Mr. Kuffer
Concepts of Physics
Like different musical instruments, different types of stars produce different types of sound
waves. Small stars produce a sound with a higher pitch than bigger stars, just like the 'piccolo'
produces a higher sound than the cello
Resonance in Air Columns – Pre-Lab
Resonance - a vibration of large amplitude in a mechanical or electrical
system caused by a relatively small periodic stimulus of the same or nearly
the same period as the natural vibration period of the system
I.
A resonating tube with one end closed is called a Closed-Pipe
Resonance.
a. The length of the tube can be varied by adjusting the water
level of the apparatus.
b. Increasing and decreasing the water level will cause the sound
to vary in intensity (volume).
c. When the sound is loudest, it is said to be in resonance. (Air
column is resonating with the tuning fork)
II.
How does resonance occur?
a. The vibrating tuning fork produces sound (pressure waves) in
the air column.
b. The alternating high and low pressure meets the water
boundary and is reflected (inverted) toward the top of the
column.
c. If the reflected high pressure wave reaches the same moment
the tuning fork produces the wave, Constructive Interference
occurs (reinforcing the wave). This produces a standing wave,
and resonance is achieved.
III. Open-Pipe Resonators
a. The properties of the open-pipe resonators are opposite of
those of the closed-pipe resonators.
b. The difference, of course, is the lack of a water boundary at
the bottom of the tube. The wave reflection is therefore from
the open end of the tube (not inverted).
c. In the end, a low pressure wave is reflected where a high
pressure wave would have been reflected if it were a closedpipe resonator (as discussed above).
* An open-pipe resonator can be used to produce resonance. It is
simply a matter of phase, and therefore a matter of tube length
(ℓ).
2
IV.
Resonance length in a closed tube
Figure A
Figure B
a. How much of a wavelength (above: Figure B)?
b. When can resonance occur in a closed-pipe resonator?
c. What specific lengths (ℓ) are necessary?
d. Can you predict the location of resonance points? How?
V.
Finding Speed of sound using resonance
When a tuning fork with a frequency of 392 Hz is used with a
closed-pipe resonator, the loudest sound is heard when the column
is 21.0 cm and 65.3 cm long. The air temperature is 27 ºC. What is
the speed of sound at this temperature?
3
Understanding Resonance Lab
Objectives:
1.
Identify where the air column is resonating with the tuning fork.
2. Measure the length of the air column after identifying the resonance.
3. Calculate the wavelength using the separation between the resonance points.
4. Calculate the speed of sound in the room (____ C) based on the temperature in
the room.
5. Calculate the speed of sound based on the frequency of the tuning fork (known)
and the wavelength (measured).
6. Determine the absolute error in your calculation for the speed of sound.
V= f
Temp
Tuning Fork
Frequency
(Hz)
R1
(m)
R2
(m)

2(R2 – R1)
(m)
Experimental
Speed of
Sound
(m/s)
Actual
Speed of
Sound
(m/s)
Absolute %
error
392
.210
.653
.886
347.3
344.62
.77%
Equations:
V= λ ƒ
V = 331 + (Temperature in ˚C x .6)
Lab Set-Up
4
Published: June 12, 2006
In that old battle of the wills between young people and their keepers, the young have
found a new weapon that could change the balance of power on the cellphone front: a
ring tone that many adults cannot hear.
James Estrin/The New York Times
David Herzka, a freshman at Roslyn High School on Long
Island, shared the ring tone with friends.
Multimedia
Audio: The High-Pitched Ring Tone (mp3)
In settings where cellphone use is forbidden — in class, for
example — it is perfect for signaling the arrival of a text
message without being detected by an elder of the species.
"When I heard about it I didn't believe it at first," said Donna
Lewis, a technology teacher at the Trinity School in
Manhattan. "But one of the kids gave me a copy, and I sent it
to a colleague. She played it for her first graders. All of them
could hear it, and neither she nor I could."
The technology, which relies on the fact that most adults
gradually lose the ability to hear high-pitched sounds, was
developed in Britain but has only recently spread to America
— by Internet, of course.
5
Recently, in classes at Trinity and elsewhere, some students have begun testing the
boundaries of their new technology. One place was Michelle Musorofiti's freshman
honors math class at Roslyn High School on Long Island.
At Roslyn, as at most schools, cellphones must be turned off during class. But one
morning last week, a high-pitched ring tone went off that set teeth on edge for anyone
who could hear it. To the students' surprise, that group included their teacher.
"Whose cellphone is that?" Miss Musorofiti demanded, demonstrating that at 28, her ears
had not lost their sensitivity to strangely annoying, high-pitched, though virtually
inaudible tones.
"You can hear that?" one of them asked.
