Homework Chapter 17!
!
!
17.24!
17.25!
17.29!
17.35!
17.39!
17.44!
17.48!
17.49!
!
page 1
17.24!
A sound wave from a police siren has an intensity of 100 W/m2 at a certain point. A second sound
wave from a nearby ambulance has an intensity level that is 10 dB greater than the police at the
same point. What is the sound level of the sound wave due to the ambulance?!
Solution!
The intensity level of the police siren is!
⎛ 100 W/m 2 ⎞⎟
⎛I ⎞
⎟⎟ = 140 dB !
β = 10 log ⎜⎜ ⎟⎟⎟ = 10 log ⎜⎜⎜
⎜⎝ Io ⎟⎠
⎜⎝ 10−12 W/m 2 ⎟⎠
The ambulance siren has the intensity level of 150 dB.!
!
page 2
17.25!
The power output of a certain public address speaker is 6 W. Suppose it broadcasts equally in all
directions.!
(a)!Within what distance from the speaker would the sound be painful to the ear?!
(b)!At what distance from the speaker would the sound be barely audible?!
Solution!
(a)!Being painful means an intensity level of 120 dB. This means an intensity of!
⎛I ⎞
120 dB = 10 log ⎜⎜⎜ ⎟⎟⎟
⎝ Io ⎟⎠
⇒ I = 1 W/m 2 !
The area over which the 6 W of power produces this intensity is!
I =
P
A
⇒ A=
P
6W
=
= 6 m2 !
I
1 W/m 2
The radius of a sphere that produces this area is!
A = 4πR 2
⇒ R=
A
=
4π
6 m2
= 0.691 m !
4π
(b)!Being barely audible means an intensity level of 0 dB. This means an intensity of 10-12 W/m2.
The area of this intensity is!
I =
P
A
⇒ A=
P
6W
=
= 6 ×1012 m 2 !
−12
I
10
W/m 2
The radius of a sphere that produces this area is!
A = 4πR 2
!
⇒ R=
A
=
4π
6 ×1012 m 2
= 0.691×106 m = 691 km !
4π
page 3
17.29!
A family ice show is held at an enclosed arena. The skaters perform to music with level 80 dB.
This level is too loud for your baby who yells at 75 dB.!
(a)!What total sound intensity engulfs you?!
(b)!What is the combined sound level?!
Solution!
(a)!The sound intensity due to the 80 dB sound is!
⎛I ⎞
80 dB = 10 log ⎜⎜⎜ 80 ⎟⎟⎟
⎝ Io ⎟⎠
⇒ I 80 = 10−4 W/m 2 !
The sound intensity due to the 75 dB sound is!
⎛I ⎞
75 dB = 10 log ⎜⎜⎜ 75 ⎟⎟⎟
⎝ Io ⎟⎠
⇒ I 75 = 10−4.5 W/m 2 !
The total intensity is!
I = 10−4 W/m 2 + 10−4.5 W/m 2 = 1.3162 ×10−4 W/m 2 !
(b)!The combined intensity level is!
!
⎛ 1.3162 ×10−4 W/m 2 ⎞⎟
⎛I ⎞
⎟⎟ = 81.193 dB !
β = 10 log ⎜⎜ ⎟⎟⎟ = 10 log ⎜⎜⎜
⎟⎠
⎜⎝ Io ⎟⎠
⎜⎝
10−12 W/m 2
page 4
17.35!
A driver travels northbound on a highway at a speed of 25 m/s. A police car, driving southbound,
at a speed of 40 m/s approaches with its siren producing sound at a frequency of 2500 Hz. !
(a)!What frequency does the driver observe as the police car approaches?!
(b)!What frequency does the driver detect after the police car passes him?!
(c)!Repeat parts a and b for the case when the police car is behind the driver and traveling
northbound.!
Solution!
(a)!Here is the diagram.!
driver
north @ 25 m/s
south @ 40 m/s
police
2500 Hz
!
The driver sees an approaching source and he is also an approaching detector. The detected
frequency will be!
f '' =
vs
vs + vdriver
v + vdriver
343 + 25
f = s
f =
2500 Hz = 3036.3 Hz !
vs − v police
vs
vs − v police
343 − 40
(b)!Here is the diagram.!
south @ 40 m/s
police
driver
north @ 25 m/s
2500 Hz
!
The driver sees a receding source and he is also a receding detector. The detected frequency will
be!
f '' =
vs
vs − vdriver
v − vdriver
343 − 25
f = s
f =
2500 Hz = 2075.7 Hz !
vs + v police
vs
vs + v police
343 + 40
(c)!Here is the diagram.!
police
north @ 40 m/s
driver
north @ 25 m/s
2500 Hz
!
