Question

A pipe of length $85cm$ is closed from one end. Find the number of possible natural oscillations of the air column in the pipe whose frequencies lie below $1250Hz$. The velocity of sound in the air is $340\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right..$

• A. $6$
• B. $4$
• C. $12$
• D. $8$

Answer 1

The correct option is A

$6$

Step 1. Given Data,

Length of pipe, $L=85cm=0.85m$

Velocity of sound in air, $v=340\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$

Step 2. Fundamental frequency of closed organ pipe is,

$f=\frac{v}{4L}$ (where, $f$ is the fundamental frequency.)

$$\begin{gathered} \Rightarrow f=\frac{340}{(4\times 0.85)} \\ \Rightarrow f=100Hz. \end{gathered}$$

Step 3. To calculate the frequency:

$$\begin{gathered} f=\frac{v}{\lambda } \\ \Rightarrow f=\frac{340}{(4\times 0.85)} \\ \Rightarrow f=100Hz. \end{gathered}$$

The natural frequencies of closed Organ pipe is, $f_n=\left(2n-1\right)f$

That is for $n=1,2,3,4,5,6,7,\dots$

The value of $f_n=100Hz,300Hz,500Hz,700Hz,900Hz,1100Hz,1300Hz,\dots$

Then the possible frequency is$100Hz,300Hz,500Hz,700Hz,900Hz,1100Hz$ below $1250Hz.$

Thus the number of possible natural oscillations whose frequency is below $1250$ is $6.$

Hence, the correct option is (A).