Ch. 8 Elasticity and waves: Sound; Physics with Health Science Applications
Section 8.1
8.1. (I) What is the time required for one complete vibration of an ultrasonic
stethoscope that operates at a frequency of 2.0 MHz?
Background: We are given the frequency and asked for the time required
for one complete vibration. The relationship between frequency and time is:
f =
1
T
You should memorize this rule.
It is critical to know that 1 MHz is 1x106 Hz. You also must remember that
2x10 6
1 Hz, is 1 vibration per second. Therefore 2.0 MHz equals
.
sec
1
1
Ans. f = ; Solving for T we get T =
T
f
T=
1
1
−7
=
6 = 5x10 sec.
f 2.0x10
sec
8.2. (I) What is the frequency of a sound wave made by
a tuning fork that requires 2.44 x 10-4 sec for one complete vibration?
The relationship between frequency and time is:
1
T
1
1
4100
f = =
=
= 4,100Hz
−4
T 2.44x10 sec
sec
f =
Ans.
= 4,100 Hz
8.3. (1) What are the periods of the lowest and highest frequencies normally
audible to humans (20 and 20,000 Hz)?
f =
1
1
; Solving for T we get T =
T
f
1
1
=
= 0.05 sec
20
f
sec
1
1
Ans. b) T = =
= 5.0x10 −5 sec
20,
000
f
sec
Ans. a) T =
Ch. 8 Elasticity and waves: Sound; Physics with Health Science Applications
8.7. (I) Calculate the wavelengths of the lowest and highest frequency sounds
normally audible to humans (20 and 20,000 Hz), given the speed of sound to be
344 m/sec.
Background: The velocity of a wave is given by the following equation:
v = fλ
Solving for wavelength, we get:
λ =
v
f
m
v 344 s
Ans. a.) λ = = 20 = 17.2meters
f
s
m
344
v
s = 0.01722meters
Ans. b.) λ = = 20, 000
f
s
8.8. (I) (a) If the speed of sound in seawater is 1560 m/sec, what is the wavelength
of a 100-kHz sound made by a dolphin? (b) What is the wavelength of a 100-kHz
sound made by a bat in air? Use 344 m/s as the speed of sound in are.
The velocity of a wave is given by the following equation:
v = fλ
Solving for wavelength, we get: λ =
Ans. a.) λ =
v
f
Ans. b.) λ =
v
f
v
f
m
s = 0.0156meters = 15.6mm
=
100, 000
s
m
344
s = 3.44x10 −3 meters = 3.44mm =
100, 000
s
1560
8.9. (I) A gun is fired on a day when the speed of sound is 335 m/sec and an echo
is heard 0.75 sec later. How far away is the object that created the echo?
Background: An echo is produced when a sound wave goes out and then is
reflected back by an object. Because the echo time is the time out and back, we
Ch. 8 Elasticity and waves: Sound; Physics with Health Science Applications
must divide the echo time by two in order to determine the distance to the
reflector.
Ans. Distance = vt = 335
m 0.75 sec
i
= 126 meters
sec
2
8.10. (I) At a fireworks display a person noticed that the sound of an exploding
shell arrived 0.30 sec after the flash from the explosion. How far away from the
person was the explosion if the speed of sound is 340 m/sec? Can the time needed
for the light to arrive be neglected? (The speed of light is 3.0 x 108 m/sec.)
m
i0.30 sec = 102 meters
sec
We can ignore the time that it takes light to reach our eyes from the
fireworks because light travels at 3 x 108 meters per second. Over the small
distance between the fireworks and our eyes, it takes almost no time for the light to
reach our eyes.
Ans. Distance = vt = 340
8.11. (I) A storm in the South Pacific creates ocean waves traveling at 10 m/sec.
How long does it take these waves to reach the California coast 12,000 km
away?
Please determine the time in a) seconds and b) days.
Ans. a) Distance = vt ; Solving for time we get and substituting values:
Distance = vt
Distance 12x10 6 m
time =
=
= 1.2x10 6 sec
m
v
10
s
Ans. b) Converting time in seconds to days:
1hr
1day
1.2x10 6 secondsi
i
= 13.9days
3600 sec 24hr
It looks like the surf’s up in two weeks.
