Physics 41 Chapter 18 HW Spring 2009 Serway 7th Edition
Q: 4, 8, 14, 16,
P: 1, 3, 6, 11, 15, 16, 19, 21, 23, 28, 29, 31, 38, 41, 43, 45,
Q18.4 No. The total energy of the pair of waves remains the same. Energy missing from zones of destructive
interference appears in zones of constructive interference.
*Q18.8 The fundamental frequency is described by f1 =
v
⎛T ⎞
, where v = ⎜ ⎟
2L
⎝μ⎠
12
1
. Answer (f).
2
1
will be reduced by a factor
. Answer (e).
2
(i)
If L is doubled, then f1 ∝ L−1 will be reduced by a factor
(ii)
If μ is doubled, then f1 ∝ μ −1 2
(iii)
If T is doubled, then f1 ∝ T will increase by a factor of 2 . Answer (c).
Q18.14 Walking makes the person’s hand vibrate a little. If the frequency of this motion is equal to the natural frequency
of coffee sloshing from side to side in the cup, then a large–amplitude vibration of the coffee will build up in resonance.
To get off resonance and back to the normal case of a small-amplitude disturbance producing a small–amplitude result,
the person can walk faster, walk slower, or get a larger or smaller cup. Alternatively, even at resonance he can reduce the
amplitude by adding damping, as by stirring high–fiber quick–cooking oatmeal into the hot coffee. You do not need a
cover on your cup.
Q18.16 Beats. The propellers are rotating at slightly different frequencies.
Problems 1, 3, 6, 11, 15, 16, 19, 21, 23, 28, 29, 31, 38, 41, 43, 45,
P18.1
3. y = y1 + y2 = 3.00cos ( 4.00x − 1.60t ) + 4.00sin ( 5.0x − 2.00t ) evaluated at the given x values.
(a)
x = 1.00 , t = 1.00
y = 3.00cos( 2.40 rad ) + 4.00sin ( +3.00 rad ) = −1.65 cm
(b)
x = 1.00 , t = 0.500
y = 3.00cos( +3.20 rad ) + 4.00sin ( +4.00 rad ) = −6.02 cm
(c)
x = 0.500 , t = 0
y = 3.00cos( +2.00 rad ) + 4.00sin ( +2.50 rad ) = 1.15 cm
Two pulses traveling on the same string are described by 5
−5
and y1 =
y2 =
(3x − 4t )2 + 2
(3x + 4t − 6)2 + 2
(a) In which direction does each pulse travel? (b) At what time do the two cancel everywhere? (c) At what point do the two pulses always cancel? P18.3
(a)
y1 = f( x − vt) , so wave 1 travels in the + x direction
y2 = f( x + vt) , so wave 2 travels in the − x direction
(b)
To cancel, y1 + y2 = 0 :
5
( 3x − 4t)
2
+2
=
+5
( 3x + 4t− 6) 2 + 2
( 3x − 4t) 2 = ( 3x + 4t− 6) 2
3x − 4t= ± ( 3x + 4t− 6)
or the positive root, 8t= 6
t= 0.750 s
(at t= 0.750 s, the waves cancel everywhere)
(c)
P18.6
(a)
for the negative root,
x = 1.00 m
6x = 6
always)
Δx = 9.00 + 4.00 − 3.00 = 13 − 3.00 = 0.606 m
v 343 m s
λ= =
= 1.14 m
The wavelength is
f
300 Hz
Δx 0.606
=
= 0.530 of a w ave ,
Thus,
1.14
λ
Δφ = 2π ( 0.530) = 3.33 rad
or
(b)
(at x = 1.00 m , the waves cancel
For destructive interference, we want
where Δx is a constant in this set up.
Δx
Δx
v
λ
v
343
f =
=
= 283 Hz
2Δx 2 ( 0.606)
= 0.500 = f
P18.11 y = ( 1.50 m ) sin ( 0.400x ) cos ( 200t ) = 2A 0 sin kx cosω t
Therefore, k =
2π
λ
= 0.400 rad m λ =
and ω = 2π f so f =
ω
2π
=
2π
= 15.7 m
0.400 rad m
200 rad s
= 31.8 H z
2π rad
The speed of waves in the medium is
200 rad s
λ
ω
v=λf =
2π f = =
= 500 m s
2π
k 0.400 rad m
P18.15 y1 = 3.00sin ⎣⎡π ( x + 0.600t )⎦⎤ cm ; y2 = 3.00sin ⎣⎡π ( x − 0.600t )⎦⎤ cm
y = y1 + y2 = ⎡⎣3.00sin (π x ) cos ( 0.600π t ) + 3.00sin (π x ) cos ( 0.600π t ) ⎤⎦ cm
y = ( 6.00 cm ) sin (π x ) cos( 0.600π t )
(a)
We can take cos( 0.600π t ) = 1 to get the maximum y.
