Figureshows 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 mm^2and that of the aluminium wire is 3.0 mm 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 2.6 g cm^-3and that of steel is7.8gcm^-3

Figureshows an aluminium wire of length 60 cm joined to a steel wire o

1.	Figureshows 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 mm^2and that of the aluminium wire is 3.0 mm 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 2.6 g cm^-3and that of steel is7.8gcm^-3
Answer» Solution :`P_s=7.8(gm)/cm^3` `P_A=2.6(gm)/cm^3` `m_s=P_sA_s` `=7.8xx10^-2(gm)/cm` (m=mass per unit length) `7.8xx10^-3kg/m` `m_A=P_A` `=21.6xx10^-2xx3(gm)/cm` `=7.8xx10^-2(gm)/cm` `=7.8xx10^-3(kg)/m` A node is always placed in the joint. Since ALUMINIUM and steel rod has same mass per unit lenght, velocity of wave in both of them is same `rarr v=sqrt((T/m))` `=sqrt({40/((7.8xx10^-3))})` `=sqrt((((4xx10^4)))/7.8)` `=71.6m/s` For minimum frequency there would be MAXIMUM WAVELENGTH. for maximu wavelength minimum no. of loops are to be produced. `:.` Maximum distance of a LOOP =20cm `rarr` wavelength =`lamda=2xx20` `=40cm=0.4m`=180Hz`