Transcript Physics 1422 - Introduction

Physics 203 – College Physics I
Department of Physics – The Citadel
Physics 203
College Physics I
Fall 2012
S. A. Yost
Chapter 12
Sound – Part 2
Physics 203 – College Physics I
Department of Physics – The Citadel
Announcements
The Final Exam is next Monday: 8 – 11 AM here.
Bring a page of notes. I won’t provide equations. I
will provide constants, conversion factors, and
moments of inertia.
You can find solutions to the end-of-chapter
problems from each homework set and solutions
to all of the exams in the Course Materials section
of MasteringPhysics.
Also, see my list of equations for each exam and the
last several chapters there.
Physics 203 – College Physics I
Department of Physics – The Citadel
Quiz Question 1
If a mass hanging on the end of a light-weight spring
is pulled down and released, the oscillation period
will depend on
A The spring stiffness constant
B
The hanging mass
C
The distance it is pulled down
There may be more than one answer. Enter multiple
choices sequentially.
Physics 203 – College Physics I
Department of Physics – The Citadel
Quiz Question 2
The frequency of a wave increases. If the speed of
the wave remains constant, what happens to the
distance between successive crests of the wave?
A The distance increases
B The distance decreases
C The distance remains the same
Physics 203 – College Physics I
Department of Physics – The Citadel
Quiz Question 3
When a guitar string is vibrating in its fundamental
mode, the wavelength of the vibration is
A the length of the guitar string.
B half the length of the guitar string.
C twice the length of the guitar string.
D independent of the length of the guitar string.
Physics 203 – College Physics I
Department of Physics – The Citadel
Quiz Question 4
A standing wave is created in an organ pipe which is
closed on one end and open on the other. Which
is true?
A There is a pressure node at both ends.
B There is a displacement node at both ends
C There is a pressure node at the closed end and a
displacement node at the open end.
D There is a pressure node at the open end and a
displacement node at the closed end.
Physics 203 – College Physics I
Department of Physics – The Citadel
Quiz Question 5
If I am listening to a band at a concert, and move
twice as far from the speakers, the loudness will
A be a few decibels less than before.
B be about 100 decibels less than before.
C be about half as many decibels as before.
D be about ¼ as many decibels as before.
Physics 203 – College Physics I
Department of Physics – The Citadel
Quiz Question 6
The Doppler Effect on the frequency of a sound is
 vsnd  vobs
f '  f 
 vsnd  vsrc



where vsnd, vobs, and vsrc are the velocities of the
sound, the observer, and the source, if the sign
conventions are chosen appropriately. The correct
choice when the source and observer are
approaching each other is
A) vobs > 0, vsrc > 0
B) vobs < 0, vsrc > 0
C) vobs < 0, vsrc < 0
D) vobs > 0, vsrc < 0
Physics 203 – College Physics I
Department of Physics – The Citadel
Doppler Effect
If you are moving toward the source of a sound, you
pass through the wave crests more rapidly than if
you were standing still.
Physics 203 – College Physics I
Department of Physics – The Citadel
Doppler Shift
If you are moving toward the source of a sound, you
pass through the wave crests more rapidly than if
you were standing still.
The relative speed of the sound crests and you
would be v’ = vsnd + vobs, so the frequency is
f’ = v’/l = (vsnd + vobs)/l
with l = vsnd/f
so that
f’ =
(
vsnd + vobs
f =
vobs
)
(
vsnd
vobs + 1
)f
Physics 203 – College Physics I
Department of Physics – The Citadel
Doppler Effect
If the source is moving toward you and you are still,
the relative speed of sound is unchanged, but the
wavelength is.
Physics 203 – College Physics I
Department of Physics – The Citadel
Doppler Effect
The fire truck emits sound-wave crests with period
T = 1/f. The crest moves ahead a length
l = vsnd T in this time.
But the fire truck is moving too, so it is a distance
D = vsrc T closer to the previous crest when it emits
the next crest.
Since T = 1/f = l/vsnd ,
D = l vsrc/vsnd.
The wavelength is then
(
l’ = lD  l 1 –
vsrc
vsnd
)
Physics 203 – College Physics I
Department of Physics – The Citadel
Doppler Effect
Equivalently,
l’ 
(
vsnd – vsrc
l
vsnd
)
To get the frequency, use l = vsnd/f, l’ = vsnd/f’ and
taking the reciprocal of both sides gives
vsnd
f’ = v – v
f.
snd
src
(
)
If both the source and observer are moving, combine
the two expressions:
 vsnd  vobs
f '  f 
 vsnd  vsrc



