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Section 12.1 Review Q, pg 413:
[1] A greater frequency is perceived as a higher pitch.
[2] Echolocation can find objects about one wavelength long.
Since λ = Speed/f, the higher the frequency the smaller the
object (fish) that can be located. Thus, a dolphin can locate a
fish as small as: λ = 1450 m/s/220000 Hz = 0.006 m = 6 mm long.
[3] The compressions in a sound wave are created by the collisions of
molecules. Since the molecules in water are closer together the
collisions take place more frequently and the sound moves faster.
[4] No, because plane waves are flat and not curved. As the wave
moves farther from its source it becomes more flat, less curved,
and can approximate a plane wave.
[5] To the right
[6] To the left.
[7] The dolphin is catching up to the fish.
Ch 12 Review Q, pg 434: Q 1-12
[1] Because air molecules vibrate in a direction parallel to the
direction of wave motion.
[2] Diagrams should depict a sine curve that begins and ends at its
lowest point and that has three crests and tow troughs.
[3] Frequency is an objective measure of the rate of particle
vibration. Pitch is a subjective quality that depends on the
listener.
[4] Infrasonic waves are below 20 Hz, audible waves are between 20
and 20,000 Hz, and ultrasonic waves are greater than 20,000 Hz.
[5] Molecules that have more motion (higher temperature) can transfer
their vibrations more easily. This is less noticeable in liquids
and solids because the particles are closer together.
[6] the siren’s pitch will drop.
[7] Because their short wavelengths can image small objects.
[8] Frequency doubles; Speed remains constant.
[9] Sound travels faster through the ground.
[10] Notes that are played at the same time reach your ears at the
same time.
[11] The driver of the van.
[12] Greater than 40 kHz
Practice A, pg 415
[1a] 8.0x10-4 W/m2 [b] 1.6x10-3
W/m2 [c] 6.4x10-3 W/m2
[2] 8.91x10-3 W/m2
[3] 2.3x10-5 W
[4] 4.5 W
[5] I = P --->1.2x10-3 = 0.35 ---> r2x1.2x10-3
= 0.35
4πr2
4πr2
4π
r2 = 0.35 ----> r2 = 23.2 --->r = √23.2 =
4.8 m
(4π1.2x10-3)
Note parenthesis
Section 12.2 Review Q, pg 420:
[1] The intensity has increased by a factor of 100 (102)
[2] no; because the sensation of loudness is approximately
logarithmic in the human ear.
[3] The second tuning fork will pick up the vibrations of the
first tuning fork, and a faint sound will be heard from the
second fork. This occurs because the two forks have the
same natural frequency, which is the condition required for
resonance.
[4] 2.0x104 m
[5] a,d,f; c,e
Ch 12 Review, Q 13-23. pg 434:
[13] Intensity is power per area; decibel level is a measure of
relative intensity
[14] 90dB, 30 dB, 20 dB, 60 dB (Answers may vary slightly.)
[15] Because the threshold of hearing depends on both frequency
and intensity.
[16] When forced vibration is the same as the natural frequency
of a vibrating system.
[17] The sound intensity from the orchestra is 100 times that
from the violin.
[18] 9 machines (for a total of 10)
[19] Because intensity decreases with distance and the sound has
traveled from the source to a reflecting surface and back
(and because of imperfect reflection).
[20] The swing’s amplitude is maximized when the pushes match
the swing’s natural frequency.
[21] Vibrations could set the bridge in motion if they match one
of the bridge’s natural frequencies.
[22] 70 dB
[23] 7.96x10-2 W/m2
Practice B, pg 427
[1] 440 Hz
[2] 260 Hz, 520 Hz, 780 Hz
[3a] 82.1 Hz [b] 115 Hz [c] 144 Hz
[4] 440 m/s
Section 12.3 Review Q, pg 431:
[1] 524 Hz
[2] 346 m/s
[3] 388 Hz and 396 Hz
[4] the odd harmonics
[5] b, c
Ch 12 Review, Q 24-35, pg 435:
[24] The lowest possible frequency of a vibrating system;
They are integral multiples of the fundamental frequency.
[25a] 4.0 m [b] 2.0 m [c] 1.3 m [d] 1.0 m
[26] Because a closed end is a node, an open end is an antinode.
[27] The instruments have different harmonics present at various
intensity levels.
[28] The guitar’s body transfers the string’s vibrations to the air
more efficiently, which increases the intensity of the sound.
[29] 3 Hz
[30] To change the length of the air column, thereby changing the
fundamental frequency.
[31] Yes. This difference will equal the fundamental frequency if
the pipe is open at both ends but will equal twice the
fundamental frequency if the pipe is closed at one end.
[32] The first possible wavelength is 2L for the flute and 4L for the
clarinet. Because the speed of sound is the same in each and
v = λf, the flute’s fundamental frequency is twice the
clarinets.
[33] As temperature increases, the speed of sound in air increases.
Because f1 is proportional to V, fundamental frequency likewise
increases.
