Physics 7C
Spring 2003: Block 12 Exercises & Q/Ps
1
03 03 20
Examples
1
.A frequency generator in DL creates a sinusoidal signal:
Signal t
( )
=
A
sin
2
π
t
0.00025 s
(
)
+
π
2
⎛
⎝
⎜
⎜
⎞
⎠
⎟
⎟
.
This generator is connected to two speakers.
However, speaker 2 hasit
s wires switched, with respect to the correctly wired speaker 1.
Assume that the sound waves generated by these speakers are 1D.
(a)What are the harmonic sound wave functions for the sounds
produced by speaker 1 and for speaker 2?
(b)At time
t
= 0, what are the total phases
Φ
1
and
Φ
2
, and what is the d
ifference in total phase ∆
Φ
, at the location of the microphone?
Is
there constructive or destructive interference?
Solution
Since these are sound waves, the wavelength
λ
of the sound
waves created by speakers 1 and 2 is:
λ =
v
wave
f
=
v
wave
Τ =
340
m
s
⎛
⎝
⎜
⎞
⎠
⎟
0.00025 s
(
)
=
0.085 m .
The harmonic wave functions for the sounds coming from speakers 1
and 2 are then given by:
∆
P
1
x
,
t
(
)
=
A
sin
2
π
t
0.00025 s
(
)
–
2
π
x
0.085 m
(
)
+
π
2
⎛
⎝
⎜
⎜
⎞
⎠
⎟
⎟
;
∆
P
2
x
,
t
(
)
=
A
sin
2
π
t
0.00025 s
(
)
–
2
π
x
0.085 m
(
)
+
3
π
2
⎛
⎝
⎜
⎜
⎞
⎠
⎟
⎟
.
Note the (–) before each
x
term, as both sound waves propagate
outwards from their respective speaker (using
x
to represent any
direction outwards from the speaker).
The microphone is located
L
1
= 1.500m away from speaker 1, such
that at time
t
= 0s, the sound wave there from speaker 1 will have a
total phase
Φ
1
:
Φ
1
=
2
π
0 s
(
)
0.00025 s
(
)
–
2
π
1.500 m
(
)
0.085 m
(
)
+
π
2
= −
109.31 rad ,
and for the total phase
Φ
2
, at
t
= 0s, and
L
2
= 1.245m away from
the speaker:
Φ
2
=
2
π
0 s
(
)
0.00025 s
(
)
–
2
π
1.245 m
(
)
0.085 m
(
)
+
3
π
2
= −
87.31 rad .
frequency
generator
microphone
1.500 m
1.245 m
speaker 1
speaker 2
Physics 7C
Spring 2003: Block 12 Exercises & Q/Ps
2
03 03 20
The difference in total phase ∆
Φ
is then:
∆Φ = Φ
1
− Φ
2
= −
109.31 rad
(
)
− −
87.31 rad
(
)
= −
21.99 rad ,
and since –21.99 radians is also –7π radians, the sound waves from
speaker 1 and from speaker 2 interfere destructively at the location of
the microphone.
2
.The frequencies for the notes C
4
and G
4
are 261.63Hz and 392.00Hz, r
espectively.
These notes are played simultaneously to produce a
simple chord (try this out if you can get to a piano or keyboard).
Calculate the following:
(a)The carrier frequency (which would be perceived as pitch) of these s
uperposed notes.
(b)The subjective beat frequency of these superposed notes.
Solution
The carrier frequency is the average of two frequencies in this chord:
f
c
=
f
1
+
f
2
(
)
2
=
261.63 Hz
+
392.00 Hz
(
)
2
=
326.82 Hz .
The beat frequency is the difference between the two frequencies in this
chord.
f
b
=
f
1
−
f
2
=
261.63 Hz
−
392.00 Hz
=
130.37 Hz.
If you look up these frequencies, you will find that playing the notes C
4
and G
4
together in a chord will result in a superposed sound with a
pitch of approximately E
4
(329.63Hz), but with an amplitude
modulation of approximately C
3
(130.81Hz).
Try out some different
chords, both audibly and mathematically (
cf.
the piano key frequencies
in the following Appendix section).
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 Spring '08
 MAHMUD
 Physics, Wavelength, Hz, total phase difference, Q/Ps

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