Question

Figure (15-E10) 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-selectional area of the steel wire is $1.0m{m}^2$ and that of the aluminium wire is $3.0m{m}^2$. 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 $2.6gc{m}^{-3}$ and that of steel is $7.8gc{m}^{-3}$

Answer 1

$P_S=7.8gm/c{m}^3$

$P_A=2.6gm/c{m}^3$

$m_s=P_S{A}_S$

$=7.8\times {10}^{-2}gm/cm$

(m = mass per unit length)

$=7.8\times {10}^{-3}kg/m$

$m_A=P_A{A}_A$

$=2.6\times {10}^{-2}\times 3gm/cm$

$7.8\times {10}^{-2}gm/cm$

$7.8\times {10}^{-3}kg/m$

A node is always placed in the joint. Since aluminium and steel rod has same mass per unit length, velocity of wave in both of them is same.

$\Rightarrow v=\sqrt{\left(\frac{T}{m}\right)}$

$=\sqrt{\left\{\frac{40}{(7.8\times {10}^{-3})}\right\}}$

$=\sqrt{\left(\frac{(4\times {10}^4)}{7.8}\right)}$

$=71.6m/s$

For minimum frequency there would be maximum wavelength. For maximum wavelength minimum no. of loops are to be produced.

$\therefore$ Maximum distance of a loop = 20 cm

$\Rightarrow Wavelength=\lambda =2\times 20$

$=40cm=0.4m$

$\therefore f=\frac{v}{\lambda }=\frac{71.6}{0.4}=180Hz$