16
--,.------. ,.---------
Section 16.1
_'.
__ ._-~-- ------
-.- ._----- -- ,.--------
,--..
..•..•
Propagation of a Disturbance
Replace x by
P16.1
..
x - vt = x - 4.5t
6
to get I y=[(x-4.5t?+3]
430
Chapter 16
n
P16.5
Let u = IOnt - 3nx + -
(a)
du = IOn _ 3n dx = 0 at a point of constant phase
4
dt
:
dt
= l~ = I 3.33 m/s I
The velocity is in the I positive x-direction I·
---._~--,_'--~..
P16.18
m)sin( -o.300n+~)=-0.0548
(b)
y(O.IOO, 0)=(0.350
(c)
k= 2; =3n:
(d)
v y = ay
at = (0.350)(10n)cos(IOm-3nx+'::)
m=j-5.48
m=2nf=IOn:
.1,=10.667 m I
cm I
f=15.00
Hz I
v y. max = (10n)(0.350) = I------'.
11.0 m/s
4
...
(a)
Let us write the wave function as
y(x, t)=Asin(kx+OJt+Ij»
y(O, O)=Asinlj>=
0.020 0 m
dyl
dt 0.0 =Amcoslj>=-2.00 m/s
Also,
2n
2n
T
0.025 0 s
m=-=---=80.0n/s
A2
m
= x; +( Vi)2
80.0 n/s
= (0.020 0 m)2 +( 2.00
m/s
J
A = I 0.021 S m I
(b)
Asinlj> =
0.0200
Acoslj> -2/80.0n
=-2.51=tanlj>
Your calculator's answer tan-1 (-2.51) = -1.19 rad has a negative sine and positive cosine,
just the reverse of what is required. You must look beyond your calculator to find
Ij>=n-1.l9
rad = 11.95 rad I
(c)
vy,max=Am=0.0215 m(80.on/s)=IS.41
(d)
.1,= vxT = (30.0 m/s)(0.025
2n
k=- =
A
I
m/s I
0 s) = 0.750 m
2n
0.750 m
- 8.38/m
y(x, t)=(0.0215 m)sin(8.38x
m = 80.0n/s
rad/m+80.0nt
rad/s+1.95
rad)
I
I
Section 16.3
The Speed of Waves on Strings
The down and back distance is 4.00 m + 4.00 m = 8.00 m.
P16.21
The speed is then
v=
Now,
j1
So
P16.42
t
d,aw.1
0.800 m)
s = 40.0 m/s = vJi
= 4(8.00
[
= 0.200 kg = 5.00
4.00 m
X
10-2 kg/m
T=j1v2=(5.00XI0-2
kg/m)(40.0
m/s)2=180.0N
The linear wave equation is
= eb(x-vt)
If
y
then
at = _bveb(x-vt)
dy
ay
b(x-vt)
-=be
and
ch
and
Therefore,
P16.43
dt =
d2;
demonstrating that
v2 d2;,
dX
i(x-vt)
is a solution.
The linear wave equation is ~ ch = d2y
v2 dt2 dX2
To show that y = In [b (x - vt)] is a solution, we find its first and second derivatives with respect
to x and t and substitute into the equation.
1
b(x-vt)
-1(-bv)2
dt2 - b2(x-vt?
d2y _
(-bv)
__ ~
- (X-vt)2
1
ay
dX
= [b(x- vt)t
b
r2--
2
~=--(x-vt
ax2
b
___
1
P16.24
(x-vt
)2
so the given wave function is a solution.
d2y
(x-vt?
-
- dX2
v=~
T = j1v2 = pAv2 = pnr2v2
T = (8 920 kg/m3 )(n)(7.50
T
= I 631 N I
x
10--4
m
t (200
m/s)2
I
c£ 17
P17.4
(a)
At 9000 m, t.T = (9 150
000}_1.000C)
= -60.0°C
so
T = -30.0°C
Using the chain rule:
dv
dv dT
dv dT
dt = dT
dx dx
dt = v dT
dx = v (0.607)(_1_)
150 = ~247 '
t
VI
so
dt = (247 s) dv
v
d
f dt = (247 s) f ~
o
v;
t=(247
v
S)ln(vf Vi )=(247
+ 0.607 (-30.0)
s)ln[ 331.5
331.5+0.607(30.0)]
t = I 27.2 s I for sound to reach ground.
(b)
t=l::.=
9000
v [331.5+0.607(30.0)]
=1257sl
.
I It takes longer when the air cools off than if it were at a uniform temperature.
i!').
