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Unformatted text preview: Manage this Assignment: Print Version with Answers Homework # 12
Due: 10:00pm on Thursday, December 10, 2009
Note: You will receive no credit for late submissions. To learn more, read your instructor's Grading Policy Ear Damage from a Small Firecracker
Description: Find the sound intensity level one meter away from a ﬁrecracker. Then determine at what distance the sound of the ﬁrecracker would rupture be intense enough to rupture an eardrum. Learning Goal: To understand how to convert between different sound intensity scales and how the decibel intensity of a sound changes with distance. The decibel scale is logarithmic in intensity: . In this formula, is a reference intensity, which, for sound waves, is taken to be . This constant must be used to convert a particular physical intensity into a sound intensity level measured in decibels. Once we know the sound intensity level (in decibels) at a certain reference distance from a sound source, the decrease of intensity with distance can be accounted for by subtracting the decibel value appropriate to the ratio of the new distance to the reference distance. In this problem you will use the decibel scale to analyze a small ﬁrecracker that emits 1200 intensities denoted by Part A What is the peak intensity Hint A.1 Find at 1 m in watts per meter squared of the sound from the ﬁrecracker at a distance of 1 . in decibels at a distance of 1 m from the ﬁrecracker? are in units of watts per meter squared; intensities denoted by of peak power. To avoid confusion, are in units of decibels. Find the peak intensity Hint A.1.1 Start with geometry All of the power from the ﬁrecracker has to pass through a sphere 1 in radius. What is the area of this sphere? Express your answer numerically, in square meters, to two decimal places. ANSWER: = Express your answer numerically to one decimal place. ANSWER: = Hint A.2 Converting intensity to dB into decibels. Note that the reference intensity is . Use the formula in the introduction to convert Express in decibels to the nearest integer. 1 of 8 12/12/09 3:47 AM ANSWER: = dB Part B It takes a sound intensity of about 160 dB to rupture the human eardrum. How close must the ﬁrecracker described in the introduction be to the ear to rupture the eardrum? Hint B.1 How to approach the problem This problem can be worked from ﬁrst principles. To do so, ﬁgure out the intensity in watts per meters squared corresponding to 160 , then divide this quantity into the 1200W peak power of the ﬁrecracker. The result will be the surface of a sphere, centered on the ﬁrecracker, whose radius is the distance at which the sound intensity is 160 . of intensity An easier way to solve the problem is to ﬁgure out by what factor the distance must be changed to give another 20 above the 140 Hint B.2 just calculated for a distance of 1 . Method 1: Find the intensity corresponding to this decibel level ? What intensity corresponds to a decibel level of 160 Express your answer numerically, in watts per meter squared, to three signiﬁcant ﬁgures. ANSWER: = Now ﬁnd the radius at which the sound has this intensity for the given power. Hint B.3 Method 2: Convert change in intensity to change in distance must the distance between a sound source and a listener change so that the sound intensity level at the position of ? By what factor the listener changes by 20 Hint B.3.1 Calculate the dB increase due to changing distance by a factor If you change the distance between a sound source and a listener by a factor sound intensity level change? Express the change in dB (call it increase (decrease) in the intensity. ANSWER: = dB ) in terms of , noting that a positive (negative) change in dB corresponds to an (i.e., ) by how many decibels will the Express the factor numerically to one decimal place. ANSWER: = This result implies that at one tenth of the original distance to a sound source, the sound intensity level will have increased by 20 compared to the sound intensity level at the original distance. Use this conclusion and the result of Part A to ﬁnd the distance from the ﬁrecracker at which the sound intensity level is 140 Express the distance , in meters, to one decimal place. . 2 of 8 12/12/09 3:47 AM ANSWER: = m This result might seem to suggest that small ﬁrecrackers (e.g., "ladyﬁngers") are not particularly dangerous; however, permanent hearing loss will occur at several times this distance, or from repeated exposure above 130 . People who are at a genetic risk to develop tinitusa permanent ringing in the earshave an even lower threshold, as do those who have already developed this afﬂiction. Part C Will this ﬁrecracker produce temporary loss of hearing in someone who sets if off and stands 3 m away from the explosion? Momentary sounds above 120 produce such loss. ANSWER: yes no Interference of Sound Waves
Description: Given the distance from a loudspeaker, ﬁnd the minimum distance from a second speaker needed to experience destructive interference. Two loudspeakers, A and B, are driven by the same ampliﬁer and emit sinusoidal waves in phase. The frequency of the waves emitted by each speaker is 172 . You are 8.00 from speaker A. Take the speed of sound in air to be 344 . Part A What is the closest you can be to speaker B and be at a point of destructive interference? Hint A.1 How to approach the problem Destructive interference occurs when the difference in path lengths traveled by sound waves is a halfinteger number of wavelengths. Therefore, to apply the condition for destructive interference you need to know the wavelength of the sound, which can be easily determined given its frequency. Hint A.2 Find the wavelength of the sound wave of the sound wave emitted by the loudspeakers. Take the speed of sound in air to be 344 . Find the wavelength Hint A.2.1 Relationship between the wavelength and frequency of a periodic wave For a periodic wave the speed of propagation equals the product of the wavelength . Express you answer in meters. ANSWER: = and frequency of the wave. In symbols, Hint A.3 In general, if Find the condition for destructive interference and are the paths traveled by two waves of equal frequency that are originally emitted in phase, the condition for destructive interference is 3 of 8 12/12/09 3:47 AM where is the wavelength of the sound waves and is any nonzero odd integer. Given this condition for destructive interference and the situation described in the introduction of this problem, then, what is the value of that corresponds to the shortest distance ? ANSWER: = Express your answer in meters. ANSWER: Open Organ Pipe Conceptual Question