The driver sees an approaching source and he is a receding detector. The detected frequency will
be!
f '' =
!
vs
vs − vdriver
v − vdriver
343 − 25
f = s
f =
2500 Hz = 2623.8 Hz !
vs − v police
vs
vs − v police
343 − 40
page 5
(d)!Here is the diagram.!
driver
north @ 25 m/s
police
north @ 40 m/s
2500 Hz
!
The driver sees a receding source and he is an approaching detector. The detected frequency will
be!
f '' =
!
vs
vs + vdriver
v + vdriver
343 + 25
f = s
f =
2500 Hz = 2402.1 Hz !
vs + v police
vs
vs + v police
343 + 40
page 6
17.39!
A block with a speaker bolted to it is connected to a spring having spring constant k = 20 N/m and
it oscillates left to right. The total mass of the block and speaker is 5 kg and the amplitude of this
unit motion is 0.5 m. The speaker in the midst sound waves of frequency 440 Hz.!
(a)!Determine the highest frequencies heard by the person to the right of the speaker.!
(b)!Determine the lowest frequencies heard by the person to the right of the speaker.!
(c)!The maximum sound level heard by the person is 60 dB when the person is at it's closest
distance d = 1 m from him, what is the minimum sound level heard by the observer?!
Solution!
(a)!The highest frequency is generated through the following pattern.!
speaker
vmax
person
440 Hz
!
The speed vmax is the maximum speed of the oscillator and it is given by!
v(t) = −ωA cos(ωt) ⇒ v max = ωA !
The angular frequency of the oscillator is!
ω=
k
=
m
20 N/m
= 2 1/s !
5 kg
v max = (2 1/s)(0.5 m) = 1 m/s !
The source is approaching the detector so!
f'=
vs
343
f =
440 Hz = 441.29 Hz !
vs − v police
343 − 1
(b)!The lowest frequency is generated when the speaker is receding at the same speed.!
f'=
vs
343
f =
440 Hz = 438.72 Hz !
vs + v police
343 + 1
(c)!At 1 m, the intensity level is 60 dB. Assuming the sound spreads into a sphere, doubling the
distance to 2 m increases the area to 4 time the original area. This decreases the intensity by
4 and decreases the intensity level by 6 dB. The resulting intensity level would be 54 dB.!
!
page 7
17.44!
A tuning fork vibrating at 512 Hz falls from rest and accelerate at 9.8 m/s2. How far below the point
of release is the tuning fork when waves of frequency 485 Hz reach the release point?!
Solution!
Given the original and doppler shifted frequencies, the receding source’s speed was!
f'=
vs
f
vs + v police
⇒
485 Hz =
343
512 Hz ⇒ v = 19.095 m/s !
343 + v
Falling to this speed requires a distance of !
v f2 − vi2 = 2a(x f − xi ) ⇒ 19.0952 = 2(9.8)Δy
⇒ Δy = 18.603 m !
At this distance is when the sound is emitted. At the speed of sound, it take another !
t=
∆y
19.095 m
=
= 0.055671 s !
v
343 m/s
to reach back to the top. After this much more time, the tuning fork falls another!
1
x f − (−18.603 m) = (−19.095 m/s)(0.055671 s) + (−9.8 m/s 2 )(0.055671)2
2
⇒ x f = −19.681 m !
The tuning fork has fallen 19.681 m.!
!
page 8
17.48!
If a salesman claims he loudspeaker is rated at 150 W, he is referring to the maximum electrical
power input to the speaker. Assume a loudspeaker with an input power of 150 W broadcasts
equally in all directions and produces sound with the level of 103 dB at a distance of 1.6 m from
the center.!
(a)!Find its sound output.!
(b)!Find the efficiency of the speaker.!
Solution!
(a)!The sound power is!
⎛I ⎞
103 dB = 10 log ⎜⎜⎜ ⎟⎟⎟
⎝ Io ⎟⎠
⇒ 1010.3 =
P /A
−12
10
⇒ P = 4π(1.6 m)210−1.7 = 0.64187 W !
(b)!The efficiency of the speaker is!
e=
!
0.64187 W
= 4.2792 ×10−3 = 0.42792% !
150 W
page 9
17.49!
Interstate highway has been built through a neighborhood in the city. In the afternoon, the sound
level in an apartment in the neighborhood is 80 dB as 100 cars pass outside the window every
minute. Late at night, the traffic flow is only five cars per minute. What is the average late-night
sound level?!
Solution!
Changing the amount of source of the sound means changing the power and the intensity of the
sound. The amount of change here a decrease by a factor of 20. This is a decrease by the
factors of 2 and 10. A decrease by a factor of 10 decrease the intensity level by 10 dB. A
decrease by a factor of 2 decrease the intensity level by 3 dB. The final intensity level would be
67 dB.
page 10