--
-----~~~-~
17.2
~~:r'~~=---~----:-----------------------~
v
~~ection
Periodic Sound Waves
~P17.8 I (a)
I
~
--~
..-,-.-~~~
The speed gradually changes from = (331 m/s)( 1 + 27°CI273°qll2 = 347 m/s to
(331 m/s) (1 + 0/273°qll2 = 331 m/s, a 4.6% decrease. The cooler air at the same
pressure is more dense.
(b)
The frequency is unchanged, because every wave crest in the hot air becomes one crest
without delay in the cold air.
(c)
The wavelength decreases by 4.6%, from v!f= (347 m/s)/(4000/s) = 86.7 mm to
(331 m/s)/(4000/s) = 82.8 mm. The crests are more crowded together when they move
slower.
P17.11
(a)
A
= 12.00
A
= 15.7
211: = 0.400 m = I---40.0 cm 1
v
= co
858 = 154.6
k = 15.7
'-----. m/s
J1m I
I
(b)
s = 2.00 cos [(15.7)( 0.0500) - (858)( 3.00 x 10-3)
(c)
vmax
=
Aco
= (2.00
J1m)(858
S-I) = 11.72 mm/s I
]
= I -0.433
J1m
I
,
,~~~._--
P18.1
y
= y\ + Y2 = 3.00cos(4.00x-1.60t)+4.00sin(5.0x= 1.00, t = 1.00
(a)
x
(b)
x = 1.00, t = 0.500
Y
= 3.00 cos (+3.20 rad) + 4.00 sin (+4.00 rad) = 1-6.02 cm I
(c)
x = 0.500, t = 0
Y
= 3.00 cos (+2.00 rad) + 4.00 sin (+2.50 rad) = 11.15 cm
~-.---------~_._-_...
P18.4
2.00t) evaluated at the given x values.
Y = 3.00 cos(2AO rad) + 4.00sin( +3.00 rad) = 1-1.65 cm
I
I
------ _.-------------_._---_._._-'~
._--~"..,....~
...•.
.'--.----.,.--'.- ..
Suppose the waves are sinusoidal.
The sum is
(4.00 cm) sin (kx - rot) + (4.00 cm)sin(kx - rot + 90.0°)
2( 4.00 cm) sin (kx - rot + 45.00)cos45.0°
= 15.66 cm
So the amplitude is (8.00 cm) cos 45.0°
P18.7
I·
We suppose the man's ears are at the same level as the lower speaker. Sound from the upper
speaker is delayed by traveling the extra distance .J L2 +
d2
L.
-
He hears a minimum when this is (2n -1 )A with n = 1, 2, 3, ...
2
Then,
.J L2
+ d2
.JL2+d2
_
I
L = (n - 1/2) v
= (n-l/2)v
I
I
L + d2 = (n_l/2)2v2
2
2
+L
I
2(n-I/2)vL
+ L2 +----
d2 _ (n - 1/2)2 v2/ 12
L=
2(n-l/2)v/I
n=I,
2, 3, ...
This will give us the answer to (b). The path difference starts from nearly zero when the man is
very far away and increases to d when L = O.The number of minima he hears is the greatest
..
(n -1/2) v
mteger solution to d ~ ---
I
.
dl
n = greatest mteger ~ -
v
P18.11
1
+2
Y = (1.50 m) sin (OAOOx) cos(200t) = 2Ao sin kx cos rot
Therefore,
and
ro
2n
k = - = 00400 rad/m
= 2nl
A
A = 004002nrad/m---
so
I=!!!.2n = 200
2n rad/s
rad -131.8
.--- Hz
The speed of waves in the medium is
115.7 m
v=AI=-2nl=-=----=
A
ro
2n
k
I
I
200 rad/s
00400
rad/m
1--500 m/s
I
-
-~
*P18.16 (a)
The resultant wave is
y
==2A sin ( kx + % ) cos ( rot -
%)
The oscillation of the sin(kx + </>/2)factor means that this wave shows alternating nodes
and antinodes. It is a standing wave.
k _1n7r
The nodes are located at
so
x
==
2k
1-
to the left by the phase difference between the
which means that each node is shifted
traveling waves.
2k
(b)
k _12k ]_ [n7r
k _12k ]~x ==!E.==
k
~x ==[ (n + 1)!E.
~2
The separation of nodes is
The nodes are still separated by half a wavelength.
(c)
-·
As noted in part (a), the nodes are all shifted by the distance
l.•.....•
__
.,....,....
~.
</>l2k
to the left.
I
...........-
~~ --_ ~
....
P18.31
'.•~,
. A/4.•..---},J2-
.-.
