1
HW 8 Solutions
Normal Modes and Resonance Frequencies
a) The string described in the problem introduction is oscillating in one of its normal modes.
Which of the following statements about the wave in the string is correct?
The wave is traveling in the +
x
direction.
The wave is traveling in the 
x
direction.
The wave will satisfy the given boundary conditions for any arbitrary wavelength
λ
i
.
The wavelength
λ
can have only certain specific values if the boundary conditions are to be
satisfied.
The wave does not satisfy the boundary condition
( )
0,
0
i
yt
=
.
b)
Which of the following statements are true?
The system can resonate at only certain resonance frequencies
f
i
and the wavelength
λ
i
must
be such that
()
0,
0
i
=
,
,0
i
yL
t
=
.
A
i
must be chosen so that the wave fits exactly on the string.
Any one of
A
i
or
λ
i
or
f
i
can be chosen to make the solution a normal mode
The key factor producing the normal modes is that there are two spatial boundary conditions,
0,
0
i
=
and
i
t
=
that are satisfied only for particular values of
λ
i
.
c)
Find the three longest wavelengths (call them
λ
1
,
λ
2
, and
λ
3
) that "fit" on the string, that is,
those that satisfy the boundary conditions at x = 0 and x = L. These longest wavelengths have the
lowest frequencies.
123
,
1,2,3,.
..
, ,
2 , ,2 /3
2
Ln
n
L
LL
λ
λλλ
==
⇒
d, e)
The frequency of each normal mode depends on the spatial part of the wave function,
which is characterized by its wavelength
λ
i
. Find the frequency
f
i
of the
i
th normal mode. Find
the three lowest normal mode frequencies
f
1
,
f
2
, and
f
3
.
,
1,2,3,.
..
2
i
i
vv
fn
n
L
=
Interference of Sound Waves
Two loudspeakers, A and B, are driven by the same amplifier and emit sinusoidal waves in
phase. The frequency of the waves emitted by each speaker is 172 Hz. You are 8.00 m from
speaker A. Take the speed of sound in air to be 344 m/s. What is the closest you can be to
speaker B and be at a point of destructive interference?
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A
B
r
1
r
2
For destructive interference,
12
1
,
1,2,3,.
..
2
rr n
n
λ
⎛⎞
−= +
=
⎜⎟
⎝⎠
.
The wavelength
344
/
2
172
vm
s
m
fH
z
==
=
, thus you are exactly four wavelengths away from speaker A. The
smallest r
2
that satisfies the condition for destructive interference is
λ
/2=1m.
Exercise 15.41
A thin, taut string tied at both ends and oscillating in its third harmonic has its shape described
by the equation
()
,
5.6
sin 0.034
sin 50
rad
rad
y xt
cm
x
t
cm
s
⎡
⎤⎡
⎤
=×
⎢
⎥⎢
⎥
⎣
⎦⎣
⎦
, where the origin is at the left
end of the string, the
x
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 Spring '11
 EIns
 Wavelength, Fundamental frequency

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