16-14
Chapter 16
EVALUATE:
The decrease in intensity level corresponds to a decrease in intensity, and this means an increase in
distance. The intensity level uses a logarithmic scale, so simple proportionality between
r
and
β
doesn&t apply.
16.57.
IDENTIFY:
The sound is first loud when the frequency
0
f
of the speaker equals the frequency
1
f
of the
fundamental standing wave for the gas in the tube. The tube is a stopped pipe, and
1
.
4
v
f
L
=
.
RT
v
M
γ
=
The sound
is next loud when the speaker frequency equals the first overtone frequency for the tube.
SET UP:
A stopped pipe has only odd harmonics, so the frequency of the first overtone is
31
3.
f
f
=
EXECUTE:
(a)
01
1
.
44
vR
T
ff
L
LM
== =
This gives
22
0
16
.
L Mf
T
R
=
(b)
0
f
EVALUATE:
(c)
Measure
0
f
and
L
. Then
0
4
v
f
L
=
gives
0
4.
vL
f
=
16.58.
IDENTIFY:
beat
.
AB
f
=−
1
2
v
f
L
=
and
FL
v
m
=
gives
1
1
.
2
F
f
mL
=
Apply
0
z
τ
=
∑
to the bar to find the
tension in each wire.
SET UP:
For
0
z
=
∑
take the pivot at wire
A
and let counterclockwise torques be positive. The free-body
diagram for the bar is given in Figure 16.58. Let
L
be the length of the bar.
EXECUTE:
0
z
=
∑
gives
lead
bar
(3 /4)
( / 2)
0.
B
FL w
L
w L
−−
=
lead
bar
3
/ 4
/2
3(185 N)/4 (165 N)/2
221 N.
B
Fw
w
=+
=
+
=
bar
lead
FFw w
+= +
so
bar
lead
165 N 185 N
221 N 129 N.
Fw w F
−
=
+
−
=
1
3
1
129 N
88.4 Hz.
2 (5.50 10 kg)(0.750 m)
A
f
−
==
×
11
221 N
115.7 Hz.
129 N
BA
beat
1
1
27.3 Hz.
f
=−=
EVALUATE:
The frequency increases when the tension in the wire increases.
Figure 16.58
16.59.
IDENTIFY:
The flute acts as a stopped pipe and its harmonic frequencies are given by Eq.(16.23). The resonant
frequencies of the string are
1
,
1, 2, 3,.
..
n
fn
f
n
The string resonates when the string frequency equals the flute
frequency.
SET UP:
For the string
1s
600.0 Hz.
f
=
For the flute, the fundamental frequency is
1f
344.0 m s
800.0 Hz.
4
4(0.1075 m)
v
f
L
=
Let
f
n
label the harmonics of the flute and let
s
n
label the harmonics of the
string.
EXECUTE:
For the flute and string to be in resonance,
f
1f
s 1s
1s
, where
600.0 Hz
nf
f
is the fundamental
frequency for the string.
4
sf
1
f
1
s
f
3
()
.
nn
n
s
n
is an integer when
f
3,
1
,
nN
N
=
=
3, 5, ± (the flute has only
odd harmonics).
fs
3
g
i
v
e
s
4
Flute harmonic 3
N
resonates with string harmonic 4 ,
1,
NN
=
3, 5, ±.
EVALUATE:
We can check our results for some specific values of
N
. For
N
=
f
3
n
=
and
3f
2400 Hz.
f
=
For
this
N
,
s
4
n
=
and
4s
2400 Hz.
f
=
For
3,
N
=
f
9
n
=
and
9f
7200 Hz,
f
=
and
s
12,
n
=
12s
7200 Hz.
f
=
Our general
results do give equal frequencies for the two objects.