Physics 203 – College Physics I
Department of Physics – The Citadel
Guitar Sound
Low A on a slightly-out-of-tune open guitar string
(Low A should be 110 Hz, not 108 Hz.)
A3
216
A2
108
324
E3
A4
432
C5
540
E5
648
G5
756
A5
864
B5
972
Physics 203 – College Physics I
Department of Physics – The Citadel
Wind Instruments
Wind Instruments produce sounds using a
vibrating column of air.
For example, consider this tube, open on the
ends.
Physics 203 – College Physics I
Department of Physics – The Citadel
Wind Instruments
Air can vibrate back and forth, but at the ends, the
pressure must be the same as it is outside the
tube.
The air can blow in and out freely, keeping the
pressure fixed. So there is no pressure variation
at the ends.
Physics 203 – College Physics I
Department of Physics – The Citadel
Wind Instruments
Drawing a red line for the difference
between the atmospheric pressure and
pressure in the pipe, the wave’s maxima
and minima would look like this:
L
node
antinode
node
Physics 203 – College Physics I
Department of Physics – The Citadel
Wind Instruments
Then the frequencies produced by the open
tube are
fN = Nv/2L = Nf1
L
antinode
node
node
Physics 203 – College Physics I
Department of Physics – The Citadel
Wind Instruments
The open tube produces frequencies
f1, 2f1, 3f1, 4f1, …
These are the fundamental, 2nd harmonic,
3rd harmonic, …
This is just as for a vibrating string.
0
Df
f1
2f1
Df
3f1
Df
4f1
Df
5f1
Df
Df = v/2L
= f1
Physics 203 – College Physics I
Department of Physics – The Citadel
Wind Instruments
If the tube is closed on one end, the frequencies
produced are different. Then the pressure
variation is a maximum at the closed end,
where the air cannot move.
L
antinode
node
Physics 203 – College Physics I
Department of Physics – The Citadel
Wind Instruments
The fundamental mode is produced with ¼
wavelength in the tube.
This means that l1 = 4L,
f1 = v/l1 = v/4L.
L
antinode
node
Physics 203 – College Physics I
Department of Physics – The Citadel
Wind Instruments
The next harmonic occurs at the next higher
frequency with an antinode on the left:
Then 3l/4 = L,
f = v/l = 3v/4L = 3f1.
L
antinode
antinode
node
Physics 203 – College Physics I
Department of Physics – The Citadel
Wind Instruments
Since f = 3f1, this is called the third harmonic.
There is no second harmonic in a tube open on
one end and closed on the other.
L
antinode
antinode
node
Physics 203 – College Physics I
Department of Physics – The Citadel
Wind Instruments
In general, for a tube closed on one end,
fN = Nv/4L = Nf1
with N = 1, 3, 5, 7, 9, …
L
antinode
antinode
node
Physics 203 – College Physics I
Department of Physics – The Citadel
Wind Instruments
Notice that the difference between successive
harmonics still corresponds to a half
wavelength, as for an open tube:
D f = v/2L
( = f3 – f1, …)
L
antinode
antinode
node
Physics 203 – College Physics I
Department of Physics – The Citadel
Wind Instruments
The closed tube produces frequencies
f1, 3f1, 5f1, 7f1, …
These are the fundamental, 3rd harmonic, 5th
harmonic, …
The even harmonics are all missing.
0
f1
Df/2
Df
3f1
5f1
Df
7f1
Df
9f1
Df
Df = v/2L
= 2f1
Physics 203 – College Physics I
Department of Physics – The Citadel
Wind Instruments
Example: Didgeridoo – the first two harmonics
produced are (analyzed using my laptop)
f1 = 74 Hz
f3 = 214 Hz
Compare 3 f1 = 222 Hz
T1 = 1/f1 = 0.0113 s
T3 = 1/f3 = 0.00467 s
Physics 203 – College Physics I
Department of Physics – The Citadel
Actual Frequencies Produced
The actual frequencies produced by an instrument
are a superposition of the possible harmonics, with
one of them usually being dominant.
Didgeridoo…
(± means slightly sharp/flat)
74 D+
148 D+
222 A+
369 F# 295 D+
Frequency (Hz)
516 C443 A+
Physics 203 – College Physics I
Department of Physics – The Citadel
Wind Instruments
We can find the speed of sound using the
frequencies and length of the tube.
Df = f3 - f1 = 140 Hz = v/2L
This works for either an open or closed tube!
L = 1.20 m,
giving v = 2LDf = 341 m/s.
This is very close to the expected result (343 m/s).
Physics 203 – College Physics I
Department of Physics – The Citadel
Vocal Harmonics
Higher harmonics can also be produced strongly
from the voice, a type of wind instrument. This is
used, for example, in Tuvan throat singing to
produce several notes simultaneously, or to
generate very high pitched sounds – up to the
16th - 18th harmonic.
There are no instruments or whistling here,
just one person singing.
Mergen Mongush, on TUVA, Voices from the
Center of Asia – Smithsonian/Folkways Records