[34] 443 Hz, 886 Hz, 1330 Hz
[35] 3.0x103 Hz
[36a] 52 cm [b] 640 Hx, 960 Hz [40a] 70.1 Hz [b] 285
[37] 5 beats per second [41] Lclosed = 1.5Lopen
[38] 20 m, 2x10-2 m [42] 1.9x10-2 m
[39] 0.20 s [43a] 5.0x104
W [b] 2.8x10-3 W
Standardized Test Prep, pg 438
1] B [2] J [3] D [4] H [5] C [6] F [7] B [8] G [9] D [10] G 11] 7 Hz
[12] 750 Hz [13] 0.33 m [14] 0.47 W/m2 [15] 5.0x10-5W
12.1 Sound Waves
1 mi = 1609 m Speed of Light = 3.0 x 108 m/s Io = 1x10-12 W/m2
1 Define: [a] Wavelength [f] Hertz
[b] Frequency [g] Doppler Effect
[c] Period [h] Transverse Wave
[d] Amplitude [i] Longitudinal Wave
[e] Cycle
2 a] State two uses of the Doppler Effect
b] Identify all terms in the equation V/c =
Δf/f
3 Draw & label the diagram for a:
a] Transverse Wave b] Longitudinal Wave
4 Diagram & label the
following situations: Use tuna can & washer.
a] Both the source and listener are stationary.
b] The listener is stationary and the source is
approaching.
c] The listener is stationary and the source is
receding.
d] The source is stationary and the listener is
approaching.
e] The source is stationary and the listener is
receding.
1 pt for correct labeling of Source & Listener
1 pt for diagram with incorrect placement of source
dots.
1 pt for correct placement of source
dots.
1 pt for labeling diagram 3 second snapshot
1 pt for labeling the waves as 1st-3 sec, 2nd 2
sec, 3rd 1 sec
5 Find the speed (mi/h) of a
car if the radar gun (9.375 GHz) measures a frequency
change of 2000 Hz.
12.2 Intensity & Resonance
6 Define: [a] Intensity [b] Resonance [c] Decibel
7 Write the equation for intensity in words & symbols.
8 Find the intensity of sound if you are located 3 m
from a person speaking with a
0.50 W voice.
A1 & A2, pg 415
9 Find the power of a source if your audiometer reads 8.0x10-5 W/m2 when you are 2.5 m away from the source. A3 & A4, pg 415
10 Find the distance from a 6 watt sound if the
audiometer reads
2.0x10-4 W/m2.
A5, pg 415
11 When the decibel level in the auditorium goes from 70 dB to 110 dB how much greater is the intensity? 12.1 Rev Q1, pg 420, #17 pg 434
12 You shout at your friend
3.0 m away.
If your shouting power is 6.0x10-3 W, how
loud is your voice when it reaches your
friend?
#22, pg 435
12.3 Harmonics
13 Define: [a] Fundamental Frequency [c] Timbre
[b] Harmonic Series [d] Beat
14 Draw & label the diagram for
a] a string vibrating at its fundamental frequency.
b] an open pipe vibrating at its fundamental frequency.
c] a closed pipe vibrating at its fundamental frequency.
d] a closed pipe vibrating at its 2nd harmonic.
e] a closed pipe vibrating at its 3rd harmonic.
Ans: [5] 143 mi/h [8] 4.42x10-3
[9] 6.28x10-3 W [10] 48.9 m
[11] 10,000 [12] 77 dB
1 Define: [a] Wavelength: Distance between
corresponding points on consecutive
waves
[b] Frequency: The number of events per unit time.
Ex: 70 heart beats per minute
[c] Period: The time per cycle.
Ex: Count the cycles 10 sec
Period = 10 sec/#Cycles
[d] Amplitude The Maximum displacement from rest.
Ex: The height of a crest or trough.
[e] Cycle One complete back & forth motion.
[f] Hertz
One cycle per second (sec-1)
[g] Doppler Effect The
apparent change in frequency (not amplitude)
due to the motion of the source or the listener.
[h] Transverse Wave A
wave whose vibration is perpendicular to wave travel.
[i] Longitudinal Wave A
wave whose vibration is parallel to wave travel.
[j] Intensity
Power per unit area (W/m2).
Ex: Power =
P
Area of a Sphere
4πR2
[k] Resonance
A large vibration caused by a small periodic stimulus at
the system's natural frequency.
Ex: A child moves mover her legs two feet in rhythm with a
swing and the swings in a 10 ft arc.
[l] Decibel
The loudness of sound perceived by humans.
It is a logarithmic function of intensity:
dBHi - dBLo = 10Log(IHi/ILo)
No Calculator
1 mi = 1609 m Speed of Light = 3.0 x 108 m/s Io = 1x10-12 W/m2
5 Find the speed (mi/h) of a
car if the radar gun (9.375 GHz) measures a frequency
change of 2000 Hz.