8.12. (I) How many times a minute does a boat bob up and down on ocean waves
that have a wave length of 50 m and a speed of 6.0 m/sec?
Background: We are asked to determine the frequency in vibrations per
minute.
Baby step 1. v = f λ ; solving for frequency and substituting values:
Ch. 8 Elasticity and waves: Sound; Physics with Health Science Applications
m
v
s = 0.12
f = =
λ 50m
sec
6.0
Baby step 2. Converting to number of bobs per minute:
0.12 60s
7.2
i
=
Ans.
f =
sec 1min min
8.13. (I) What is the frequency of ultrasound with a wavelength of 0.25 mm in
tissue if the speed of sound in tissue is 1540 m/sec? Please convert your answer to
megahertz. Use 1540 m/s as the speed of sound in soft tissue. Remember that 1
MHz = 1 x 106 Hz
v = f λ ; solving for frequency and substituting values:
m
v
s = 6.16x10 6 Hz
f = =
λ 0.25x10 −3 m
1540
Converting to MHz:
6
Ans. 6.16x10 Hzi
1MHz
= 6.16MHz
1x10 6 Hz
8.14. (I) If the person in Figure 8.2(a) shakes the spring up and down 3 times a
second and the crests on the spring are observed to be 0.50 m apart, what is the
speed of propagation of the wave on the spring?
Ans. v = f λ =
3
m
i0.50m = 1.50
sec
s
−7
8.15. (I) What is the sound level, β , of a sound with an intensity of 1x10
W
?
m2
Background: The equation that gives us the relationship between sound
I
level, β , and intensity I, is: β = 10 log( )
I0
This equation uses log base ten. You will press the log key on your
calculator when using log base 10. You must also remember that the
−12
reference intensity, I0 is: I 0 = 1x10
Watts
m2
Ch. 8 Elasticity and waves: Sound; Physics with Health Science Applications
W
m 2 ) = 10 log 1x10 5 = 10i5 = 50 decibels
β = 10 log(
W
1x10 −12 2
m
Ans. The sound level β is 50 decibels
1x10 −7
(
)
8.16. (I) What is the sound level β of ultrasound with an intensity of 10
W
used
m2
for deep heat treatments (diathermy)?
W ⎞
⎛
10 2
⎜
⎟
I
m
β = 10 log( ) = β = 10 log ⎜
⎟
−12 W
I0
⎜⎝ 1x10
⎟
m2 ⎠
β = 10 log (1x1013 ) = 10i13 = 130 dB
Ans.
β = 130 dB
8.17. (I) What is the intensity in watts per square meters of a 40-dB sound?
Background: In this problem we are told that beta = 40 dB and are asked to
determine the value of the intensity I.
I
)
Baby step 1:
Fill in the known values: 40 = 10 log(
−12 W
1x10
m2
Baby step 2: Notice that the intensity I is the only unknown in the problem.
To solve for I, we will first divide both sides of the equation by 10. This
yields:
I
4
=
log(
) −12 W
1x10
m2
Baby step 3: we will take the antilog of both sides of the equation. This
will get rid of the log function the right side of the equation.
I
10, 000 =
W
1x10 −12 2
m
Ch. 8 Elasticity and waves: Sound; Physics with Health Science Applications
−8
Baby step 4: Now solving for I we determine that I equals: I = 1x10
−8
Ans. I = 1x10
W
m2
W
m2
8.23. (II) (a) What frequency ultrasound should be used to see details as small as
1.0 mm in tissue? The speed of sound in soft tissue is 1540 m/sec. Is this a
maximum or minimum frequency? (b) To what depth in centimeters, is this
frequency sound effective as a diagnostic probe?
Background: The amount of detail that may be observed is determined by
the wavelength. In order to see objects that that are 1.0 mm in size, the
wavelength must be 1.0 mm or smaller.
8.23 a) Baby step 1: v = f λ ; solving for frequency and substituting
values:
m
v
s = 1.54x10 6 Hz
f
=
=
Baby step 2:
−3
λ 1.0x10 m
1540
Baby step 3:
Convert the frequency to megahertz.
