Chapter 15 solutions P15.2. Prepare: The wave is a traveling wave on a stretched string. We will use Equation 15.2 to find the
tension that corresponds to a wave speed of 180 m/s. To be able to obtain the tension from 15.2, we will first
obtain µ from the first part of the problem.
Solve: The wave speed on a stretched string with linear density µ is
For a wave speed of 180 m/s, the required tension will be
Assess: Increased wave speed must lead to increased tension, as is previously obtained. P15.5. Prepare: Two pulses of sound are detected because one pulse travels through the metal to the
microphone while the other travels through the air to the microphone. Sound travels faster through solids than
gases, so the pulse traveling through the metal will reach the microphone before the pulse traveling through the air.
We will take the speed of sound in air as 343 m/s.
Solve: The time interval for the sound pulse traveling through the air is
Because the pulses are separated in time by 11.0 ms, the pulse traveling through the metal takes Δtmetal = 0.66 ms to
travel the 4.0 m to the microphone. Thus, the speed of sound in the metal is
Assess:
We see from Table 15.1 that the speed obtained as shown is quite typical of metals.
P15.11. Prepare: We will use Equation 15.9 to find the wave speed.
Solve:
The wave speed is
P15.15. Prepare: The wave is a traveling wave. A comparison of the given wave equation with Equation 15.8
yields A = 5.2 cm,
= 5.5 rad/m, and
Solve: (a) The frequency is
to two significant figures.
(b) The wavelength is
= 72 rad/s.
to two significant figures.
(c) The wave speed
to two significant figures.
P15.24. Prepare: Light is an electromagnetic wave that travels with a speed of 3 × 108 m/s. The frequencies of
the blue and red light are 450 nm and 650 nm, respectively.
Solve: (a) The frequency of the blue light is
(b) The frequency of the red light is
Assess: A higher wavelength for the red compared to the blue light means that the frequency for the red light is
smaller than the blue light.
P15.28. Prepare: Follow Example 15.12. The intensity at the stage is
We do not know the power of the source unless we assume a distance for the given 110 dB data; so let’s assume
1.0 m for the distance from the speaker to the edge of the stage. Equation 15.13 shows that the intensity at 30 m
will be
times less if we assume that the wave is spherical.
Therefore the new intensity is
The power (or energy/time) is the intensity multiplied by the area.
We can deduce from the information given that the area of the eardrum is
Solve:
So the eardrum receives
each second during the concert.
Assess: This doesn’t seem like much energy, but prolonged exposure to loud sounds like this can damage your
hearing.
P15.31. Prepare: If the solar panel produces electrical power with efficiency
the sun is
then the power received from
This is because if we multiply the power from the sun by
we should get
If the area of the solar panel is
then the intensity of radiation from the sun is
To find the intensity at some other distance from the sun, we use Equation 15.13,
Solve: Using the previous formula, we can find the intensity of sunlight at the location of Saturn’s orbit.
Then using
we find that
that is the solar panel would receive
of power from
the sun. Finally, since the panels work with efficiency
the power which would be produced if the spacecraft
were in orbit around Saturn is
Assess: The reduction from
to
corresponds to a factor of about 100. This is reasonable since
intensity is inversely proportional to distance squared. At Saturn’s orbit, the satellite is about 10 times farther
away from the sun so the intensity of light it receives is reduced by a factor of about 100.
P15.33. Prepare: To find the power of a laser pulse, we need the energy it contains,
and the time duration
of the pulse,
Then to find the intensity, we need the area of the pulse. Its radius is
Solve: (a) Using
we find the following:
(b) Then from
we obtain
Assess: This is very intense light. Using the data from Problem 15.32, the laser light is about 400 million times
as intense as the energy from the sun.
P15.35. Prepare: Table 15.3 tells us that the intensity of a whisper at one meter is
use Equation 15.12 and ratios to find what it would be twice as far away,
Solve:
The sound intensity level is given by Equation 15.14 where
We’ll
Assess: Table 15.3 gives the sound intensity level for a whisper at 1 m as 20 dB; we expect it to be less at 2 m,
and 14 dB is just about what we expect.
P15.38. Solve: (a) The intensity of a uniform spherical source of power
a distance r away is
Thus, the intensity at the position of the microphone is
(b) The sound intensity may be determined from Equation 15.14 as follows:
Using
Assess: This is quite loud, as you may expect (compare with values in Table 15.3).
P15.40. Prepare: We will use Equation 15.14 to relate the intensity level to the intensity.
Solve: If we solve Equation 15.14 for I, we have:
Now plugging in
for
we get
and plugging in
for
we get
The ratio of the latter to the former is
Assess: This intensity ratio of
which represents the smallest increase perceptible to our ears, is much greater
than the minimum frequency ratio we can hear. For example, the ratio of the frequency of the musical note
middle C# to the frequency of middle C is only 1.06 and this is not even the smallest frequency ratio which can
be heard.
