(i) A cylindrical metal tube has a length of 60 cm and is open at both ends. Find the frequencies between…
(i) A cylindrical metal tube has a length of 60 cm and is open at both ends. Find the frequencies between 1000 Hz and 2000 Hz at which the air column in the tube can resonate. (Take, speed of sound in air = 340 m/s)
(ii) Find the greatest length of an organ pipe open at both ends that will have its fundamental frequencies in the normal hearing range (30-2000 Hz). (Take, speed of sound in air = 340 ms$^{-1}$)
(ii) Two successive resonance frequencies in an open organ pipe are 1944 Hz and 2592 Hz, respectively. Find the length of tube. (Take, speed of sound in air is 324 ms$^{-1}$)
Solution
Sol. (i) For open pipe, fundamental frequency,
$f=\dfrac{v}{2l}=\dfrac{340}{2\times0.6}=283.33\ \text{Hz}$
Possible frequencies of open pipe = $nf$
where, $n=1,2,3,\dots$
$\therefore$ Possible frequencies of open pipe =
(284, 568, 852, 1136, 1420, 1704, 1988) Hz
Frequencies between 1000 and 2000 Hz
$=1136\ \text{Hz},\ 1420\ \text{Hz},\ 1704\ \text{Hz},\ 1988\ \text{Hz}$
(ii) Fundamental frequency, $f=\dfrac{v}{2l}\Rightarrow l=\dfrac{v}{2f}$
Hence, $l$ will be maximum, if $f$ is minimum
$\therefore\quad l_{\max}=\dfrac{340}{2\times30}=5.6\ \text{m}$
(iii) The difference in successive frequencies of a pipe (open or closed) = $\dfrac{v}{2l}$
$\Rightarrow\ 2592-1944=\dfrac{324}{2l}\ \Rightarrow\ l=\dfrac{1}{4}\ \text{m}=0.25\ \text{m}$
Answer: 0.25 m