Question

An open organ pipe has a length of $10\text{cm}$ . If the speed of sound in air is $340\text{m/s}$ and the audible range is $(20-20,000\text{Hz})$, then choose the correct option(s):

• A. The fundamental frequency of vibration of this pipe is $1700\text{Hz}$.
• B. $11$ is the highest harmonic of such tube in audible range.
• C. $6800\text{Hz}$ is the third overtone frequency.
• D. The fundamental frequency of vibration of this pipe is $1200\text{Hz}$.

Answer 1

Answer: (A), (B), (C)

The correct options are
A The fundamental frequency of vibration of this pipe is $1700\text{Hz}$.
B $11$ is the highest harmonic of such tube in audible range.
C $6800\text{Hz}$ is the third overtone frequency.

Given that,
Length of pipe, $L=10\text{cm}\text{or}0.1\text{m}$
Speed of sound, $v=340\text{m/s}$

Mode of vibration of air column in an open organ pipe is given by

$f_n=\frac{nv}{2L}$ where $n=1,2,\mathrm{3...}$

Fundamental frequency, $f_1=\frac{v}{2L}$
From the data given in the question,
$f_1=\frac{340}{2\times 10\times {10}^{-2}}=1700\text{Hz}$

For maximum audible frequency,
$f_n=20000\text{Hz}$
So
$f_n=\frac{nv}{2L}$
$\Rightarrow n=\frac{20000\times 2\times 0.1}{340}=11.76\simeq 11$

For third overtone or fourth harmonic,

$f_4=\frac{4v}{2L}=\frac{4\times 340}{2\times 0.1}=6800\text{Hz}$

Thus, options (a) , (b) and (c) are the correct answers.