Resonance Set 1

Standing Waves & Resonance (AP1)
1.
1.
A pipe of length 80 cm is closed at one end and open at the other. Sound is created in the pipe at four different
frequencies. The diagram show the locaiton of notes (N) and antinodes (A) in the pipe for the four modes. The
table has an entry for wave speed of the second overtone. Complete the tale of frequencies, wavlengths, and
speeds for the four modes.
Frequency
Wavelength Wave Speed
Fundamental
106 Hz
4*80cm =
320 cm
340 m/s
First overtone
(second harmonic)
320 Hz
(4/3)*80 =
107 cm
340 m/s
Second overtone
(third harmonic)
530 Hz
(4/5)*80 =
64 cm
Third overtone
(fourth harmonic)
740 Hz
(4/7)*80 =
46 cm
340 m/s
340 m/s
A pipe of length L is open at both ends. Sound is created in the pipe at four different frequencies. The diagram
show the locaiton of notes (N) and antinodes (A) in the pipe for the four modes. The table tot he right has an entry
for wave speed of the second overtone. Use the given information to find the length L of the pipe. Then complete
the table of frequencies, wavelengths, and wave speedf for the four modes.
L = 1.5*40cm = 60. cm (use 2nd overtone data)
Frequency
Wavelength Wave
Speed
Fundamental
2*60cm =
280 Hz
120 cm
340 m/s
First overtone
(second harmonic)
570 Hz
60 cm
340 m/s
Second overtone
(third harmonic)
850 Hz
40. cm
340 m/s
110 Hz
(1/2)*60cm
= 30 cm
Third overtone
(fourth harmonic)
340 m/s
2.
Some pipes on a pipe organ are open at both ends, others are closed at one end. For pipes that play lowfrequencies, there is an advantage to using pipes that are closed at one end. What is that advantage?
If the pipe of open at both end, the fundamental frequency will have a wavelength where half the wave must fit in the
pipe (the wavelength will be twice the length of the pipe). But if one end is closed, then the findamental frequency will
have a wavelength where only a quarter of the wave will fit into the pipe (the wavelength will be four times the length of
the pipe). This will alow for a much longer wavelength and a much lower sound.
3.
One open organ pipe and an organ pipe that is closed at one end both have lengths of 0.52 m at 20 oC. What is the
fundamental frequency of each pipe? (330Hz; 170Hz)
vsound  (331)
293
 343m / s
273
open:

There is half a wave in the pipe if both ends
Closed:
are open. Therefore the wavelength is 2*0.52m.
closed. Therefore the wavelength is 4*0.52m
v  f
vf
343  2(0.52) f
343  4(0.52) f
f  330Hz
4.

There is a quarter of a wave in the pipe if one end is
f  165Hz  170 Hz
Blowing across the top of a glass bottle will produce a hollow sounding sound. If some water is put in the glass
bottle, the bottle will now produce a different sound when blowing across the top. Will this new sound be higher
or lower than the original and why does this happen?
The wavelength will decrease (be smaller), therefore, the frequency will increase (or be higher) because the
speed of sound is the same.
5.
An empty soda bottle is 18.0 cm deep at 20oC.
a. What resonant frequency would you expect to hear from blowing across the top of the bottle (assume
one end is closed)? (476Hz)
v  f
343  4(0.18) f
f  476Hz
b.

What would the resonant frequency be if the bottle were one-third full of soda? (715Hz)
vf
343  4(0.12) f
f  715Hz
6.
When you speak after breathing helium, in which the speed of sound is much greater than in air, you voice sounds
quite different. The frequencies emitted by your vocal cords so not change since they are determined by the mass
and tensio of your vocal cords. So what does change when you vocal tract is filled with helium rather than air?
V = f. If frequency does not change, but the velocity does, then the wavelength of the sound must change.
7.
The A string on a violin has a fundamental frequency of 440 Hz. The length of the vibrating portion is 32 cm, and it
has a mass of 0.35 g. Under what tension must the string be placed? (86N)
f  440Hz
l  .32m
m  .35 103 kg
m .35 103
 
 1.09 103 kg/m
l
.32

There is half a wave on the
string. So the wavelength is
2*32cm = 64 cm = 0.64m
v
v  f

8.
T

v  (0.64)(440)
281.6 
v  281.6m / s
T  86.N
T
1.09 103
A guitarist can play a “harmonic“ by lightly touching a string in the exact center and plucking the string. The result
is a clear note one octive higher than the fundamental frequency eventhough the string is not pressed to the
fingerboard. Why does this happen?
Eventhough the string is not held tightly, it still is not able to move like it normally would. This
creates a node. The string now has a node at the end, in the middle, and at the other end. This leads
to a whole wave being on the string (instead of one half a wave). Since this is the second largest wave
that will fit, it is the second harmonic.
9.
Given open and closed pipes of the same length, which would have the lowest natural frequency: (1) the open
pipe, (2) the closed pipe, or (3) both would have the same low frequency? Why?
(3) Both would have the same lowest natural frequency.
As shown, both pipes contain half a wave. If the wavelength
of the pipe is the same for both pipes (2*length) and the speed
of sound is the same for each pipe, then the frequency must be
the same for each pipe.