At x = 0.250 cm ,
ymax = ( 6.00 cm ) sin ( 0.250π ) = 4.24 cm
(b)
At x = 0.500 cm ,
ymax = ( 6.00 cm ) sin ( 0.500π ) = 6.00 cm
(c)
Now take cos( 0.600π t ) = −1 to get ymax :
(d)
At x = 1.50 cm ,
ymax = ( 6.00 cm ) sin ( 1.50π )( −1) = 6.00 cm
The antinodes occur when
x=
But k =
and
2π
λ
= π so
nλ
( n = 1, 3, 5, K )
4
λ = 2.00 cm
x1 =
λ
= 0.500 cm as in (b)
4
3λ
= 1.50 cm as in (c)
x2 =
4
5λ
= 2.50 cm
x3 =
4
*P18.16
(a)
φ⎞
φ⎞
⎛
⎛
y = 2A sin ⎜ kx + ⎟ cos⎜ ω t − ⎟
2⎠
2⎠
⎝
⎝
The resultant wave is
The oscillation of the sin(kx + φ/2) factor means that this wave shows alternating nodes and
antinodes. It is a standing wave.
φ
nπ φ
kx + = nπ so
The nodes are located at
−
x=
2
k 2k
φ
which means that each node is shifted
2k
to the left by the phase difference between the
traveling waves.
(b)
(c)
π φ ⎤ ⎡ nπ φ ⎤
π λ
⎡
Δx = ⎢( n + 1) − ⎥ − ⎢
− ⎥ Δx = =
k 2
k 2k ⎦ ⎣ k 2k ⎦
⎣
The nodes are still separated by half a wavelength.
The separation of nodes is
As noted in part (a), the nodes are all shifted by the distance φ/2k to the left.
19. In the arrangement shown, an object can be hung from a string (with linear mass density μ =0.002 00 kg/m) that passes over a light pulley. The string is connected to a vibrator (of constant frequency f), and the length of the string between point P and the pulley is L = 2.00 m. When the mass m of the object is either 16.0 kg or 25.0 kg, standing waves are observed; however, no standing waves are observed with any mass between these values. (a) What is the frequency of the vibrator? (Note: The greater the tension in the string, the smaller the number of nodes in the standing wave.) (b) What is the largest object mass for which standing waves could be observed? P18.19
(a)
(b)
Let n be the number of nodes in the standing wave resulting from the 25.0-kg mass. Then
n + 1 is the number of nodes for the standing wave resulting from the 16.0-kg mass. For
2L
v
, and the frequency is f = .
standing waves, λ =
n
λ
n Tn
2L μ
Thus,
f=
Thus,
n+1
=
n
Therefore,
4n + 4 = 5n , or n = 4
Then,
f=
Tn
=
Tn+ 1
4
2( 2.00 m
f=
and also
)
n + 1 Tn+1
2L
μ
( 25.0 kg) g = 5
(16.0 kg) g 4
( 25.0 kg) ( 9.80 m
s2
0.002 00 kg m
)=
350 H z
The largest mass will correspond to a standing wave of 1 loop
( n = 1)
so
350 H z =
1
2( 2.00 m
(
m 9.80 m s2
)
)
0.002 00 kg m
m = 400 kg
yielding
P18.21 dNN = 0.700 m = λ/2
λ = 1.40 m
T
(1.20 × 10 ) ( 0.700)
f λ = v = 308 m s =
−3
(a)
T = 163 N
(b)
With one-third the distance between nodes, the frequency
is f 3 = 3 ⋅ 220 Hz = 660 Hz
FIG. P18.21
23. Review problem. A sphere of mass M is supported by a string that passes over a light horizontal rod of length L (Fig. P18.29). Given that the angle is θ and that f represents the fundamental frequency of standing waves in the portion of the string above the rod, determine the mass of this portion of the string. P18.23 In the fundamental mode, the string above the rod has only two nodes,
at A and B, with an anti-node halfway between A and B. Thus,
λ
2
= AB =
A
L
2L
or λ =
.
