Figure shows an aluminium wire of length 60 cm joined to a steel wire of length 80 cm and stretched between…
Figure shows an aluminium wire of length 60 cm joined to a steel wire of length 80 cm and stretched between two fixed supports. The tension produced is 40 N. The cross-sectional area of the steel wire is 1.0 mm2 and that of the aluminium wire is 3.0 mm2. What could be the minimum frequency of a tuning fork which can produce standing waves in the system with the joint as a node? The density of aluminium is 2.6 g/cm3 and that of steel is 7.8 g/cm3.
Solution
Sol. According to the question,
Steel Aluminium
$A_s = 1\ \mathrm{mm}^2$ $A_a = 3\ \mathrm{mm}^2$
$\rho_s = 7.8\ \mathrm{g/cm}^3$ $\rho_a = 2.6\ \mathrm{g/cm}^3$
$L_s = 80\ \mathrm{cm}$ $L_a = 60\ \mathrm{cm}$
Let $p$th harmonic of steel wire coincides with $q$th harmonic of aluminium wire.
The
$n = \frac{p}{2L_s}\sqrt{\frac{T}{\rho_s A_s}} = \frac{q}{2L_a}\sqrt{\frac{T}{\rho_a A_a}}$
so
$\displaystyle \frac{p}{q} = \frac{L_s}{L_a}\sqrt{\frac{\rho_s A_s}{\rho_a A_a}} = \frac{80}{60}\sqrt{\frac{7.8\times 1}{2.6\times 3}} = \frac{4}{3}$
4th harmonic of steel wire coincides with 3rd harmonic of aluminium wire, the minimum frequency of tuning fork can be determined by taking $p=4$ or $q=3$.
$\displaystyle n_{\rm min} = \frac{p}{2L_s}\sqrt{\frac{T}{\rho_s A_s}}$
$= \frac{4}{2\times 0.8}\sqrt{\frac{40}{7.8\times 10^3\times 1\times 10^{-6}}}$
$= 2.5\times 71.6 = 179\ \mathrm{Hz}$
Answer: $179\ \mathrm{Hz}$