I~'
A
e
The wavelength is
N
A
A
N
A
A
N
A AA
N
A
N
-I~I
L
LAA
~I
r=t2:]Je
N
A
N
---.
v
343 m/s -1.31 m
A == == 261.6/s
7
.1.•. ' N e
~I.1
=:f21j
~ A-Ie
.•--.</'
•....
J..
},J2 N
J../4
},J2
},J2
le
A
FIG. P18.31
so the length of the open pipe vibrating in its simplest (A-N-A) mode is
dAtoA
_. __
==~A==I 0.656 m I
A closed pipe has
(N-A) for its simplest resonance,
(N-A-N-A) for the second,
and
(N-A-N-A-N-A)
for the third.
----
J:I3!e, the~pi_pe_l_e_ng_t_h._is
__
5dN_~~:~:==_%.~_.3_1
m)_==I_I_.6_4_m_1
.
*P18,39 CallL the depth of the well and v the speed of sound.
Thenfor some integer n
L ==(2n - I)~
==(2n -1)~
4
and for the next resonance
4h
L==[2(n+1)-I]AZ
==(2n+I)~==
4
Thus,
(2n-1)(343
(2n+I)(343
4J;
m/s) _ (2n+1)(343
-
4(51.5s-l)
" an mteger so I utIOn
.
and we reqUIre
to
==(2n -1)(343 m/s)
4(51.5 S-I)
m/s)
4(60.0s-l)
2n + 1 ==-2n - 1
-60.0
51.5
..
111.5 ==6 .56 , so th e b est
TheequatIOngIves
n ==-17
Thenthe results
fi"ttmg
. n ==7.
mteger IS
L==
[2(7)-1](343
4(51.5sl)
L ==
[2(7) + 1](343 m/s) ==21.4 m
4 (60.0 Si)
and
m/s)
4( 60.0 S-I)
m/s) ==21.6 m
suggestthat we can say
the depth of the well is (21.5 ± 0.1) m. The data suggest 0.6-Hz uncertainty in the frequency
measurements, which is only a little more than 1%.
--., ...-...-
---...
Ch
P35.3
The experiment is most convincing if the wheel turns fast enough to pass outgoing light through
one notch and returning light through the next: t = 2£
C
so
(0
c8 _ (2.998x108)[2n/(720)]
2(11.45 x 103)
= 2£ -
=1 114 rad/s I
The returning light would be blocked by a tooth at one-half the angular speed, giving another data
point.
P35.12
(a)
(b)
_
Ag1ass
-
vg1as,
-
(c)
n = 632.8
1.50nm =1422
Aair
_ cair
n = 3.00
--
P35.19
At entry, .
or
n1
----,---
m/s = 2.00 x 108 m/s = I 200 Mm/s I
-
--~-----_.
sin 81 = n2 sin 82
1.00 sin 30.0°
82
X 108
1 <;n
nm I
= 1.50 sin 82
= 19.5°
The distance h the light travels in the medium is given by
= --2.00 cm
cos 8
h
2
h = 2.00 cm
cos19.5°
or
2.12 cm
FIG. P35.19
The angle of deviation upon entry is
The offset distance comes from sin a = !:..:
h
P35.21
d
= (2.21 cm) sin 10.5° = I 0.388 cm I
Applying Snell's law at the air-oil interface,
nair
sin 8 = noil sin 20.0°
yields
Applying Snell's law at the oil-water interface
nw
yields
I
sin 8'
= noil sin 20.0°
8' = 22.3° I
FIG. P35.21
*P3S.34 For total internal reflection,
1.50 sin e1
P3S.36
sinec =
= 1.33 (1.00)
nair
npipe
or
= 1.00 = 0.735
1.36
2.00 /lID
Geometry shows that the angle of refraction at the end is
1>
= 90.0° - ec = 90.0° - 47.3° = 42.7°
Then, Snell's law at the end,
1.00 sine = 1.36 sin 42.7°
~r
FIG. P35.36
gives
The 2-/lm diameter is unnecessary information.
=-t
Ch
P36.4
(1)
The first image in the left mirror is 5.00 ft behind the mirror, or 110.0 ft 1from the position
of the person.
(2)
The first image in the right mirror is located 10.0 ft behind the right mirror, but this location
is 25.0 ft from the left mirror. Thus, the second image in the left mirror is 25.0 ft behind the
mirror, or I 30.0 ft Ifrom the person.
(3)
The first image in the left mirror forms an image in the right mirror. This first image is
20.0 ft from the right mirror, and, thus, an image 20.0 ft behind the right mirror is formed.
