A church organ consists of pipes, each open at one end, of different lengths. The minimum length is 30 mm and the longest is 4 m. Calculate the frequency range of fundamental notes. (speed of sound = 340 m s⁻¹). (Ans: 21 Hz to 2833 Hz)
Data Given:
Speed of sound = v = 340 m s⁻¹
Minimum length = l₁ = 30 mm = 0.03 m
Longest length = l₂ = 4 m
To Find:
Frequency range (f₁ = ? and f₂ = ?)
Solution:
The formula for fundamental frequency for closed pipe at minimum length (l = 0.03 m)
f₁ =
putting values
f₁ =
f₁ = 2833 Hz -----------Ans. (1)
For maximum length (l₂ = 4 m)
f₂ =
putting values
f₂ =
f₁ = 21.25 Hz
f₁ = 21 Hz -----------Ans. (2)
Hence the frequency range is from 21 Hz to 2833 Hz.
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