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-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 cm⁻³ and that of steel is 7⋅8 g cm⁻³.

Answer 1

Given:
Length of the aluminium wire (L_a)= 60 cm = 0.60 m
Length of the steel wire (L_s)= 80 cm = 0.80 m
Tension produced (T) = 40 N
Area of cross-section of the aluminium wire (A_a) = 1.0 mm²
Area of cross-section of the steel wire (A_s) = 3.0 mm²
Density of aluminium (ρ_a) = 2⋅6 g cm⁻³
Density of steel (ρ_s) = 7⋅8 g cm⁻³

$$\begin{gathered} \mathrm{Mass}\mathrm{per}\mathrm{unit}\mathrm{length}\mathrm{of}\mathrm{the}\mathrm{steel},m_{\mathrm{s}}={\rho }_{\mathrm{s}}\times A_s \\ =7.8\times {10}^{-2}\mathrm{gm}/\mathrm{cm} \\ =7.8\times {10}^{-3}\mathrm{kg}/\mathrm{m} \\ \mathrm{Mass}\mathrm{per}\mathrm{unit}\mathrm{length}\mathrm{of}\mathrm{the}\mathrm{aluminium},m_{\mathrm{A}}={\rho }_{\mathrm{A}}{A}_{\mathrm{A}} \\ =2.6\times {10}^{-2}\times 3\mathrm{gm}/\mathrm{cm} \\ =7.8\times {10}^{-2}\mathrm{gm}/\mathrm{cm} \\ =7.8\times {10}^{-3}\mathrm{kg}/\mathrm{m} \end{gathered}$$

A node is always placed at the joint. Since aluminium and steel rod has same mass per unit length, the velocity of wave (v) in both of them is same.
Let v be the velocity of wave.
$$\begin{gathered} \Rightarrow v=\left(\frac{T}{m}\right) \\ ={\sqrt{\left\{\frac{40}{7.8\times {10}^{-3}}\right\}}}^{-} \\ =\sqrt{\left(\frac{4\times {10}^4}{7.8}\right)} \\ \Rightarrow v=71.6\mathrm{m}/\mathrm{s} \end{gathered}$$
For minimum frequency, there would be maximum wavelength.
For maximum wavelength, minimum number of loops are to be produced.
∴ Maximum distance of a loop = 20 cm
$$\begin{gathered} \Rightarrow \mathrm{Wavelength},\lambda =2\times 20=40\mathrm{cm} \\ \mathrm{Or}\mathrm{\lambda }=0.4\mathrm{m} \\ \therefore \mathrm{Frequency},f=\frac{v}{\lambda }=\frac{71.6}{0.4}=180\mathrm{Hz} \end{gathered}$$