Question

Consider the situation shown in figure. The wire which has a mass of $2.00\text{g}$ and length $20\text{cm}$, oscillates in its second harmonic and sets the air column in the closed tube into vibrations in its fundamental mode. Assuming that the speed of sound in air is $340{\text{ms}}^{-1}$, find the tension in the wire.

• A. $5.6\text{N}$
• B. $2.9\text{N}$
• C. $11.6\text{N}$
• D. $28\text{N}$

Answer 1

The correct option is B $2.9\text{N}$

Let $n$ be the frequency of the tuning fork, $T$ be the tension in the string.
Given,
Length of the wire $l=20\text{cm}\text{or}0.2\text{m}$
Mass of the wire $m=2\text{g}=2\times {10}^{-3}\text{kg}$
Length of the tube, $L=1\text{m}$

So, linear mass density $\left(\frac{m}{l}\right)={10}^{-2}\text{kg/m}$

Frequencies of oscillation of the wire are given by
$f_{nw}=\frac{n}{2\ell }\sqrt{\frac{T}{\mu }}$

For second harmonic, $f_{2w}=\left(\frac{2}{2\ell }\right)\sqrt{\frac{T}{\mu }}$

Fundamental frequency of vibration of the air column of a closed organ pipe

$f_{1t}=\frac{v}{4L}=\frac{340}{4\times 1}=85\text{Hz}$

From the data given in the question,
$f_{1t}=f_{2w}$
$f_{1t}=\frac{n}{2\ell }\sqrt{\frac{T}{\mu }}$

$\Rightarrow T=\frac{f_{1t}^2\times 4{\ell }^2\times \mu }{n^2}$

$\Rightarrow T=\frac{{85}^2\times 4\times (0.2{)}^2\times {10}^{-2}}{4}=2.9\text{N}$

Thus, option (b) is the correct answer.