16. Sound and Light Waves
16.18 Since sound wave amplitude increases with piston displacement, low-frequency speakers need
bigger displacements for comparable power output. Mechanical design requires a bigger system
to produce the greater displacements.
b) Wavelength:
closed at one end
16.19 The wave corresponding to the fundamental
frequency is sketched hi each tube.
open organ pipe
AI =4L
Frequency:
/i = -2- = ~ .\ L = —~
AI
4L
4/1
This is half the answer of part a): L — 2.6 m
close one end
16.24 /i = 3.1 kHz
16.25 a) Pressure amplitude P* is given by:
p*
_
^SQ
=
In the open pipe case:
Ai =2L
closed pipe case:
=
! = 41
So, since the wavelength is inversely proportional to frequency, it follows that the frequency
is cut in half when the pipe is closed.
3
0 p* = 6.3 mg/m ; v0 = 1.7 mm/s; s0 = 53 nm
\
16.21/At 0° C the speed of sound is 330 m / s. Speed
of the jet craft,
_
o f
P*[povs2irfrl
10.2 Pa_
3
(1.3 kg / m ) (330 m / s) 27r (103 Hz)
3.8 x 10~6 m
b)
so = (3.8 x 10-6 m) (1""OHz) = 76 x 10~6 m
50 Hz
16.27 The chamber must have a displacement node
at each end.
Vacuum chamber cylinder
F mdamental
f i km '
«jet
- (0.82) (330 m/s)
1000 m
1
"
L = 3.5 m
"
In general the wavelength of a standing wave is:
= 970^
h
An = 2L \
16.22 L = 1.3 m; 196 Hz; 327 Hz; 458 Hz
X _ 2L
AS - -33
16.23 a) Wavelength is AI = 2L.
n
r
\ —
—- L
A4
f-
open organ pipe
Frequency is related to wavelength and the
speed of sound vs by:
\ f
Frequency:
1L
2(32.7 Hz)
vs _nvs
if vs = 340 m/s, then the fundamental frequency is
Thus L is:
2/i
e
An/n - V. -> /„ - — - —
VIA
— 5,2 in
340 m / s
2(3.5 m)
The next harmonic has A = f £ and
/3 = 5/0 = 5 (65.4 Hz) = 327 Hz
The third harmonic has frequency:
. /4 = 7/0 = 7(65.4 Hz) =45 8 Hz
16.24 The ear canal acts like a closed pipe. Thus the fundamental frequency is related
to the length by
v
'
u
»
340 m/s
damental frequency is about 3 kHz in this model.
*f Q_16Jj6^The pipe is "open at one end", so it is presumably closed at the other end. Thus
in the fundamental mode, A = 4£ and /„ = vs/X, so
2
v, = \f = 4Lf = 4 (25 x 10~ m) (340 Hz) = 340 m/s
Using the given formula:
So:
Thus
T = (546°C) (1.0303 -1) = 16.5°C
But we only have 2 sig fig, so T = 17°C.
16.28 The refractive index is n — c/v, so
c 3.0 x 108 m/s
= 1.25x 10s m/s
2.4
The speed of light in diamond is 1.3 x 10s m/s.
16.30 In 1 y light travels a distance:
d = ct = (3.00 x 10a m/s) (365.25 d)
*
3
= 9.47 x 1015 m
So 1 light year equals 9.47 xlO15 m.
Tf-
Cjl6.32/rhe radio communications between astronauts and mission control travelled at the
speecTof light In each communcation, signals had to travel twice the Earth/Moon distance.
Thus
d
4xlO«m _ 8 ^ _ 2 4 2
- C-23TIo5^-3S~2-'S
The time delay experienced by the astronauts was 3 s.
1634 Let's assume that gaps and teeth have equal width. If the wheel rotates so that the
315
16. Sound and Light Waves
where air density po and the speed of sound vs
are given typical values:
L6.40 /visible = 0.066 W/m2
16.41 Spherical loudspeaker
po
1.3kg
=
m
vs = 330 m / s
SILD
= 10 dB Iog10
+20dBlog 1 0 (^
SILD
=
Now
radius.
R= 10.0 cm = .100 m
us0R = (27r)(3.0 x 102 Hz)
x(5.0 x 10~5 m)(.10 m)
frequency,
/ = 3.0x 102Hz
= 9.43 xlO 3 —
s
amplitude,
s0 = .050 mm = 5.0 x 10~5 m
So
SILD
SIL is given by:
= 10dBlog 10
I.
2(iO-ia -2
= 53 dB
Where the reference intensity is
Yes, this is audible.
/ref = 10-12 W/m2
16.42 5.6 m
The difference in SIL at the speaker
and at distance D(SILo) is:
SILD - SILR
"^^x
16.43/The explosion releases 1.0 x 107 J in 1 second,
so the power of the explosion is:
= I
-10dBlog10
10dBlog10
According to the inverse square law, intensity is
inversely proportional to distance squared.
of the energy is energy is converted to
lund waves (the other half presumably is converted to light and mechanical deformation).