"Adults are not supposed to be able to hear that," said another, according to the teacher's
account.
She had indeed heard that, Miss Musorofiti said, adding, "Now turn it off."
The cellphone ring tone that she heard was the offshoot of an invention called the
Mosquito, developed last year by a Welsh security company to annoy teenagers and
gratify adults, not the other way around.
It was marketed as an ultrasonic teenager repellent, an ear-splitting 17-kilohertz buzzer
designed to help shopkeepers disperse young people loitering in front of their stores
while leaving adults unaffected.
The principle behind it is a biological reality that hearing experts refer to as presbycusis,
or aging ear. While Miss Musorofiti is not likely to have it, most adults over 40 or 50
seem to have some symptoms, scientists say.
While most human communication takes place in a frequency range between 200 and
8,000 hertz (a hertz being the scientific unit of frequency equal to one cycle per second),
most adults' ability to hear frequencies higher than that begins to deteriorate in early
middle age.
"It's the most common sensory abnormality in the world," said Dr. Rick A. Friedman, an
ear surgeon and research scientist at the House Ear Institute in Los Angeles.
6
But in a bit of techno-jujitsu, someone — a person unknown at this time, but probably not
someone with presbycusis — realized that the Mosquito, which uses this common adult
abnormality to adults' advantage, could be turned against them.
The Mosquito noise was reinvented as a ring tone.
"Our high-frequency buzzer was copied. It is not exactly what we developed, but it's a
pretty good imitation," said Simon Morris, marketing director for Compound Security,
the company behind the Mosquito. "You've got to give the kids credit for ingenuity."
British newspapers described the first use of the high-frequency ring tone last month in
some schools in Wales, where Compound Security's Mosquito device was introduced as a
"yob-buster," a reference to the hooligans it was meant to disperse.
Since then, Mr. Morris said his company has received so much attention — none of it
profit-making because the ring tone was in effect pirated — that he and his partner,
Howard Stapleton, the inventor, decided to start selling a ring tone of their own. It is
called Mosquitotone, and it is now advertised as "the authentic Mosquito ring tone."
David Herzka, a Roslyn High School freshman, said he researched the British
phenomenon a few weeks ago on the Web, and managed to upload a version of the highpitched sound into his cellphone.
He transferred the ring tone to the cellphones of two of his friends at a birthday party on
June 3. Two days later, he said, about five students at school were using it, and by
Tuesday the number was a couple of dozen.
"I just made it for my friends. I don't use a cellphone during class at school," he said.
How, David was asked, did he think this new device would alter the balance of power
between adults and teenagers? Or did he suppose it was a passing fad?
"Well, probably it is," said David, who added after a moment's thought, "And if not, I
guess the school will just have to hire a lot of young teachers."
Kate Hammer and Nate Schweber contributed reporting for this article.
Next Article in Technology (2 of 15) »
7
Hearing: the Young and the Old
Name:______________
Period:_____________
1. What is the “new weapon” on the cell phone front?
2. How can 1st graders use this weapon while adults are unable?
3. Like many “advances” in technology, the ring tone is an offshoot of a
development made by Welsh Security. What was the intent of their
‘Mosquito’?
4. The technical term for an aging ear is _________________. Most
adults from age ________ to _________ have symptoms of this
disorder.
5. Most human adults can hear frequencies between __________ and
___________ hertz.
6. Explain what a ‘hertz’ is…
7. How old was the pioneer of the first pirated ring tone that only young
adults can hear? (Most likely, provide an age range) _____________
8. What is the “authentic” ring tone now advertised as? (provide the
name of the ring tone) ________________________________
9. Have you heard of this?
10. Have you used it?
11. In class?
8
The ear has three major regions: the outer ear, the middle ear and the inner ear. The
process of hearing begins when sound waves make it to the middle ear, causing the ear drum
to vibrate.
Hearing and the Ear
http://health.howstuffworks.com/hearing.htm
1.
What are the three major regions of the ear?
a. ___________
b. ____________
c. ____________
2. The vibration of ear drum is transferred to the ___________________.
3. The Ossicles are also known as the _________________,
____________________, and the ____________________.
4. What amplifies the sound waves to the Oval Window? __________________
__________________________________________________________________
5. The Oval Window then vibrates moving fluid in the ____________________.
6. Specialized hair cells send _________ ___________ to the brain.
7. The brain interprets the signal as _______________.
9
CONCEPTS OF PHYSICS
Mr. Kuffer
Chapter 26: Sound
Formulas:
v=fλ
vair = 331 m/s + (0.6m/s x____ ° C)
1. a) How does the speed of sound vary with the temperature of air?
b) Compare the speed of sound in air for two sounds, one with a frequency of 200 Hz. And the
other with a frequency of 400 Hz.
c)
Compare the speed of sound in air with its speed in water and steel.
