Question

An aluminum wire of length $60\text{cm}$ is joined to a steel wire of length $80\text{cm}$ and stretched between two fixed supports. The tension produced is $40\text{N}$. Cross-sectional area is $1{\text{mm}}^2$ (steel) and $3{\text{mm}}^2$ (aluminum). Minimum frequency of the tuning fork which can produce standing waves with the joint as a node is nearly equal to

(density of $Al=2.6{\text{g cc}}^{-1}$ and density of steel $=7.8{\text{g cc}}^{-1})$

• A. $90\text{Hz}$
• B. $145\text{Hz}$
• C. $180\text{Hz}$
• D. $250\text{Hz}$

Answer 1

The correct option is C $180\text{Hz}$

frequency of wire oscillating is $f=\frac{n}{2l}\sqrt{\frac{T}{\mu }}=\frac{n}{2l}\sqrt{\frac{T}{A\rho }}$
Since frequency remains same for both the wires,
$\therefore \frac{n}{2l_1}\sqrt{\frac{T}{A_1{\rho }_1}}=\frac{P}{2l_2}\sqrt{\frac{T}{A_2{\rho }_2}}$
$\Rightarrow \frac{n}{P}=\frac{4}{3}$

Taking the frequency expression for the aluminium wire and for minimum frequency $P=3$
$f=\frac{3}{2\left(0.6\right)}\sqrt{\frac{40}{{10}^{-6}\times 2.6\times {10}^3\times 3}}$
$=\frac{1}{0.4}\sqrt{\frac{40\times {10}^3}{2.6\times 3}}$
$=\frac{100}{0.4}\sqrt{\frac{2}{3.9}}=180\text{Hz}$