An organ pipe is 116 cm long. Determine the fundamental and first three audible overtones if the pipe is (a) closed at one end, and (b) open at both ends.

(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

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;

Wavelength for first overtone or audible frequency;

L = Node ----->  Node  +  Node ----> Antinode

L =          λ/2                 +        λ/4

L = 3λ/4

λ = 4L/3

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;

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

f=ma newton's second law 7x8=56n