Question

An electronically driven loudspeaker is placed near the open end of a resonance column apparatus. The length of air column in the tube is $40\text{cm}$. The frequency of the loudspeaker can be varied between $20\text{Hz}-2\text{kHz}$. Choose among the following frequencies at which the column will resonate.

[Speed of sound in air $=320{\text{ms}}^{-1}$ ]

• A. $600\text{Hz}$
• B. $800\text{Hz}$
• C. $1000\text{Hz}$
• D. $1400\text{Hz}$

Answer 1

Answer: (A), (C), (D)

$1400\text{Hz}$

Resonance column apparatus is equivalent to a closed organ pipe.
Here, $\ell =40\text{cm}\text{or}0.4\text{m}$
Speed of sound in air, $v_{air}=320\text{m/s}$
Modes of vibration in a closed organ pipe are given by
$f_n=\frac{(2n-1)v_{air}}{4\ell }$

Substituting the given data,
$f_n=\frac{(2n-1)\times 320}{4\times 40\times {10}^{-2}}=200(2n-1)\text{Hz}\text{where}n=1,2,3,\mathrm{4....}$

Thus we get,

Fundamental mode $\left(f_1\right)=200\text{Hz}$
First overtone $\left(f_2\right)=600\text{Hz}$
Second overtone $\left(f_3\right)=1000\text{Hz}$
Third overtone $\left(f_4\right)=1400\text{Hz}........$

Hence, options (a) , (c) and (d) are the correct answers.