Sound and Hearing
169
16.36.
I
DENTIFY
:
Destructive interference occurs when the path difference is a half integer number of wavelengths.
S
ET
U
P
:
(
) (
)
344 m s, so
λ
344 m/s
172 Hz
2.00 m.
v
v f
=
=
=
=
If
8.00 m
A
r
=
and
B
r
are the distances of the
person from each speaker, the condition for destructive interference is
(
)
1
2
λ
,
B
A
r
r
n
−
=
+
where
n
is any integer.
E
XECUTE
:
Requiring
(
)
1
2
λ
0
B
A
r
r
n
=
+
+
>
gives
(
) (
)
1
2
8.00 m
2.00 m
4,
A
n
r
λ
+
> −
= −
= −
so the smallest
value of
B
r
occurs when
4,
n
= −
and the closest distance to
B
is
(
)
(
)
1
2
8.00 m
4
2.00 m
1.00 m.
B
r
=
+ −
+
=
E
VALUATE
:
For
1.00 m,
B
r
=
the path difference is
7.00 m.
A
B
r
r
−
=
This is 3.5
.
λ
16.37.
I
DENTIFY
:
Compare the path difference to the wavelength.
S
ET
U
P
:
(
) (
)
344 m s
860 Hz
0.400 m
v f
λ
=
=
=
E
XECUTE
:
The path difference is 13.4 m
12.0 m
1.4 m.
−
=
path difference
3.5.
λ
=
The path difference is a half
integer number of wavelengths, so the interference is destructive.
E
VALUATE
:
The interference is destructive at any point where the path difference is a halfinteger number of
wavelengths.
16.38.
I
DENTIFY
:
beat
1
2
.
f
f
f
=
−
.
v
f
λ
=
S
ET
U
P
:
344 m/s,
v
=
Let
1
6.50 cm
λ
=
and
2
6.52 cm.
λ
=
2
1
λ
λ
>
so
1
2
.
f
f
>
E
XECUTE
:
2
2
1
1
2
2
2
1
2
1
2
1
1
(
)
(344 m/s)(0.02
10
m)
16 Hz.
(6.50
10
m)(6.52
10
m)
v
f
f
v
λ
λ
λ
λ
λ λ
−
−
−
⎛
⎞
−
×
−
=
−
=
=
=
⎜
⎟
×
×
⎝
⎠
There are 16 beats per
second.
E
VALUATE
:
We could have calculated
1
f
and
2
f
and subtracted, but doing it this way we would have to be
careful to retain enough figures in intermediate calculations to avoid roundoff errors.
16.39.
I
DENTIFY
:
beat
.
a
b
f
f
f
=
−
For a stopped pipe,
1
.
4
v
f
L
=
S
ET
U
P
:
344 m/s.
v
=
Let
1.14 m
a
L
=
and
1.16 m.
b
L
=
b
a
L
L
>
so
1
1
.
a
b
f
f
>
E
XECUTE
:
2
1
1
1
1
(
)
(344 m/s)(2.00
10
m)
1.3 Hz.
4
4
4(1.14 m)(1.16 m)
b
a
a
b
a
b
a
b
v
v L
L
f
f
L
L
L L
−
⎛
⎞
−
×
−
=
−
=
=
=
⎜
⎟
⎝
⎠
There are 1.3 beats per
second.
E
VALUATE
:
Increasing the length of the pipe increases the wavelength of the fundamental and decreases the
frequency.
16.40.
I
DENTIFY
:
beat
0
.
f
f
f
=
−
.
2
v
f
L
=
Changing the tension changes the wave speed and this alters the frequency.
S
ET
U
P
:
FL
v
m
=
so
1
,
2
F
f
mL
=
where
0
.
F
F
F
=
+ Δ
Let
0
0
1
.
2
F
f
mL
=
We can assume that
0
/
F
F
Δ
is very
small. Increasing the tension increases the frequency, so
beat
0
.
f
f
f
=
−
E
XECUTE
:
(a)
(
)
1/ 2
0
beat
0
0
0
0
1
1
1
1 .
2
2
F
F
f
f
f
F
F
F
mL
F
mL
⎛
⎞
⎡
⎤
Δ
⎜
⎟
=
−
=
+ Δ
−
=
+
−
⎢
⎥
⎜
⎟
⎣
⎦
⎝
⎠
1/ 2
0
0
1
1
2
F
F
F
F
⎡
⎤
Δ
Δ
+
=
+
⎢
⎥
⎣
⎦
when
0
/
F
F
Δ
is small. This gives that
beat
0
0
.
2
F
f
f
F
⎛
⎞
Δ
=
⎜
⎟
⎝
⎠
(b)
beat
0
0
2
2(1.5 Hz)
0.68%.
440 Hz
F
f
F
f
Δ
=
=
=
E
VALUATE
:
The fractional change in frequency is onehalf the fractional change in tension.
16.41.
I
DENTIFY
:
Apply the Doppler shift equation
L
L
S
S
.
v
v
f
f
v
v
⎛
⎞
+
=
⎜
⎟
+
⎝
⎠
S
ET
U
P
:
The positive direction is from listener to source.
S
1200 Hz.
f
=
L
1240 Hz.
f
=
E
XECUTE
:
L
0.
v
=
S
25.0 m/s.
v
= −
L
S
S
v
f
f
v
v
⎛
⎞
=
⎜
⎟
+
⎝
⎠
gives
S
L
S
L
(
25 m/s)(1240 Hz)
780 m/s.
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 Spring '06
 Buchler
 Physics, m/s, PMAX, Δf

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