Question

An aluminium wire of length $60cm$ is joined to a steel wire of length $80cm$ and stretched between two fixed supports. The tension produced is $40N$. Cross-sectional area is $1m{m}^2$ (steel) and $3m{m}^2$ (aluminium). The minimum frequency of the tuning fork which can produce standing waves with the joint as a node is _____________
(Density of $Al=2.6gc{c}^{-1}$ and density of steel = $7.8gc{c}^{-1}$)

• A. $90Hz$
• B. $145Hz$
• C. $180Hz$
• D. $250Hz$

Answer 1

Answer: (C)

$180Hz$

$f=\frac{n}{2l}\sqrt{\frac{T}{\mu }}=\frac{n}{2l}\sqrt{\frac{T}{A\rho }}$

Since frequency will remain the same

$\therefore \frac{n}{2l_1}\sqrt{\frac{T}{A_1{\rho }_1}}=\frac{p}{2l_2}\sqrt{\frac{T}{A_2{\rho }_2}}$

$Or\frac{n}{p}=\frac{l_1}{l_2}=\frac{p}{2l_2}\sqrt{\frac{T}{A_2{\rho }_2}}$

Or $\frac{n}{p}=\frac{l_1}{l_2}$ that is $\frac{n}{p}=\frac{4}{3}$

$f=\frac{3}{2\left(0.6\right)}\sqrt{\frac{40}{{10}^{-6}\times 2.6\times {10}^3\times 3}}=\frac{1}{0.4}\sqrt{\frac{40\times {10}^3}{2.6\times 3}}=\frac{100}{0.4}\sqrt{\frac{2}{3.9}}=180Hz$