A closed organ pipe of length λ = 100 cm is cut into two unequal pieces. The fundamental frequency of the new closed organ pipe piece is found to be same as the frequency of first overtone of the open organ pipe piece. Determine the length of the two pieces and the fundamental tone of the open pipe piece. Take velocity of sound = 320 m/s.
Text Solution
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(20, 80 cm, 200 Hz)
= λ 1 , 
= λ 2
λ 1 = 
= λ 1 , λ 2 = 4 λ 2
f 1 = f 2
= 
λ 1 = 4 λ 2
λ 1 + λ 2 = 100
5 λ 2 = 100
λ 2 = 20 cm
λ 1 = 80 cm
Fundamental frequency of open pipe.
f = 
= 
= 
= 200 Hz
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