P15.42. Prepare: This is a Doppler effect problem, so we refer to Equation 15.16.
We are given f0 = 2200 Hz, f = 2300 Hz, and we use v = 343 m/s. We want to know vs.
Solve: First solve for vs and then plug in the numbers.
+
Assess: An examination of our final equation shows that the closer f0 and f are to each other, the smaller vs is,
as we would expect.
+
P15.47. Prepare: If the carpenter hits the nail twice per second then the frequency is 2 Hz and the period is
0.50 s. However, it takes half a period (0.25 s) for the hammer to go from the nail to the upraised position—and
this same half a period for the sound to get to you at 343 m/s, the speed of sound in 20°C air.
Solve:
Assess: This calculation was for the minimum distance; since the carpenter continues regular pounding, it could
also have taken 3/2 or 5/2, etc., of a period for the sound to reach you. However the larger distances might make
it hard to see the carpenter’s swings.
P15.56. Prepare: Since the problem mentions maximum acceleration, we will use Equation 14.17,
We are also given the speed and frequency of the wave from which we can find the wavelength
using
Solve: (a) To find the time between the earthquake and first detection we simply divide the distance traveled by
the speed of the wave,
(b) This was a longitudinal wave since it traveled from east to west and caused the ground to vibrate in an east–
west direction.
(c) The wavelength was
(d) The maximum horizontal displacement of the ground, that is,
for A:
can be obtained by solving Equation 14.17
Assess: It is interesting to note that even though the wave traveled incredibly fast—
is about 16,000 mph,
the ground moved much slower. The maximum speed of the ground, using Equation 14.15, is given by the following:
which is a snail’s pace.
P15.58. Solve: The difference in the arrival times for the P and S waves is
This will be reported as 1200 km to two significant figures.
Assess: d is approximately one-fifth of the radius of the earth and is reasonable.
P15.66. Prepare: Knowing the speed and wavelength of the wave (periodic disturbance) moving along the
string, we can determine the frequency of the disturbance. This frequency tells us the rate at which a point on the
string is moving up and down (i.e., its linear frequency). Knowing the linear frequency, we can determine the
angular frequency which can be combined with the amplitude information to determine the maximum
translational speed of any point on the string.
Solve: The frequency of the disturbance and hence the linear frequency of the oscillating particle on the string
is determined by
The maximum transverse speed of any point on the string is determined by
Assess: It is important to distinguish between the speed of the wave along the string and the transverse speed of
a point on the string.
P15.76. Prepare: The bat’s chirping frequency is altered by the Doppler effect. The frequency is increased as
the bat approaches and it decreases as the bat recedes.
Solve: The bat must fly away from you, so that the chirp frequency observed by you is less than 25 kHz. From
Equations 15.16,
Assess:
This is a rather large speed, 100 m/s ≈ 230 mph. This is not possible for a bat.
Properties of Ocean Waves
Description: Calculate the speed and amplitude of ocean waves.
A fisherman notices that his boat is moving up and down periodically without any horizontal motion,
owing to waves on the surface of the water. It takes a time of 2.50
highest point to its lowest, a total distance of 0.650
spaced a horizontal distance of 6.20
for the boat to travel from its
. The fisherman sees that the wave crests are
apart.
Part A
How fast are the waves traveling?
Hint
A.1
How to approach the problem
Calculate the period of the ocean waves, using the fisherman's observations. Then, use the period
and wavelength to calculate the speed of the waves.
Hint
A.2
Calculate the period of the waves
Calculate the period
Hint
A.2.1
of the ocean waves.
Definition of period
The period of a wave is the time it takes for one full wavelength to pass a particular point. This is
also the time it takes to go from one crest to the next, or from one trough to the next.
Express your answer in seconds using three significant figures.
ANSWER:
=
Hint
A.3
Equation for the speed of a wave
The speed of a wave is given by
= 6.20
, where
is the frequency of the waves and
is the wavelength. The frequency is simply the reciprocal of the period, or
.
Express the speed
in meters per second using three significant figures.
ANSWER:
=
.
Express the speed
in meters per second using three significant figures.
ANSWER:
=
Part B
What is the amplitude
Hint
B.1
of each wave?
Definition of amplitude
The amplitude of a wave is the vertical distance from the top of the crest to the neutral position,
halfway between the crest and trough. Equivalently, the amplitude is the vertical distance from the
bottom of the trough to the neutral position.
Express your answer in meters using three significant figures.
