Answer:
The 9th harmonic is 1,010.67 Hz.
Explanation:
Given;
length of the pipe, L = 1.23 m
speed of sound at 0°C = 331.5 m/s
A pipe closed at one end is known as a closed pipe;
The wavelength of the sound for the first harmonic is calculated as;
L = Node ------ > Antinode
[tex]L = \frac{\lambda }{4} \\\\\lambda = 4L[/tex]
First harmonic: [tex]F_0 = \frac{V}{\lambda} = \frac{V}{4L}[/tex]
The wavelength of the sound for the first harmonic is calculated;
L = Node ----> Node + Node ------> Antinode
[tex]L = \frac{\lambda}{2} + \frac{\lambda}{4} = \frac{2\lambda+\lambda}{4} = \frac{3\lambda}{4} \\\\\lambda = \frac{4 L}{3}[/tex]
Second harmonic: [tex]F_1 = \frac{V}{\lambda} = \frac{3V}{4L}[/tex]
F₁ = 3F₀
The increment from F₀ to F₁ is 1 to 3; (odd number).
(1, 3, 5, 7, 9, 11, 13, 15, 17, 19..... n+2, where n is odd number)
The 9th harmonic = F₈
F₈ = 15F₀
[tex]F_8= 17(F_0) = 15 (\frac{V}{4L} )\\\\F_8 = 15(\frac{331.5}{4\times 1.23} )\\\\F_8 = 1,010.67 \ Hz[/tex]
Therefore, the 9th harmonic is 1,010.67 Hz.
The frequency of the 9th harmonic if the pipe is closed at one end is 605.48 Hz.
Based on the given information,
• The length (l) of the shortest pipe given is 1.23 m.
The speed of sound at 0 degree C is,
[tex]V = 331\sqrt{1+\frac{T}{273} } \\V = 331\sqrt{1+\frac{0}{273} } \\V = 331 m/s[/tex]
Now the condition is that the pipe is closed at one end. The formula to use in the given case is,
[tex]fn = \frac{nv}{4l} \\[/tex]
Here l is the length of the organ pipe.
Now the frequency of the 9th harmonic will be,
[tex]f9 = \frac{9v}{4l} \\\\f9 = \frac{9*331}{4*1.23} \\f9 = 605.48 Hz[/tex]
Thus, the frequency of the 9th harmonic if the pipe is closed at one end is 605.48 Hz.
To know more about:
https://brainly.com/question/15085005