2. Calculate the speed of sound in air at…….
a) 20 ° C
c) 12 ° C
e) 21 °C
b) 33.3 °C
d) 8 ° C
f) 4 °C
3. At 0 ° C, sound travels in air at 331.5 m/s . What is the frequency of a sound wave whose
wavelength is 2.00 m?
4. A soldier hears the sound of the firing of a distant cannon 6.00 seconds after seeing the flash. If
the temperature is 20 °C, how far is the soldier from the cannon?
5. A girl hears her echo return from a wall 10.0 seconds after she makes a sound. If it is 13.5 ° C,
how far from the wall is the girl?
6. A bullet having a velocity of 680 m/s hits a target 510 m away. If the temperature of the air is 18
°C, how long after the bullet is fired will the shooter hear the bullet hit the target?
7. A miner drops a stone into a mine shaft 122.5 m deep. If the temperature is 21.3 ° C, how soon
after dropping the stone will the miner hear the splash?
8. Children often make a tin can string “telephone”. Explain the sound principles involved.
9. How many times faster is the speed of light than the speed of sound at 0 °C?
10. You look up and see a helicopter pass directly overhead. 3.10 seconds later you hear the sound of
the engines. If the air temperature is 23.6 ° C, how high was the helicopter flying?
11. On a day when the air temperature is 10.5 ° C, you use a whistle of frequency 2.5 X 10 4 Hz to
call your dog. What is the wavelength of the sound produced by the whistle?
12. A hobo is listening with his ear on the track for an oncoming train. When the train is 1.65 km
away, what is the time lag between the sound heard by the hobo and his friend, who is standing
nearby? (Assume the air temperature is 25.0 °C)
10
The Doppler Effect
As a wave source moves, the wave pattern also changes. As you can see from the
diagram below, the frequency is greater on the left while the wavelength is less. On the
right, the wavelength increases while the frequency decreases. This change in frequency
associated with the relative motion between a sound source and a listener is called the
Doppler Effect.
For a source of frequency, ƒ moving toward an observer, we can readily determine the
new frequency ƒ ` that the observer hears. During the period (T) of the source, the wave
moves out a distance vT (where v represents the speed of sound). During that same
period, the source moves at a different speed vs (speed of the source). The source,
therefore, moves a distance vsT. The difference between these two sources in the new
wavelength, λ`:
λ` = (v-vs)T
ƒ ` = v/ λ` = v / (v-vs) T
Knowing that ƒ = 1/T, we can say…
ƒ ` = ƒ / (1- vs / v)
(Source approaching)
If the source is moving away, the wavelength is increased. Note the expression stays the
same with the exception of the sign.
ƒ ` = ƒ / (1 ± vs / v)
11
Doppler Lab
DATA
Trial
#1
Trial
#2
Radius
(m)
Time for
Angular
10
Velocity
Revolutions (rad)/s
(s)
.65
8.76
1.5
13.83
#33
Linear
Velocity
(m/s)
Frequency
approaching
your ear
(Hz)
#34
#35
Frequency
away
from your
ear
(Hz)
#36
#37
#38
#39
Speed of Sound = 340 m/s
Frequency of the source = 440 Hz.
#33. What is the angular velocity in Trial #1?
a. 7.16 (rad)/s b. 3.56 (rad)/s c. 12.3 (rad)/s
d. 2.77 (rad)/s
#34. What is the linear velocity in Trial #1?
a. 2.99 m/s
b. 4.66 m/s c. 7.88 m/s d. 5.44 m/s
#35. What is the approaching frequency in Trial #1?
a. 454.78 Hz
b. 446.11 Hz
c. 450.88 Hz
d. 433.88 Hz
#36. What is the receding frequency in Trial #1?
a. 435.66 Hz
b. 428.66 Hz
c. 434.05 Hz
d. 446.33 Hz
#37. What is the linear velocity in Trial #2?
a. 5.66 m/s
b. 2.77 m/s
d. 6.81 m/s
c. 9.77 m/s
#38. What is the approaching frequency in Trial #2?
a. 445.32 Hz
b. 448.99 Hz
c. 430.44 Hz
d. 432.23 Hz
#39. What is the receding frequency in Trial #2?
a. 426.44 Hz
b. 428.33 Hz
c. 431.36 Hz
d. 449.0 Hz
12
ƒ ` = ƒ / (1 ± vs / v)
The Doppler Effect
ƒ ` = ƒ / (1- vs / v)
(Source approaching)
ƒ ` = ƒ / (1+ vs / v)
(Source moving away)
If the source is moving away, the wavelength is increased. Note the
expression stays the same with the exception of the sign.
