Question

Two consecutive overtones produced by a narrow air column closed at one end open at the other are $750Hz$ and $1050Hz.$ Then the fundamental frequency from the column in $Hz$ is :

Answer 1

Formula of frequency at any harmonics or overtone

$f=\frac{(2n+1)v}{4L}$ ,

here $(2n+1)$ denote harmonic and $n$ denote overtone.

at any $nth$ overtone we will find frequency

it is given by $f_1=\frac{(2n+1)v}{4L}$$=750$

for next consecutive overtone replace $n$ by $(n+1)$

we get $f_2=\frac{\left(2\right(n+1)+1)v}{4L}$$=\frac{(2n+3)v}{4L}$$=1050$

divide both equations:

$\frac{2n+1}{2n+3}$$=\frac{750}{1050}$

on solving we get $n=2$

putting it in any equation above we will get value of $\frac{v}{L}$$=600$

now

fundamental frequency is $f_0=\frac{v}{4L}$

so, $f_0=\frac{600}{4}$$=150Hz$