The sound waves have power:
Psound
=
'=
/ oc -s
5.0 x 10 W
Assuming spherical wave fronts, the intensity is
Thus,
,-
20dBlog 10 (\
At the speaker, intensity is related to frequency
and displacement amplitude, so by Eqn. 16.16:
J = -p0vs(us0)2
(.5)(1.0X107W)
6
Jsound
4-Trr2
where r is the distance at which we wish to fine
me intensity, r = no iu
5.0 x 106 W
47r(110 m)2
= 33 W/m2
I =
116
Sound intensity in decibels is the SIL:
SIL
10dBl 0glo ( 10 _ 12 7 w/m2 )
-
Lea & Burke
Physics: The Nature of Things
16.46
= 1.5 x
16.47 SLL is given by:
32.88
= 10dBloglo —
,
-'ref
=
140 dB
where reference intensity
L6.44 130 dB; 130 Pa; 2.5 x 10~4 m; 500 W
16.45 Signal withfrequency/ = 250 Hz.
area A
Intensity is proportional to inverse square of distance:
I oc -=•2
r
The difference in two sound intensity levels
Tube of diameter
10 cm. = d
filled with
air
Amplifier
5.0 W of power
A5IL -
Intensity of the sound wave generated: The
power incident on any cross sectional area of
the tube, A, is P = 5.0 W. Cross-sectional area
is
Thus intensity is
=
636.6
5.0 W
f (.1 m)2
W
If r2 = 2r1; then
W
= 640-^
2
1= P ,
AS7L = 20 dBlog10 ( i ) = -6 dB
16.48 20%
16.49 The speed of light is the same in all frames
•P.= ^2Ip0vs
Using air density po = 1.2 kg / m3 and speed of
sound va = 340 m / s
= N/2(637W/m2)(1.2 kg/m3)(340 m/s)
= 720 Pa
Intensity is related to displacement amplitude
so by:
1
c = 3 x 108 m / s
16.50 If light is blue-shifted, then the object is approaching. If light is redshifted, then the object
is receding. Jupiter is spinning with respect tc
Earth. The blueshifted side is spinning towards
Earth.
16.51
T = period
'^Wvr 2 / 2 ^
'
637 W/m 2
2(1.2 kg/m3)(340 m / s )
=
Vrrf
In terms of distance:
Displacement amplitude and pressure amplitude
of wave: Intensity is related to pressure amplitude P, by:
P.
V-'ref/
-(I)'
A
I
- 10dBlog10 (A") _ 10dBloglo (£-
1.1 mm
Above the threshold of pain.
vwT
If the man with whistle bicycles toward the friend at 10.0 m/s, then we have a source
with velocity 3 m/s (in the air frame) toward the observer, and the observer has a velocity
-7 m/s in the air frame. Thus the Doppler shifted frequency is (equation 16.22a):
1 + 7/330
/ = (2.00 kHz)
= (2.00 kHz) 1.0306 = 2.06 kHz
1 - 3/330
If we use the approximate formula 16.22b, we get
/ = (2.00 kHz) l -
= (2.00 kHz)
= 2. 06 kHz
which is the same result to 3 sig fig.
f 16.50 Following the method of Example 16.9:
1+
Thus:
(3.8T7)2
=
3.8T72 - 1
= 0.87524
3.87T2 +1
c
Thus:
v, = 0.87524 (2.998 x 10s m/s) = 2.624 x 10s m/s
16.58 U&ve leaving A at time t = 0reaches0 after a time Ai = r/uw = tt
leaving B at time t = T travels a shorter distance rj and arrives at time ta where:
Since s = veT -C r, we may expand the square root:
= ri/l - 2- cos 5 + f-) 2 ~ r f l - -cos d
V
r
XT-/
\ r
Thus:
A
The time interval between the arrival times of waves from A and B is:
r f
a
\ r
s
veT
tg—t^ = TH
11
cost)}
= T
cos9 = T
cos9 ••
vw \ r
I vw
vw
vw
S
v. • vu
The observed frequency is (time between passage of wavefronts)"1, so
/=-
:
-
/0
equation
16.6p^Mfe divide the problem into two parts: outgoing sound and returning sound.
B
/.- "
/ _^ JZ
&-?/
Outgoing: ^fe have bom a moving source (the torpedo) and a moving observer (the
submarine), so we use equation 16.22. The speed of sound in water is 1531 m/s (Table
16.1), much greater than the speed of either source or observer, so let's use 16.22b. Thus:
(16-32) m/sN
/
16 \
1531 m/s )~ \ ' 1531 /
Returning: The reflected pulse is Doppier shifted again. Both source and observer are
moving opposite the direction that the wave is moving. Now the submarine is the source
and the torpedo is the observer. Thus:
Thus the frequency received by me torpedo is:
2
- 12-3kHz
16.62, 64, 66 See solutions manual
16.68 When light is incident at the critical angle, the angle of refraction is 90°. Thus
SnelFs law becomes:
m this case, HI — 1.31 (ice) and n2 = 1.00 (air), so:
am*.. IS .ig.. 0.78338
HI l.ol
Thus
0e = sin-1 0.76336 = 0. 8685 rad = 49. 8°
From Example 16.13, 9a = 41° for glass. So glass is a better material for light pipes since
more of the light is totally internally reflected.
16.70 At the first side, ni = 1 (air) and n-i = n (glass), so the angle of refraction <p is
given by:
sin# = nsin<£
(1)
At the far side of the slab, n\ = n and n2 = 1 . The angle of incidence is <b (see diagram)
so
n3in(p = sin0/
(2)
Comparing equations 1 and 2, we see that 9f = 9. Thus the light emerges at the same
angle that it enters. However it has been displaced, and emerges from the slab at point 3.
The undisplaced ray intersects the side of the slab at point C. We want to find the distance
T
cL
Jt
ft