Question

Two identical wires, one made of iron and the other of aluminium are stretched alongside on a sonometer board by equal stretching forces. [Density of iron $=7.5\text{gm/cc}$, density of aluminium $=2.7\text{g/cc}$]. Find the frequency of the lowest harmonic for which both wires vibrate in unison, given that the length of the wires is $1\text{m}$, their diameters $1\text{m}m$ and tension $75\pi$.

Answer 1

Velocity of transverse wave on a string is given by
$v=\sqrt{\frac{T}{\mu }}$

where $\mu =$ linear mass density $=\mu \times A$

Frequency of standing waves on string, $f=\frac{nv}{2l}$

Thus, for same frequency of vibration,
$\frac{n_{Fe}}{n_{Al}}=\frac{\sqrt{\frac{T}{{\rho }_{Al}{A}_{Al}}}.\frac{1}{2l}}{\sqrt{\frac{T}{{\rho }_{Fe}{A}_{Fe}}}.\frac{1}{2l}}$

Given, $A_{Al}=A_{Fe}$ (diameters equal)
${\rho }_{Al}=2.7\text{g/cc};{\rho }_{Fe}=7.5\text{g/cc}$
Also, tension & lengths of the wires are equal.

$\Rightarrow \frac{n_{Fe}}{n_{Al}}=\sqrt{\frac{{\rho }_{Fe}}{{\rho }_{Al}}}=\sqrt{\frac{7.5}{2.7}}=\frac{5}{3}$

Hence, for lowest common frequency of vibration,
Fifth harmonic of $Fe$ = third harmonic of $Al$ wire.

Using $n_{Fe}=5;$
$f=\frac{5}{2\times 1}\sqrt{\frac{75\pi \times 4}{\pi \times {10}^{-6}\times 7.5\times {10}^3}}=500\text{Hz}$