A pipe of \(1 m\) length is closed at one end. Taking the speed of sound in air as \(320 \mathrm{~ms}^{-1}\)…
A pipe of \(1 m\) length is closed at one end. Taking the speed of sound in air as \(320 \mathrm{~ms}^{-1}\) the air column in the pipe cannot resonate for the frequency (in Hz)
80
560
160
240
Solution
Given data for the closed organ pipe:
Velocity, \(v=320 \mathrm{~ms}^{-1}\)
Length, \(l=1 m\)
For an organ pipe whose one end is closed, only odd harmonics containing odd multiples of the fundamental frequency are present.
Fundamental frequency (Resonance Frequency) of the closed pipe can be given as:
\(n_1=\frac{v}{4 l}\)
Second mode, \(n_2=3 n_1\)
Third mode, \(n_3=5 n_1\)
Fourth mode, \(n_4=7 n_1\)
Fifth mode, \(n_5=9 n_1\)
So, \(n_1=\frac{320}{4}=80 \mathrm{~Hz}\)
\(n_2=3 \times 80=240 \mathrm{~Hz}\)
\(n_3=5 \times 80=400 \mathrm{~Hz}\)
\(n_4=7 \times 80=560 \mathrm{~Hz}\)
\(n_5=9 \times 80=720 \mathrm{~Hz}\)
So according to the result, option (B) can not be the multiple of the resonance frequency.
So, the correct option will be (\(B\))