Answer:
(a) when the pipe is closed at one end;
F₀ = 73.92 Hz
F₁ = 221.76 Hz
F₂ = 369.6 Hz
F₃ = 517.44
(b) when the pipe is open at both ends;
F₀ = 147.85 Hz
F₁ = 295.7 Hz
F₂ = 443.55 Hz
F₃ = 591.4 Hz
Explanation:
Given;
length of the pipe, L = 116 cm = 1.16 m
speed of sound in air, v = 343 m/s
The formula below will be used to determine the different frequencies of the pipe at different wavelengths
[tex]F = \frac{V}{\lambda}[/tex]
where;
F is frequency
λ is wavelength
(a) when the pipe is closed at one end;
Wavelength for fundamental frequency;
L = Node -------> Antinode
L = λ/4
λ = 4L
the fundamental mental frequency;
[tex]F_0 = \frac{V}{4L} = \frac{343}{4 \times 1.16} = 73.92 \ Hz[/tex]
Wavelength for first overtone or audible frequency;
L = Node -----> Node + Node ----> Antinode
L = λ/2 + λ/4
L = 3λ/4
λ = 4L/3
[tex]F_1 = \frac{V}{4L/3} = 3(\frac{V}{4L} ) = 3F_0 = 3 (73.92) = 221.76 \ Hz[/tex]
Thus, the next two audible frequencies will be multiple of next consecutive odd numbers after 3 (i.e 5 and 7);
F₂ = 5F₀ = 5(73.92) = 369.6 Hz
F₃ = 7F₀ = 7(73.92) = 517.44 Hz
(b) when the pipe is open at both ends;
Wavelength for fundamental frequency;
L = Antinode ----> Node + Node -----> Antinode
L = λ/4 + λ/4
L = λ/2
λ = 2L
the fundamental frequency;
[tex]F_0 = \frac{V}{2L} = \frac{343}{2\times 1.16} = 147.85 \ Hz[/tex]
The first three audible overtones will be multiples of the 3 consecutive positive integers after 1 (i.e 2, 3, 4)
F₁ = 2F₀ = 2(147.85) = 295.7 Hz
F₂ = 3F₀ = 3(147.85) = 443.55 Hz
F₃ = 4F₀ = 4(147.85) = 591.4 Hz
