Question

An aluminium wire of length L = 60 cm and of cross-sectional area $0.01c{m}^2$ is connected to a steel wire of the same cross-sectional area. The compound wire is loaded with a block of mass 10 kg as shown in the figure. The distance $L_2$ from the joint to the supporting pulley is 86.6 cm. Transverse waves are set up in the composite wire at its lowest frequency, such that the joint in the wire is a node. What is the total number of nodes observed at the frequency excluding the two at the ends of the wire? Density of aluminium is $2.6gc{m}^{-3}$ and that of steel is $7.8gc{m}^{-3}$.___

Answer 1

Velocities on Al and steel wires are different because $\mu$ is different for both of them. So, $\lambda$ is also different as f is same for both.
${\mu }_1={\mu }_{steel}=\rho 1\times Area=7.8\times 103\times {10}^{-6}$
$=7.8\times {10}^{-3}kg{m}^{-1}$
${\mu }_2={\mu }_{Al}=\rho 2\times Area$
$=2.6\times {10}^3\times {10}^{-6}=2.6\times {10}^{-3}kg{m}^{-1}{f}_1=f_2$
$\Rightarrow \frac{n_1}{2L_1}\sqrt{\frac{1}{{\mu }_1}}=\frac{n_2}{2L_2}\sqrt{\frac{1}{{\mu }_2}}=\frac{n_1}{0.866\times \sqrt{7.8}}=\frac{n_2}{0.6\times \sqrt{2.6}}$
$\Rightarrow \frac{n_1}{n_2}=\frac{5}{2}$
Therefore, lowest frequency oscillation causes 2 loops in Al and 5 loops in steel.

So, total no. of nodes = 8, required no. of nodes = 6