ANSWER:
=
The fisherman does not simply move up and down as the waves pass by. In fact, the motion of the
fisherman will be roughly circular with both upward and forward components (with respect to the
direction of the wave) as the wave rises and downward and backward components as the wave falls.
The water that comprises the ocean wave itself moves in this same way. Thus, an ocean wave is not a
purely transverse wave; it also has a longitudinal component.
± Find the Wavelength
Description: ± Includes Math Remediation. Calculate the wavelengths of two different EM waves in
both meters and nanometers.
Assume the following waves are propagating in air.
Part A
Calculate the wavelength
Hint
A.1
for gamma rays of frequency
How to set up the problem
Recall the formula
.
Express your answer in meters.
ANSWER:
=
= 5.50×1021
.
ANSWER:
=
Part B
Now express this gamma-ray wavelength in nanometers.
Hint
B.1
Relation between meters and nanometers
.
Express your answer in nanometers.
ANSWER:
=
Part C
Calculate the wavelength
Hint
C.1
for visible light of frequency
How to set up the problem
Recall the formula
.
Express your answer in meters.
ANSWER:
=
Part D
Now express this visible wavelength in nanometers.
Hint
D.1
Relation between meters and nanometers
.
Express your answer in nanometers.
ANSWER:
=
= 6.30×1014
.
ANSWER:
=
± Musical Logs
Description: ± Includes Math Remediation. Simple questions using the properties of logarithms in
sound problems.
Learning Goal: To learn the properties of logarithms and how to manipulate them when solving sound
problems.
The intensity of sound is the power of the sound waves divided by the area on which they are incident.
Intensity is measured in watts per square meter, or
.
The human ear can detect a remarkable range of sound intensities. The quietest sound that we can hear
has an intensity of
, and we begin to feel pain when the intensity reaches 1
. Since the intensities that matter to people in everyday life cover a range of 12 orders of magnitude,
intensities are usually converted to a logarithmic scale called the sound intensity level
measured in decibels (
). For a given sound intensity
,
, which is
is found from the equation
,
where
.
The logarithm of
if
, written
, then
, tells you the power to which you would raise 10 to get
. It is easy to take the logarithm of a number such as
can directly see what power 10 is raised to. That is,
, because you
.
Part A
What is the value of
Hint
A.1
?
What is 1,000,000 equivalent to?
Powers of 10 can be determined simply by counting the number of zeros after the one. Since
1,000,000 has six zeros after the one,
.
Express your answer as an integer.
ANSWER:
=
. So,
ANSWER:
=
Part B
If a speaker gives a sound intensity of
that point?
Hint
B.1
at a certain point, what is the sound intensity level
at
Find the number inside the logarithm
To calculate the
portion of the formula for
, you must first find the value of
. What is the value of
?
Express your answer numerically.
ANSWER:
=
Hint
B.2
Finding the logarithm
Once you know the value of
for any
. For example,
, you can find
. Recall that
,
.
Express your answer in decibels.
ANSWER:
=
One important rule for manipulating logarithms is the product rule:
.
Consider a simple example to illustrate this rule. You know from the previous discussion that
. Since
, you could rewrite the original logarithm using the
product rule:
.
Part C
The power in the speaker from Part A is doubled, bringing the sound intensity to
. This leads to
equations is equivalent to
. Which of the following
?
ANSWER:
Then, the sound intensity level
can be written as
.
Since
, we get
.
There's nothing special about
. The calculation done here can be carried out for any number. For
instance, assume you are given an initial intensity
, with sound intensity level
. If the intensity is doubled to
, the intensity level will be
, since
In summary, the product rule allows you to break down the logarithm of any number into two simpler
parts. The first step is to write the number in scientific notation. (Usually, intensities will be quoted to
you already in scientific notation.) If you need to find the logarithm of
down into the sum of two simpler logarithms:
, you can break it
.
But what is the value of the original logarithm? You know that the second logarithm is equal to 7, so the
total value is
. Since 4.36 is between 1 and 10, its logarithm will be between 0 and 1.
Thus, the logarithm of
is between 7 and 8. When you write a value in scientific
notation, you always have a number between 1 and 10 multiplied by 10 to some power. That power tells
you the integer part of the logarithm, as it did in the example of
.
Part D
Without actually calculating the logarithm, determine what two integers the value of
falls between.
ANSWER:
The value of
falls between
With this technique, you can also estimate the sound intensity level. Recall that to find
multiply the logarithm of
by 10
interval containing
. Thus, to find what interval
, and then multiply its ends by 10
you
falls within simply find the
.
Part E
Without actually calculating any logarithms, determine which of the following intervals the sound
intensity level of a sound with intensity
Hint
E.1
falls within.
Find
What is the value of
for this part?
Express your answer to three significant figures.
