Physics - Wave on String Question with Solution | TestHub

Two successive frequencies of closed pipe

$\frac{n\mathrm{v}}{4l}=220$....(i)

$\frac{\left(n+2\right)\mathrm{v}}{4l}=260$......(ii)

Dividing (ii) by (i), we get

$\frac{n+2}{n}=\frac{260}{220}=\frac{13}{11}$

$11n+22=13n$

$n=11$

So,$11\frac{\mathrm{v}}{4l}=220$

$\frac{\mathrm{v}}{4l}=20$

So fundamental frequency is 20 Hz.