Parasaurolophus was a dinosaur whose distinguishing feature was a
hollow crest on the head. The 1.4...
Parasaurolophus was a dinosaur whose distinguishing feature was a
hollow crest on the head. The 1.4 m long hollow tube in the crest
had connections to the nose and throat, leading some investigators
to hypothesize that the tube was a resonant chamber for
vocalization
If you model the tube as an open-closed system, what are the first
three resonant frequencies? (Use 350 m/s for the speed of
sound.)
Enter your answers using two significant figures in ascending order
separated by commas.
f = _______Hz, ________Hz, ________Hz
Solutions
Expert Solution
Concepts and reason
The concepts required to solve the given problem are frequency, wavelength, and velocity.
Initially, derive the expressions for the first three resonant frequencies by using the relation between velocity, frequency, and wavelength. Finally, calculate the first three resonant frequencies by using the expressions of the frequencies of open closed pipe.
Fundamentals
The lowest frequency which is produced by the oscillation of an object is called as fundamental frequency.
The fundamental frequency of open closed pipe is,
f=4Lv
Here, f is the frequency of sound, v is the velocity of sound, and L is the length of the tube.
The velocity of the sound in air is,
v=fλ
Here, λ is the wavelength.
For the first harmonic of the closed pipe, the length of the tube is same as the length of ¼ wavelength.
L=4λλ=4L
The velocity of the sound in air is,
v=f1λ
Here, f1 is the first resonant frequency.
Substitute 4L for λ .
v=f1(4L)f1=4Lv
For the third harmonic of the closed end pipe, the length of the tube is,
L=43λλ=34L
Since 3/4 of a wavelength fit into the tube.
The velocity of the sound in air is,
v=f2λ
Here, f2 is the second resonant frequency.
Substitute 34L for λ .
v=f2(34L)f2=4L3v
For the Fifth harmonic of the closed end pipe, the length of the tube is,
L=45λλ=54L
Since 5/4 of a wavelength fit into the tube.
The velocity of the sound in air is,
v=f3λ
Here, f3 is the third resonant frequency.
Substitute 54L for λ .
v=f3(54L)f3=4L5v
The first resonant frequency is,
f1=4Lv
Substitute 350 m/s for v and 1.4 m for L.
f1=4(1.4m)350m/s=62.5Hz
The second resonant frequency is,
f2=4L3v
Substitute 350 m/s for v and 1.4 m for L.
f2=4(1.4m)3(350m/s)=187.5Hz
The third resonant frequency is,
f3=4L5v
Substitute 350 m/s for v and 1.4 m for L.
f3=4(1.4m)5(350m/s)=312.5Hz
Ans:
The first three resonant frequencies are 62 Hz, 190 Hz, and 310 Hz.
(1) Parasaurolophus was a dinosaur whose distinguishing feature
was a hollow crest on the head. The 1.2m long hollow tube in the
crest had connections to the nose and throat, leading some
investigators to hypothesize that the tube was a resonant chamber
for vocalization. (Figure 1)
If you model the tube as an open-closed system, what are the
first three resonant frequencies? (Use 350 m/s for the speed of
sound.)
Enter your answers using two significant figures in ascending
order...
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