Question

A $50\text{cm}$ long wire having a mass of $20\text{g}$ is fixed at two ends and is vibrated in its fundamental mode. A $75\text{cm}$ long closed organ pipe, placed with its open end near the wire, is set up into resonance in its first overtone mode by the vibrating wire. Find the tension in the wire. Speed of sound in air is $340\text{m/s}$.

• A. $2446\text{N}$
• B. $4264\text{N}$
• C. $4624\text{N}$
• D. $4552\text{N}$

Answer 1

The correct option is C $4624\text{N}$


Given that
Length of wire, $l_w=50\text{cm}=50\times {10}^{-2}\text{m}$
Mass of wire, $m=20\text{g}=20\times {10}^{-3}\text{kg}$
Linear density of the wire, $\mu =\frac{m}{l_w}=\frac{20\times {10}^{-3}}{50\times {10}^{-2}}=\frac{2}{50}=\frac{1}{25}\text{kg/m}$

So fundamental frequency of the wire is
$f_w=f_1=\frac{1}{2l_w}\sqrt{\frac{T}{\mu }}=\frac{1}{2\times 50\times {10}^{-2}}\sqrt{\frac{T}{\left(\frac{1}{25}\right)}}-----\left(1\right)$

Length of closed organ pipe, $l_P=75\text{cm}=75\times {10}^{-2}\text{m}$
Speed of sound, $v=340\text{m/s}$

Resonance frequency of closed organ pipe is given by
$f_n=\frac{(2n-1)v}{4l_P}$ for $n=1,2,\mathrm{3..}$
First overtone of closed organ pipe is
$f_P=\frac{3v}{4l_P}=\frac{3\times 340}{4\times 75\times {10}^{-2}}-----\left(2\right)$

Since both wire and organ pipe are in resorance,
$f_w=f_P$
$\Rightarrow \frac{1}{2\times 50\times {10}^{-2}}\sqrt{\frac{T}{\left(\frac{1}{25}\right)}}=\frac{3\times 340}{4\times 75\times {10}^{-2}}$
$\Rightarrow \sqrt{T}=\frac{3\times 340}{4\times 75\times {10}^{-2}}\times \frac{2\times 50\times {10}^{-2}}{5}$
$\Rightarrow \sqrt{T}=68$
$\Rightarrow T=4624\text{N}$