An open organ pipe having fundamental frequency (n) is in unison with a vibrating string. If the tube is…
An open organ pipe having fundamental frequency (n) is in unison with a vibrating string. If the tube is dipped in water so that $75 \%$ of the length of the tube is inside the water then the ratio of fundamental frequency of the air column of dipped tube with that of string will be (Neglect end corrections)
$1: 1$
$2: 1$
$2: 3$
$3: 2$
Solution
$\mathrm{n}_{\text {open }}=\frac{\mathrm{v}}{2 \mathrm{~L}}.... (i)$
When dipped in water, pipe becomes closed at one end and open at the other.
Length available for resonance is
$\begin{aligned}
l_1 & =25 \% \times \mathrm{L} \\
& =\mathrm{L} \times \frac{25}{100} \\
& =\mathrm{L} / 4
\end{aligned}$
$\therefore \quad \mathrm{n}_{\text {closed }}=\frac{\mathrm{v}}{4 l_1}=\frac{\mathrm{v}}{4 \times \frac{\mathrm{L}}{4}}=\frac{\mathrm{v}}{\mathrm{L}}.... (ii)$
Comparing (i) and (ii),
$\therefore \quad \frac{\mathrm{n}_{\text {closed }}}{\mathrm{n}_{\text {open }}}=\frac{\left(\frac{\mathrm{v}}{\mathrm{L}}\right)^{\prime}}{\left(\frac{\mathrm{v}}{2 \mathrm{~L}}\right)}=\frac{2}{1}$