Question

A pipe of $90\text{cm}$ is closed from one end. Find the number of possible oscillations of air column in the pipe whose frequency lies below $1300\text{Hz}$. (Give your answer as an integer)
[Speed of sound in air is $340\text{m/s}$].

Answer 1

Given that,
Length of pipe, $L=90\text{cm}\text{or}0.9\text{m}$
Frequency, $f=1300\text{Hz}$
Velocity of sound, $v=340\text{m/s}$

For the closed organ pipe
Frequency $\left(f_n\right)=(2n-1)\frac{v}{4L}$ [where $n=1,2,...$]

From the data given in the question,

$(2n-1)\frac{v}{4L}<1300$
$\Rightarrow (2n-1)<\frac{1300\times 4L}{v}$
Substituting the given data, we get

$(2n-1)<\frac{1300\times 4\times 0.9}{340}$
$\Rightarrow 2n-1<13.76$
$\Rightarrow 2n<14.76$
$\Rightarrow n<7.38$
For, $n=1,f_1=\frac{1\times 40}{4\times 0.9}=94.4\text{Hz}$

$n=2,f_2=\frac{3\times 340}{4\times 0.9}=283.3\text{Hz}$

$n=3,f_3=\frac{5\times 340}{4\times 0.9}=472.2\text{Hz}$

$n=4,f_4=\frac{7\times 340}{4\times 0.9}=661.08\text{Hz}$

$n=5,f_5=\frac{9\times 340}{4\times 0.9}=849.9\text{Hz}$

$n=6,f_6=\frac{11\times 340}{4\times 0.9}=1038.4\text{Hz}$

$n=7,f_7=\frac{13\times 340}{4\times 0.9}=1227.7\text{Hz}$

So we have seven possiblities $n=1,2,3,4,5,6\text{and}7$.