V = Δf = V = 2000
Hz
c f 3.0x108m/s 10x109 Hz
V = 2000 Hz x 3.0x108m/s = 6000x10-1 =
600 m/s = 60 m/s
10x109 Hz 10 10
60 m x 3600 s x 1 mi = 6x36 = 3x36 = 108 mi
(Calculator Ans: 143)
s 1 h 2000 m 2 h
8 Find the intensity of sound if you are located 3.0 m
from a person speaking with a 0.50 W voice.
I = P = 0.50 W = 0.50 W = 0.50 = 0.50x10-2
= 5.0x10-3 W/m2
4πR2
4π32m2
12x9 108
(Calculator Ans: 4.42x10-3 W/m2)
9 Find the power of a source if your audiometer reads
8.0x10-5 W/m2 when
you are 2.5 m away from the source.
I = P ---> 8.0x10-5
W/m2 = P
4πR2
4π2.52m2
8.0x10-5 W/m2 x 4π2.52m2
= P
8.0x10-5x12x32m2
= 8x10-5x12x9 = 72x10-5x10
= 720x10-5 = 7.2x10-3
W = P
(Calculator Ans: P = 6.28x10-3
W)
10 Find the distance from a 6 watt sound if the
audiometer reads
2.0x10-4 W/m2.
I = P ---> 2.0x10-4
W/m2 = 6
4πR2
4πR2
2.0x10-4
W/m2 x R2 = 6
4π
R2 =
6 = 6x104 = 60000 = 60000 =
10000 = 2500 m2
4π x 2.0x10-4
4π2 (8π)
24 4
____
R = √2500 = 50 m (Calculator: 48.9 m)
11 When the decibel level in the auditorium goes from 70
dB to 110 dB how much greater is the intensity?
dBHi - dBLo = 10 Log(IHi/ILo) OR
dBHi - dBLo = X & IHi/ILo =
10x
10
110 - 70 = X --->X = 4 & IHi/ILo = 104
---> IHi = ILox10
10
12 You shout at your friend 3.0 m away. If your shouting
power is
6.0x10-3
W, how loud is your voice when it reaches your friend?
Given: P = 6.0x10-3 W Since
dB = 10Log(I/Io)
R = 3.0 m We need to find I
Io = 1x10-12 W/m2 I = P = 6x10-3 =
6x10-3 = 6.0x10-5
Find: dB = ? 4πR2
4π32
12x9
(Calculator Ans: 5.31x10-5)
dB = 10Log(I/Io)
dB = 10Log(6.0x10-5/10-12)
dB = 10 Log(6.0x107)
dB = 10(Log(6) + Log(107)
dB = 10(0.6 + 7) = 10(7.6) = 76 dB (Calculator Ans: 77 dB)
1 mi = 1609 m Speed of Light = 3.0 x 108 m/s Io = 1x10-12 W/m2
5 Find the speed (mi/h) of a
car if the radar gun (9.375 GHz) measures a frequency
change of 2000 Hz.
V = Δf = V =
2000 Hz
c f 3.0x108m/s 9.375x109
Hz
V = 2000 Hz x 3.0x108m/s =
6000x10-1 = 600 m/s = 64 m/s
9.375x109 Hz
9.375 9.375
64 m x 3600 s x 1 mi = 143 mi
s 1 h 1609 m
h
8 Find the intensity of sound if you are located 3 m
from a person speaking with a 0.50 W voice.
I = P = 0.50 W = 4.42x10-3 W/m2
4πR2 4π32m2
9 Find the power of a source if your audiometer reads
8.0x10-5 W/m2 when you are 2.5 m
away from the source.
I = P ---> 8.0x10-5 W/m2 =
P
4πR2 4π2.52m2
8.0x10-5 W/m2 x 4π2.52m2
= P
P = 6.28x10-3 W
10 Find the distance from a 6 watt sound if the
audiometer reads
2.0x10-4 W/m2.
I = P ---> 2.0x10-4 W/m2 =
6
4πR2 4πR2
2.0x10-4 W/m2 x R2 =
6
4π
R2 = 6 = 6x104 = 60000 =
2387 m2 Be sure to put 8π in ()
4π x 2.0x10-4 4π2 (8π)
____
R = √2387 = 48.9 m
11 When the decibel level in the auditorium goes from 70
dB to 110 dB how much greater is the intensity?
dBHi - dBLo = 10 Log(IHi/ILo) OR
dBHi - dBLo = X & IHi/ILo =
10x
10
110 - 70 = X --->X = 4 & IHi/ILo = 104
---> IHi = ILox10
10
12 You shout at your friend 3.0 m away. If your shouting
power is
6.0x10-3 W, how loud is your voice when it
reaches your friend?
Given: P = 6.0x10-3 W Since dB = 10Log(I/Io)
R = 3.0 m We need to find I
Io = 1x10-12 W/m2 I = P = 6x10-3 = 0.006 = 5.31x10-5
Find: dB = ? 4πR2 4π32 113
dB = 10Log(I/Io)
dB = 10Log(5.31x10-5/10-12)
dB = 10 Log(5.31x107)
dB = 77.2 Decibels--SigFig-->77 dB