Question

A closed pipe has length 0.6m. The air inside the pipe is maintained at temperature ${27}^{\circ }C$. Calculate the fundamental frequency and the frequency of the next two overtones.
(given velocity of sound in air at $0^{\circ }C=330m{s}^{-1}$)

• A. 144 Hz, 288 Hz, 432 Hz
• B. 144 Hz, 432 Hz, 720 Hz
• C. 144 Hz, 216 Hz, 288 Hz
• D. None of these

Answer 1

The correct option is B

144 Hz, 432 Hz, 720 Hz

$v_1=330m{s}^{-1}$
$v_2=?$
$T_1=0^{\circ }C\Rightarrow 273K$
$T_2={27}^{\circ }C\Rightarrow 300K$
$\frac{v_1}{v_2}=\sqrt{\frac{T_1}{T_2}}$
$\frac{330}{v_2}=\sqrt{\frac{273}{300}}=0.954\therefore v_2=\frac{330}{0.954}=345.9m{s}^{-1}$
Fundamental frequency of a closed pipe= $f=\frac{v}{4l}$
$f=\frac{345.9}{4\times 0.6}=144Hz$
The first overtone which is the third harmonic will have frequency = $3f=432Hz$
The second overtone is the fifth harmonic
This will have frequency $5f=5\times 144=720Hz$