Question

A steel wire of length $116\text{cm}$ is connected to an aluminum wire of length $150\text{cm}$ and stretched between two fixed supports. The tension produced is $52\text{N}$. The cross-section of each wire is $2{\text{mm}}^2$. If a transverse wave is set up in the wire, find the lowest frequency for which standing waves with a node at the joint are produced. (density of aluminum $\left({\rho }_{al}\right)=2.6{\text{g/cm}}^3$ and density of steel $\left({\rho }_s\right)=7.8{\text{g/cm}}^3)$.

• A. $50\text{Hz}$
• B. $100\text{Hz}$
• C. $150\text{Hz}$
• D. $200\text{Hz}$

Answer 1

The correct option is B $100\text{Hz}$

Given that
${\rho }_{al}=2.6{\text{g/cm}}^3,{\rho }_s=7.8{\text{g/cm}}^3$
$A_a=A_s=2{\text{mm}}^2=2\times {10}^{-6}{\text{m}}^2$
Tension, $T=52\text{N}$
Length of aluminium wire, $L_{al}=150\text{cm}$
Length of steel wire, $L_s=116\text{cm}$
Let ${\mu }_{al}$ be the linear mass density of aluminium, and ${\mu }_s$ be the linear mass density for steel.

So, ${\mu }_{al}={\rho }_{al}\times A_{al}=(2.6\times {10}^3{\text{kg/m}}^3)\times (2\times {10}^{-6}{\text{m}}^2)=5.2\times {10}^{-3}\text{kg/m}$
and ${\mu }_s={\rho }_s\times A_s=(7.8\times {10}^3{\text{kg/m}}^3)\times (2\times {10}^{-6}{\text{m}}^2)=15.6\times {10}^{-3}\text{kg/m}$

Now for wave speed,
$v_{al}=\sqrt{\frac{T}{{\mu }_{al}}}=\sqrt{\frac{52}{5.2\times {10}^{-3}}}=100\text{m/s}$
and
$v_s=\sqrt{\frac{T}{{\mu }_s}}=\sqrt{\frac{52}{15.6\times {10}^{-3}}}\approx 58\text{m/s}$

For node at the joint, let us consider number of loops in aluminium is $n_{al}$ and in steel is $n_s$.
For lowest frequency, $f_{al}=f_s$
$\Rightarrow \frac{n_{al}{v}_{al}}{2L_{al}}=\frac{n_s{v}_s}{2L_s}$
$\Rightarrow \frac{n_{al}}{n_s}=\left(\frac{v_s}{v_{al}}\right)\times \left(\frac{L_{al}}{L_s}\right)=\frac{58}{100}\times \frac{150}{116}=\frac{3}{4}$

Therefore, for minimum frequency,
$f_{al}=\frac{3\times v_{al}}{2L_{al}}=\frac{3\times 100}{2\times 150\times {10}^{-2}}=100\text{Hz}$