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Parasaurolophus was a dinosaur whose distinguishing feature was a hollow crest on the head. The 1.5-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? Assume a speed of sound of 350 m/s.

Answer :

boffeemadrid

Answer:

58.33 Hz

175 Hz

291.67 Hz

Explanation:

L = Length of tube = 1.5 m

v = Speed of sound in air = 350 m/s

The first resonant frequency is given by

[tex]f_1=\dfrac{v}{4L}\\\Rightarrow f_1=\dfrac{350}{4\times 1.5}\\\Rightarrow f_1=58.33\ Hz[/tex]

The first resonant frequency is 58.33 Hz

The second resonant frequency is given by

[tex]f_2=3\dfrac{v}{4L}\\\Rightarrow f_2=3\dfrac{350}{4\times 1.5}\\\Rightarrow f_2=175\ Hz[/tex]

The first resonant frequency is 175 Hz

The third resonant frequency is given by

[tex]f_3=5\dfrac{v}{4L}\\\Rightarrow f_3=5\dfrac{350}{4\times 1.5}\\\Rightarrow f_3=291.67\ Hz[/tex]

The first resonant frequency is 291.67 Hz

The first three resonance frequencies are 58.3Hz, 175Hz and 291.Hz respectively.

Data given;

  • velocity v = 350 m/s
  • length of the pipe = 1.5m

Resonance Frequency

For an organ pipe with a closed end, the resonance frequency is given as

[tex]f(n)= \frac{nV}{4L} [/tex]

and the possible values of n are 1, 3, 5, 7, 9...

First Resonance Frequency

For the first frequency, the values are

[tex]f(1)=\frac{1*350}{4*1.5}\\ f(1) = 350/6 = 58.3Hz[/tex]

Second Resonance Frequency

For the second frequency, the values are

[tex]f(3) = \frac{3*350}{4*1.5}\\ f(3)= 1050/6\\ f(3) = 175Hz[/tex]

Third Resonance Frequency

For the third frequency, the values are

[tex]f(5) = \frac{5*350}{4*1.5} \\ f(5) = 1750/6 = 291.6[/tex]

From the calculation, the first three resonance frequencies are 58.3Hz, 175Hz and 291.6Hz respectively.

learn more about resonance frequencies here;

https://brainly.com/question/3292140

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