Experiment 11
Velocity of Sound in Air – Resonance Tube
(Physics Laboratory Experiments – J. D. Wilson, DC Heath and Co.)
Systems generally have one or more natural vibrating frequencies. When a system
is driven at a natural frequency, there is a maximum energy transfer and the vibrational
amplitude increases to a maximum. When a system is driven at a natural frequency, we
say that the system is in resonance (with the driving source) and refer to the particular
frequency at which this occurs as a resonance frequency.
From the relationship between the frequency f, the wavelength λ, and the wave
speed v, which is λf = v, it can be seen that if the frequency and wavelength are known,
the wave speed can be determined. And, if the wavelength and speed are known, the
frequency can be determined.
Objective:
To measure the velocity of sound in air at room temperature.
Theory:
L=3λ/4
L=λ/4
Figure 1
Air columns in pipes or tubes of fixed lengths have particular resonant
frequencies. The interference of the waves traveling down the tube and the reflected
waves traveling up the tube produces (longitudinal) standing waves which must have a
node at the closed end of the tube and an antinode at the open end of the tube.
The resonance frequencies of a pipe or tube depend on its length L. As seen in
Figure 1, only a certain number of wavelengths or “loops” can be “fitted” into the tube
length with the node-antinode requirements. Since each loop corresponds to one half
wavelength, resonance occurs when the length of the tube is nearly equal to an odd
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number of quarter wavelengths (i.e. L =
λ
,L =
3λ
5λ
nλ
,L =
, etc.) or in general L =
4
4
4
4
4L
where n = 1, 3, 5,…., and λ =
. Incorporating the frequency f and the speed vs through
n
the general relationship λf = v, or f = v/λ, we have:
fn =
nv
4L
where n = 1, 3, 5, …….
Hence, an air column (tube) of length L has particular resonance frequencies and will be
in resonance with the corresponding odd-harmonic driving frequencies.
As can be seen from the above equation, the three experimental parameters
involved in the resonance condition of an air column are f, v, and L. To study the
resonance in this experiment, the length L of an air column will be varied for a given
driving frequency. The length of the air column will be achieved by changing the position
of the movable piston in the tube as seen in Figure 2.
Speaker – Connect to power supply
Resonance tube
Movable piston
Microphone – Connect to voltage sensor
Figure 2
As the piston is removed, increasing the length of the air column, more
wavelength segments will fit into the tube, consistent with the node-antinode
requirements at the ends. The difference in the tube lengths when successive antinodes
are at the open end of the tube and resonance occurs is equal to a half wavelength; for
example:
∆L = L2 − L1 =
3λ λ λ
− =
4 4 2
as seen in Figure 1.
When an antinode is at the open end of the tube, a loud resonance tone is heard.
Hence, the tube lengths for antinodes to be at the open end of the tube can be determined
by moving the piston away from the open end of the tube and “listening” for successive
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resonances. No end correction is needed for the antinode occurring slightly above the end
of the tube in this case, since the differences in tube lengths for successive antinodes is
equal to λ/2.
If the frequency of the driving source is known and the wavelength is determined
by measuring the difference in tube length between successive antinodes, ∆L =
2∆L, the speed of sound in air vs can be determined from vs =λf.
λ
2
or λ =
The speed of sound in air is temperature dependent and is given to a good
approximation over the normal temperature range by:
v s = 331.5 m / s + 0.6Tc m / s
where Tc is the air temperature in degrees Celsius. The equation above shows that the
speed of sound in air at 0 C is 331.5 m/s and increases by 0.6 m/s for each degree of
temperature increase. For example, at 20 C, the speed of sound is 343.5 m/s.
Questions
1. Weak background noises from a room set up the fundamental standing wave in a
cardboard tube of length L = 67.0 cm with two open ends. Assume that the speed
of sound in the air within the tube is 343 m/s. What frequency do you hear from
the tube if you jam your ear against one end of the tube?
With your ear closing one end, the fundamental frequency is given by
f = v/4L = 343/(4)(0.67) = 128 Hz
2. Weak background noises from a room set up the fundamental standing wave in a
cardboard tube of length L = 67.0 cm with two open ends. Assume that the speed
of sound in the air within the tube is 343 m/s. What frequency do you hear from
the tube if you move your head away enough so that the tube has two open ends?
With both ends open, the fundamental frequency is given by
f = v/2L = 343/(2)(0.67) = 256 Hz
3. The audible frequency range for normal hearing is about 20 Hz to 20 kHz. What
are the wavelengths of sound waves at these frequencies at 20 C?
λ=v/f
for 20 Hz, λ = 343/20 = 17.15 m
for 20 kHz, λ = 343/20000 = 1.715 cm
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4. The shortest wavelength emitted by a bat is about 3.3 mm. What is the
corresponding frequency?
f = v / λ = 343/0.0033 = 104 kHz
5. Diagnostic ultrasound of frequency 4.5 MHz is used to examine tumors in soft
tissue. What is the wavelength in air of such a sound wave?
λ = v / f = 343/(4.5x106) = 76.2 µm
6. Diagnostic ultrasound of frequency 4.5 MHz is used to examine tumors in soft
tissue. If the speed of sound in tissue is 1500 m/s, what is the wavelength of this
wave in tissue?
λ = v / f = 1500/(4.5x106) = 333.3 µm
7. A conical loudspeaker has a diameter of 15.0 cm. At what frequency will the
wavelength of the sound it emits in air be equal to its diameter?
λ = 15.0 cm
f = v / λ = 343/0.15 = 2.286 kHz
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