Ans. a) The frequency is 1.54 MHz.
1.54x10 6 Hzi
1MHz
= 1.54MHz
1x10 6 Hz
8.23 b) Background: As of the year 2010, the maximum imaging depth is
approximately 200 wavelengths. We are told that the wavelength is 1.0 mm.
Therefore:
1.0mm
200 wavelengthsi
= 200 mm = 20 cm
wavelength
Ans. b) 20 cm
Ch. 8 Elasticity and waves: Sound; Physics with Health Science Applications
Sections 8.4 and 8.5
8.32.
An Ambulance with a siren emitting a steady frequency of 1200 Hz is
moving at 120 km/hr. What frequency is observed by a stationary observer
as the ambulance a) approaches? b) As it moves away? Use 344 m/s as the
speed of sound in air.
Background: We will use the following equation:
⎛
⎞
V
fobserved = ⎜
femitted
⎟
V
±
V
⎝
source ⎠
V is the speed of the wave, in this problem it is the speed of sound in
air.
Vsource is the speed of the object that is emitting the wave. In this
problem it is the speed of the ambulance siren.
femitted is the frequency that is being emitted by the moving sound
source. In this problem it is the frequency of the siren.
The ± Vsource means that under some circumstances we add the speed
of the moving object to the speed of the wave, and under different
circumstances, we subtract the speed of the moving object from the
speed of the wave. In particular, when the sound emitting object is
moving toward the listener, we must subtract the speed of the source
from the speed of the wave.
When we add or subtract values, they must have the same set of units.
Therefore, we will first convert 120 km/hr to m/s so your units are all of the
same kind.
The speed of the siren is: 120
km 1000m 1hr
m
i
i
= 33.3
hr km 3600 sec
s
a) When the object is approaching, you must subtract its velocity
from the speed of the wave.
fobserved =
344
m
s
m
m⎞
⎛
⎜⎝ 344 − 33.3 ⎟⎠
s
s
i1200Hz = 1, 330Hz
Ch. 8 Elasticity and waves: Sound; Physics with Health Science Applications
8.32.b) When the object is moving away, you must add its velocity to
the speed of the wave.
m
344
s
fobserved =
i1200Hz = 1, 094Hz
m
m⎞
⎛
⎜⎝ 344 + 33.3 ⎟⎠
s
s
Ans. 8.32a)
1,330 Hz
Ans. 8.32.b) rounded to 3 significant digits is 1,090 Hz
8.33. (I) A low-flying jet aircraft approaches a stationary observer on the ground at
a speed of 1000 km/hr on a day when the speed of sound is 345 m/sec. (a) If the
engine emits a sound of frequency 4000 Hz, what frequency does the stationary
observer receive as the plane a) approaches? (b) As the plane flies away?
⎛
⎞
V
fobserved = ⎜
femitted
⎝ V ± Vsource ⎟⎠
Try this one on your own and check your answers in the back of the text. If you
are stuck, ask in class.
8.35. (II) The speed of sound in tissue is 1540 m/sec. An ultrasonic wave sent into
blood will be partly reflected back toward the source by blood cells. If the
returning echo has a frequency 400 Hz higher than the original 2.0-MHz frequency,
what is the velocity of the blood? (Note that because there is an echo involved,
there are two Doppler shifts here.) Use a doppler angle of 0˚.
In problems where the sound is reflected, we will use the following rule:
fD =
2 f0Vscatterer Cos(θ )
Vpropagation − Vscatterer Note that the Cos (0) is 1 and the velocity of
the scatterer, (the blood), is very small compared to the speed of sound in tissue.
Therefore, the above equation reduces to:
fD =
2 f0Vblood
Vpropagation
Ch. 8 Elasticity and waves: Sound; Physics with Health Science Applications
Solving for the velocity of the blood yields:
Vblood =
fDVpropagation
2 f0
=
m
s = 0.154 m
2i2x10 6 Hz
s
400Hzi1540
The velocity of the blood is approximately 15
cm
.
s