Question

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

• A. $4$
• B. $6$
• C. $8$
• D. $12$

Answer 1

The correct option is B $6$


Given: $f=1250\text{Hz},l=85\text{cm}=0.85m,v=340\text{m/s}$

For closed organ pipe,

$f=\frac{(2n+1)v}{4l}(where,n=0,1,\mathrm{2.....})$

$\frac{(2n+1)v}{4l}<1250$

$(2n+1)<1250\times \frac{4\times 0.85}{340}$

$(2n+1)<12.5$

$2n<11.50$

$n<5.25$

$\therefore n=0,1,2,\mathrm{3......5}$

So we have $6$ possibilities.

Hence, option $\left(C\right)$ is correct.