Practice Problems!!!
1. A train is moving towards an observer at 31 m/s. It emits a sound
from the horn with a frequency of 305 Hz. What is the frequency
detected by the observer?
2. Same situation as above, only now the train is moving away from the
observer. Now what frequency is detected by the observer?
3. Amy Q., a “Concepts of Physics” student, is driving toward the
Ingomar fire hall at 40 mi/hr (18m/s), when the siren goes off. If the
siren has a true frequency of 680 Hz, what frequency will Amy Q.
hear as she is moving towards the fire hall? As she moves away?
4. A clown (named Jilly Reeps) on a unicycle, playing a tuba is moving
towards you at 12 m/s and you are moving toward her at 15 m/s. If
she is playing a note with a frequency of 440Hz, what frequency do
you hear?
5. Same situation as above, what frequency will you hear as you recede?
13
Energy and Information Transfer by waves
Waves: Energy, Power, and Intensity
The energy that is transferred by a wave is representative of the amplitude.
Therefore the maximum amplitude (height) of a wave is representative of
the maximum (total) energy of the wave.
We know waves oscillate back and forth much like a mass on a spring. They
are both examples of simple harmonic motion. Therefore the properties of
waves can be represented by the same fundamentals used with springs.
What we know:
W = ½ k x2
(from Hooke’s Law)
Apply this to a transverse wave and…
W = ½ k y2
eqn 1
Were y is the vertical displacement of the wave
We also know:
ƒ = 1/(2Π) √k/m
eqn 2
This equation simply quantifies the relationship between ƒ, k, and mass. The
frequency of a wave on a spring depends upon the spring itself (k) and the
mass being displaced (m). Note the above equation represents the frequency
in terms of k and m.
14
If we solve for k, we can eliminate that variable by substitution.
ƒ = 1/(2Π) √k/m
2Πƒ = √k/m
(2Πƒ)2 = k/m
(2Πƒ)2 m = k
Make a substitution for k into eqn 1
W = ∆E = ½ (4Π2 ƒ
∆E = 2Π2 ƒ
2
2
m) y2
m y2
∆E = 2Π2 ƒ2 m y2
eqn 3
The rate at which energy propagates along the string is the power (as in
mechanics). Therefore… (As you know, energy is measured in Joules)
P=E/t
eqn 4
As you know, Power is measured in Watts
A more useful representation of the power within a wave is intensity.
Intensity is the power flowing through a unit area within a wave.
I = P/A
eqn 5
Measured in Watts per square meter
W / m2
15
Sound Levels
The unit of sound intensity measurement is the decibel, dB. The decibel unit is one tenth
the size of the bell (named in the honor of the inventor of the telephone, Alexander
Graham Bell). The intensity level LI in decibles is defined to be ten times the logarithm
of the ratio of the two intensities I and Io; that is,
LI (in decibels) = 10 log10 (I / Io)
Where Io is the reference intensity. For example, if the intensity I exceeds the reference
intensity Io by a factor of 4, the intensity level of I is 6dB above Io.
LI = 10 log10 (4) = 10 (0.6) = 6 dB
16
Waves: Energy, Power, and Intensity
Practice Problems
1. A typical adult ear has a surface are of 2.1 x 10-3 m2. The sound
intensity during a normal conversation is about 3.2 x 10-6 W/m2 at the
listener’s ear. Assume the sound strikes he surface of the ear
perpendicularly. How much power is intercepted by the ear?
2. The average sound intensity inside a busy restaurant is 3.2 x 10-5
W/m2. How much energy goes into each ear (area 2.1 x 10-3 m2) during
a one hour meal?
3. At a distance of 3.8 m from a siren, the sound intensity is 3.6 x 10-2
W/m2. Assuming that the siren radiates sound uniformly in all
directions, find the total power radiated.
4. Suppose that sound is emitted uniformly in all directions by a public
address system. The intensity at a location 22 m away from the sound
source is 3.0 x 10-4 W/ m2. What is the intensity at a spot that is 78
m away?
5. When a helicopter is hovering 1450 m directly overhead, an observer
on the ground measures a sound intensity I. Assume that sound is
radiated uniformly from the helicopter and that ground reflections
are negligible. How far must the helicopter fly in a straight line
parallel to the ground before the observer measures a sound intensity
of ¼I?
6. A dish of lasagna is being heated in a microwave oven. The effective
area of the lasagna that is exposed to the microwaves is 1.6 x 10-2 m2.
The mass of the lasagna is 0.25 kg, and its specific heat capacity is
3400 J/ (kg · ºC). The temperature rises by 80.0 ºC in 7 minutes.
What is the intensity of the microwaves in the oven?
17