Question

Find the number of possible natural frequencies of an air column in an open pipe frequencies of which lie below $1250Hz$. The length of the pipe is $85cm$ .
(speed of sound $ = 340m{s^{ - 1}}$ )
A) $3$
B) $4$
C) $5$
D) $6$

Answer 1

Hint:We can use the formula for fundamental frequency of an open organ pipe in terms of length of pipe to find the number of possible natural frequencies of an air column. We will, then, filter the frequencies which are below given frequency, from the set of found frequencies.

Complete step by step answer:
In an open organ pipe, half of wavelength will be the perfect divisor of the length of pipe, so,
$l = \dfrac{{m\lambda }}{2}$
Where $\lambda $ is the wavelength of wave,
$l$ is length of pipe,
$m$ is the factor by which half of wavelength divides length of pipe, it means, $m = 1,2,3,...$
On simplification, we get,
$\lambda = \dfrac{{2l}}{m}$
Now, formula for velocity of sound wave is,
$v = f\lambda $
Where $v$ is velocity of sound, $f$ is its frequency and $\lambda $ is its wavelength.
Now on putting expression of wavelength we get,
$v = f\dfrac{{2l}}{m}$
On putting values in standard notation, we get,
$340 = f \times \dfrac{{2 \times 0.85}}{m}$ ( $v = 340m{s^{ - 1}}$ given in the question )
On evaluating, we get,
\[f = \dfrac{{340m}}{{1.7}}\],
So we can say,
${f_m} = \dfrac{{340m}}{{1.7}}$
So we get,
${f_1} = \dfrac{{340 \times 1}}{{1.7}}$
On evaluation, we get,
${f_1} \approx 200Hz$
Similarly,
${f_2} = 400Hz$
${f_3} = 600Hz$
${f_4} = 800Hz$
${f_5} = 1000Hz$
${f_6} = 1200Hz$
All these frequencies are below given frequency $1250Hz$
So number of possible natural frequencies is $6$

So the correct answer is option (D).

Note:The given pipe was an open organ pipe from both ends, so we used the formula $l = \dfrac{{m\lambda }}{2}$. Same is for closed pipe from both ends, but if the pipe would have been opened from one end and closed from one end, so we had to use the formula as $l = \dfrac{{(2n + 1)\lambda }}{4}$