case, f1 = 100 Hz.

The lowest frequency that resonates in a

tube is called “the fundamental

frequency.”

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Sound Intensity

I = P/A

Units: W/m2

Example:

Each second, 4.0 x 10-6 J of sound energy lands

on an area A = 2.0 x 10-4 m2.

What is the sound intensity at that location?

I = P/A

= (4.0 x 10-6 W) /(2.0 x 10-4 m2)

= 2.0 x 10-2 W/m2

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Spherically-Symmetrical Sound Sources

Sound sources that broadcast equal amounts of energy each second in

all directions are called “spherically-symmetric” sound sources.

If the power output of the source of sound is P, and the listener’s ear is

a distance r away, located on the surface of a sphere whose surface area

is 4πr2, the ear of the listener receives sound intensity I = P/4πr2.

I = P/4πr2

Example:

A spherically-symmetric speaker has a power

output of 200 watts.

What is the intensity 4.0 meters away?

4π(4.0)2 = 201.06 m2

I = P/4πr2

= 200/201.06

= 1.00 W/m2

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Example A:

How far from a spherically-

symmetric 40-watt sound source

would the sound intensity be one

milli-watt per square meter?

1.0 x 10-3 = 40 / (4πr2)

r = 56.42 m

Example B:

At a certain distance from a

spherically-symmetric sound

source the intensity is 2.0 W/m2.

What is the intensity five times

farther away?

r2 is 25 times larger, so I is 1/25th

as much:

I = (1/25) 2.0

= 0.080 W/m2

The Threshold of Human Hearing

The least sound intensity the average healthy

human ear can detect is called “The Threshold

of Human Hearing,” and is Io = 1.0 x 10-12 W/m2

Below this intensity, the ear hears nothing,

hence the subscript zero: Io

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Example:

1.5 meters away from a whisper the sound

intensity is 2.0 x 10-11 W/m2.

(a) Treating the whisperer as a spherically-

symmetric sound source, what is its output

power P?

I = P/(4πr2)

P = 4πr2 I

= 4πr(1.5)2 2.0 x 10-11

= 5.65 x 10-10 W

(b) How far from the whisperer would the

whisper be inaudible, i.e., I = Io?

I = P/(4πr2)

1.0 x 10-12 = 5.65 x 10-10/4πr2

r = 6.71 m