This image in the right mirror also forms an image in the left mirror. The distance from this
image in the right mirror to the left mirror is 35.0 ft. The third image in the left mirror is,
thus, 35.0 ft behind the mirror, or I 40.0 ft 1from the person ..
P36.6
For a concave mirror, both R and/are
1 1
----q
/
---
------.-
R
/ = - = 10.0 cm
2
We also know that
(a)
positive.
p
1
1
3
10.0 cm
40.0 cm
40.0 cm
and
I q = 13.3 cm I
.,
M =!l.
13.3 cm
cm = 1-0.333
p = _ 40.0
---The image is 13.3 cm in front of the mirror, I real, and inverted
(b)
1- 1
-------
1
q
p
/
-
1
1
1
10.0 cm
20.0 cm
20.0 cm
I q = 20.0 cm I
and
M = !l.
20.0 cm
cm = I---1.00
p = _ 20.0
I
The image is 20.0 cm in front of the mirror, I real, and inverted
Cc)
.!.=~_~=
q /
p
1
_
10.0 cm
1
I No image is formed
Ca)
=0
I.
The rays are reflected parallel to each other.
M=-4=-!l.
q=4p
p
p
q - p = 0.60 m = 4p - P
1
1 1
-=-+-=--+-/
p q
Cb)
1
1
0.2 m
0.8 m
1
M =+-=--
q
2
p
Iq
I.
q = infinity
Thus,
P36.13
I.
= 0.2 m
q
= 0.8 m
/ = 1160 mm I
p = -2q
1+ p = 0.20 m = -q + P = -q - 2q
p
q = -66.7 mm
1 2
1
-+-=-=----+-----p q R 0.133 m
= 133 mm
1
-0.0667
m
R=I-267 mm
1
....------.-~-
333
Image Formation
p36.22
n1
+ n2
p
q
:= _n2_-_n_1
R
1.00
1.50
1
---+-:=20.0 cm
q
12.0 cm
1.00
1.50
1
+-:=--10.0 cm
q
12.0 cm
(a)
(b)
or
or
1.00
1.50
1
--+-:=--3.0 cm
q
12.0 cm
(c)
1.00 1.50 1.50 - 1.00
-+-:=---:=p
q
6.00 cm
becomes
or
1
12.0 cm
q :=
1.50
[(1.00/12.0 cm)-(1.00/20.0
cm)]
q:=
1.50
[(1.00/12.0 cm)-(1.00/1O.0
cm)]
q:=
1.50
[(1.00/12.0 cm)-(1.00/3.0
:=I 45 0 cm
==
[I-90.0
I -6 00
~:=
I
.
cm)]
cm
.
cm
I
.'
F or a convergmg
. 1ens, j' IS posItive.
.. UTvve use -1 + -:=-.
1 1
p q f
P36.29
1
1
40.0 cm
40.0 cm
1 1
1
=---=------
(a)
f
q
P
20.0 cm
M :=_:£ :=- 40.0 := -1.00
.
p
40.0~1
I q:= 40.0 cm I
I
The image is r-e-al-,-in-v-e-rt-ed-I,
and located 40.0 ~m past the lens.
-I
(b)
-=---=-----f
q
p
1
20.0 cm
o
I q := infinity I
20.0 cm
I No image I is formed. The rays emerging from the lens are parallel to each other.
1 1
1
=---=------
(c)
f
q
p
20.0 cm
M:= _:£:=
p
(-20.0):=
1
1
10.0 cm
20.0 cm
12.00
10.0
---
The image is I upright, virtual
-~
-:.-=1
1qf 32.0
1 _.,18.00
1f..•cm-cm
+
- 6.40
=-: + cm I
=
M
=-:£=8.00
cmis=1-0.250
Since
f> I0,f the
lens
1convergingI I.
p
32.0 cm
I
-
Iq:= -20.0 cm I
I
I and
20.0 cm in front of the lens.
(c)
P36.37
(a)
The image distance is:
q=d-p
1
1
1
p
q
f
- + - :=-
Thus,
becomes
This reduces to a quadratic equation:
which yields:
Since
f
<:!:..,
4
1
1
1
-+--=p d-p
f
p2
P
+ (-d)
=
p
+ fd = 0
224
d±.Jd2-4fd
. - - ± - d
~2
fd
both solutions are meaningf~l and the two solutions are not equal to each
other. Thus, there are two distinct lens positions that form an image on the screen.
(b)
The smaller solution for p gives a larger value for q, with a 1real, enlarged, inverted image
The larger solution for p describes a 1real, diminished, inverted image I.
I.