cosθ
cosθ
Since the fundamental frequency is f, the wave speed in this segment of
string is
θ
v = λ f=
Also, v =
T
μ
=
T
=
m AB
L
2Lf
.
cosθ
TL
m cosθ
M
where T is the tension in this part of the string. Thus,
2Lf
=
cosθ
B
4L2 f2
TL
TL
or
=
2
m cosθ
cos θ m cosθ
T
F
θ
and the mass of string above the rod is:
m =
T cosθ
4Lf2
[Equation 1]
Mg
FIG. P18.29
Now, consider the tension in the string. The light rod would rotate about point P if the string exerted any vertical
force on it. Therefore, recalling Newton’s third law, the rod must exert only a horizontal force on the string. Consider a
free-body diagram of the string segment in contact with the end of the rod.
Mg
∑ Fy = T sin θ − M g = 0 ⇒ T = sin θ
Then, from Equation 1, the mass of string above the rod is
Mg
⎛ M g ⎞ cosθ
m =⎜
=
.
⎟
2
⎝ sin θ ⎠ 4Lf
4Lf2 tan θ
P18.29
(a)
For the fundamental mode in a closed pipe, λ = 4L , as
in the diagram.
A
λ /4
v
But v = f λ , therefore L =
4f
So, L =
(b)
343 m s
4 ( 240 s−1 )
L
= 0.357 m
A
N
A
λ /2
For an open pipe, λ = 2L , as in the diagram.
So, L =
N
343 m s
v
=
= 0.715 m
2 f 2 ( 240 s−1 )
FIG. P18.29
31. The fundamental frequency of an open organ pipe corresponds to middle C (261.6 Hz on the chromatic musical scale). The third resonance of a closed organ pipe has the same frequency. What are the lengths of the two pipes? P18.31
The wavelength is
λ=
v 343 m s
=
= 1.31 m
f 261.6 s
so the length of the open pipe vibrating in its simplest (A-N-A) mode is
dA to A =
1
λ = 0.656 m
2
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.
Here, the pipe length is
5dN
to A
=
5λ 5
= ( 1.31 m ) = 1.64 m
4 4
38. A tuning fork with a frequency of 512 Hz is placed near the top of the pipe shown in Figure 18.19a. The water level is lowered so that the length L slowly increases from an initial value of 20.0 cm. Determine the next two values of L that correspond to resonant modes. 34
v
v
=
= 0.167 m .
: L1 =
P18.38 The length corresponding to the fundamental satisfies f =
4 f 4( 512)
4L
3v
5v
and f =
.
Since L > 20.0 cm , the next two modes will be observed, corresponding to f =
4L2
4L3
3v
5v
= 0.502 m and L3 =
= 0.837 m .
or L2 =
4f
4f
P18.41
(a)
(b)
f =
5 100
v
=
= 1.59 kHz
2L ( 2)( 1.60)
Since it is held in the center, there must be a node in the center as well as antinodes at the
ends. The even harmonics have an antinode at the center so only the odd harmonics are
present.
(c)
f =
3 560
v′
=
= 1.11 kHz
2L ( 2)( 1.60)
P18.43
f ∝v∝ T
540
= 104.4 Hz
600
f new = 110
Δf = 110/ s − 104.4/ s = 5.64 beats s
45. A student holds a tuning fork oscillating at 256 Hz. He walks toward a wall at a constant speed of 1.33 m/s. (a) What beat frequency does he observe between the tuning fork and its echo? (b) How fast must he walk away from the wall to observe a beat frequency of 5.00 Hz? ( v + vs) the beat frequency is f = f′ − f .
P18.45 For an echo f′ = f
b
( v − vs)
Solving for fb .
gives fb = f
(a)
(b)
vs =
( 2vs)
( v − vs)
fb = ( 256)
when approaching wall.
2( 1.33)
=
( 343 − 1.33)
1.99 H z beat frequency
When he is moving away from the wall, vs changes sign. Solving for vs gives
fbv
( 5)( 343)
=
= 3.38 m s .
2 f − fb ( 2)( 256) − 5