# The fundamental frequency of a closed tube is the same frequency as the 2nd harmonic of a stretched string. The mass per unit length of the string is 6.10x10^-4 kg/m. The tension in the string is 48.0N...?

## What is the ratio of the length of the string to the length of the closed tube Ls/Lt?

Aug 20, 2017

When a tune of fundamental frequency is produced in one end closed tube of length $l$, then only one node at closed end and only one antinode at the open end are formed as shown in above figure. If $l a m {\mathrm{da}}_{c}$ be the wave length of the tune produced then $l = l a m {\mathrm{da}}_{c} / 4$

$\implies l a m {\mathrm{da}}_{c} = 4 l$

And the frequency of the tune will be ${f}_{c} = \frac{V}{l} a m {\mathrm{da}}_{c} = \frac{V}{4 l} \ldots . . \left[1\right]$,
where V is the velocity of sound $= 340 m \text{/} s$

When second harmonic is produced in streched string of length $L$ then there exist three nodes and two antinodes in the streched string as shown in Second figure. The frequency of this second harmonics ${f}_{s}$ is given by the following relation

${f}_{s} = \frac{p}{2 L} \sqrt{\frac{T}{m}}$

where

$p \to \text{harmonic formed} = 2$

$L \to \text{length of the streched string}$

$T \to \text{tension of the stretched string} = 48 N$

$m \to \text{mass per unit length "6.1xx10^-4kg"/} m$

By the given condition

${f}_{c} = {f}_{s}$

$\implies \frac{V}{4 l} = \frac{p}{2 L} \sqrt{\frac{T}{m}}$

$\implies \frac{L}{l} = \frac{2 p}{V} \sqrt{\frac{T}{m}}$

$\implies \frac{L}{l} = \frac{2 \times 2}{340} \sqrt{\frac{48}{6.1 \times {10}^{-} 4}} \approx 3.3$