Description: A conceptual question in which the fundamental frequency of an organ pipe of a certain length is given. Questions deal with the new fundamental frequency after the organ pipe is cut in half, closed off, ﬁlled with helium. Also asks to compare closed and open pipes. An open organ pipe (i.e., a pipe open at both ends) of length Part A If the organ pipe is cut in half, what is the new fundamental frequency? Hint A.1 Fundamental wavelength in an open pipe has a fundamental frequency . For a wave in an open pipe, the fundamental wavelength is the longest wave that "ﬁts" in the pipe, with an antinode (point of maximum wave amplitude) at both open ends. Hint A.2 Fundamental frequency , wavelength , and frequency , , so the frequency of a wave is given by . Thus, the fundamental frequency and the fundamental wavelength are inversely proportional. ANSWER: For speed of sound Part B 4 of 8 12/12/09 3:47 AM After being cut in half in Part A, the organ pipe is closed off at one end. What is the new fundamental frequency? Hint B.1 Fundamental wavelength in a closed pipe For a wave in a closed pipe, the fundamental wavelength is the longest wave that "ﬁts" in the pipe, with an antinode (point of maximum wave amplitude) at the open end and a node (point of zero wave amplitude) at the closed end. ANSWER: The fundamental frequency of a halflength closed pipe is equal to that of a fulllength open pipe. Part C The air from the pipe in Part B (i.e., the original pipe after being cut in half and closed off at one end) is replaced with helium. (The speed of sound in helium is about three times faster than in air.). What is the approximate new fundamental frequency? Hint C.1 The role of the speed of sound , frequency , and wavelength , In an organ pipe, the wave speed is the speed of sound. For speed of sound . Thus, the frequency of a wave is directly proportional to the wave speed. ANSWER: Based on your answer to this part, can you explain why you have a high pitched voice after inhaling the helium from a helium balloon? Problem 16.8
Description: A person, with his ear to the ground, sees a huge stone strike the concrete pavement. A moment later two sounds are P the concrete, and heard from the impact: one travels in the air and the other in roblem 16.18 they are t apart. (a) How far away did the impact... A person, with Youear to the ground, sees a hugetwo new stereo ampliﬁers. One is ratedmomentper channel and the other is from theP_2 Description: his are trying to decide between stone strike the concrete pavement. A at P_1 later two sounds are heard rated at impact: one travels in the air dB, howother in the concrete, and they are 0.54 apart. be when both are producing sound at their... per channel. (a) In terms of and the much louder will the more powerful ampliﬁer You are trying to decide between two new stereo ampliﬁers. One is rated at 110 Part A channel. How far away did the impact occur? The speed of sound in the concrete is 3000 per channel and the other is rated at 180 . per Part A Express your answer using two signiﬁcant ﬁgures. In terms of , how much louder will the more powerful ampliﬁer be when both are producing sound at their maximum levels? ANSWER: Express your answer using two signiﬁcant ﬁgures. = 5 of 8 12/12/09 3:47 AM CHAPTER 16: Sound
8.
The
two
sound
waves
travel
the
same
distance.
The
sound
will
travel
faster
in
the
concrete,
and
thus
take
a
shorter
time.
The
speed
of
sound
in
concrete
is
obtained
from
Table
16‐1
as
3000
m/s.
Compare
the
two
power
output
ratings
using
the
definition
of
decibels.
This
would
barely
be
perceptible.
18.
36.
(a )
The
length
of
the
tube
is
one‐fourth
of
a
wavelength
for
this
(one
end
closed)
tube,
and
so
the
wavelength
is
four
times
the
length
of
the
tube.
(b )
If
the
bottle
is
one‐third
full,
then
the
effective
length
of
the
air
column
is
reduced
to
14
cm.
41.
The
speed
of
sound
will
change
as
the
temperature
changes,
and
that
will
change
the
frequency
of
the
organ.
Assume
that
the
length
of
the
pipe
(and
thus
the
resonant
wavelength)
does
not
change.
50.
To
operate
with
the
first
harmonic,
we
see
from
the
figure
that
the
thickness
must
be
half
of
a
wavelength,
so
the
wavelength
is
twice
the
thickness.
The
speed
of
sound
in
the
quartz
is
given
by
analogous
to
Eqs.
15‐3
and
15‐4.
57.
Beats
will
be
heard
because
the
difference
in
the
speed
of
sound
for
the
two
flutes
will
result
in
two
different
frequencies.
60.
(a )
To
find
the
beat
frequency,
calculate
the
frequency
of
each
sound,
and
then
subtract
the
two
frequencies.
(b )
The
speed
of
sound
is
343
m/s,
and
the
beat
frequency
is
3.821
Hz.
The
regions
of
maximum
intensity
are
one
“beat
wavelength”
apart.
62.
The
moving
object
can
be
treated
as
a
moving
“observer”
for
calculating
the
frequency
it
receives
and
reflects.
The
bat
(the
source)
is
stationary.
,
and
the
Then
the
object
can
be
treated
as
a
moving
source
emitting
the
frequency
bat
as
a
stationary
observer.
66.
The
wall
can
be
treated
as
a
stationary
“observer”
for
calculating
the
frequency
it
receives.
The
bat
is
flying
toward
the
wall.
Then
the
wall
can
be
treated
as
a
stationary
source
emitting
the
frequency
bat
as
a
moving
observer,
flying
toward
the
wall.
,
and
the
...
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This note was uploaded on 04/03/2010 for the course PHYSICS 100 taught by Professor Physics during the Spring '10 term at DeVry Bellevue.
 Spring '10
 physics
 Physics, Work

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