ANSWER:
=
Hint
E.2
Find the interval containing
In the previous part, you found that
into?
. What interval would
fall
ANSWER:
falls between
To find the sound intensity level, you simply multiply the logarithm by 10
. Do the same to
the two ends of your interval, and you will have the interval into which the sound intensity level
falls.
ANSWER:
falls between
To find the sound intensity level, you simply multiply the logarithm by 10
. Do the same to
the two ends of your interval, and you will have the interval into which the sound intensity level
falls.
ANSWER:
The sound intensity level falls between
.
You may be wondering why you should bother learning how to estimate logarithms, when your
calculator can easily find the value. Estimating is an important way to check answers. It would be
very easy to make an error in sign as you enter numbers into your calculator. Such an error could
result in you identifying a sound as being 200
(roughly the sound level if you were to stick your
head inside of a jet engine as it took off) instead of 40
(the sound level of a quiet conversation).
A calculator will almost always give you the correct answer, if you give it correct instructions.
Having a feel for how logarithms work will allow you to be more certain of answers your calculator
gives you.
In the introduction, the logarithm of
was defined as the power to which you would raise 10 to get
. This can be written as
.
You can use this formula to solve equations like
must also be true that
. Since
is equal to 4.3, it
.
This formula now allows you to replace the left side of the equation with
, giving
.
Part F
Use this technique to find a formula for the intensity
the reference intensity
. There will be a quantity (10
you enter your answer, leave off the unit
Hint
F.1
of a sound, in terms of the sound level
and
) involved in your calculations. When
.
How to approach the problem
The formula for sound intensity level relates
form that you were given,
,
, and
—just the variables you want! The
,
is similar to the type of equations solved above. In fact, if you divide both sides by 10, then you
have an expression in the same form used in the introduction, that is,.
. Next, you
,
is similar to the type of equations solved above. In fact, if you divide both sides by 10, then you
have an expression in the same form used in the introduction, that is,.
. Next, you
can use the technique above to solve for the quantity inside the logarithm, in this case
you have this, multiply both sides by
Express your answer in terms of
to isolate the intensity
and
. Once
.
.
ANSWER:
=
± The Decibel Scale
Description: ± Includes Math Remediation. Basic questions which address the logarithmic nature of
decibel scale.
Learning Goal: To understand the decibel scale.
The decibel scale is a logarithmic scale for measuring the sound intensity level. Because the decibel
scale is logarithmic, it changes by an additive constant when the intensity as measured in
changes by a multiplicative factor. The number of decibels increases by 10 for a factor of 10 increase in
intensity. The general formula for the sound intensity level, in decibels, corresponding to intensity
,
where
is a reference intensity. For sound waves,
refers to the logarithm to the base 10.
is taken to be
. Note that
Part A
What is the sound intensity level
reference intensity (i.e.,
Hint
A.1
, in decibels, of a sound wave whose intensity is 10 times the
)?
Evaluate the logarithm
If
, the argument of the log function in the definition of the dB scale is simply 10.
What is the value of
?
Express your answer numerically.
ANSWER:
0
1
is
What is the value of
?
Express your answer numerically.
ANSWER:
0
1
= 2.30
10
Express the sound intensity numerically to the nearest integer.
ANSWER:
=
Part B
What is the sound intensity level
, in decibels, of a sound wave whose intensity is 100 times the
reference intensity (i.e.
Hint
B.1
)?
Evaluate the logarithm
If
, the argument of the log function in the definition of the dB scale is 100. What is
the value of
?
Express your answer numerically.
ANSWER:
2
4.60
= 10
20
Express the sound intensity numerically to the nearest integer.
ANSWER:
=
The answers to Parts A and B demonstrate the essence of the dB scale--every 10
factor of 10 increase in the sound intensity level, so an increase in 20
increase in intensity. A graph of
vs.
indicates a
indicates a factor of
is shown below to support this idea.
increase in intensity. A graph of
vs.
is shown below to support this idea.
The ability of the ears of animals (including humans) to be sensitive to 1
animals) and yet not be permanently damaged by 120
remarkable.
(or even less for some
(a million million times the intensity) is
One often needs to compute the change in decibels corresponding to a change in the physical intensity
measured in units of power per unit area. Take
(i.e.,
to be the factor of increase of the physical intensity
).
Part C
Calculate the change in decibels (
, and
Hint
C.1
,
, and
) corresponding to
,
.
How to approach the problem
To find the increase take
in the general formula to be the initial intensity and then take
the factor of increase:
. Then,
measured in decibels is
to be
and the change in intensity
.
Give your answers, separated by commas, to the nearest integer--this will give an accuracy of 20%,
which is good enough for sound.
ANSWER:
,
,
=
ANSWER:
,
,
=