A 1.80-m-long uniform bar that weighs 648 N is suspended in a horizontal position by two vertical wires that are attached to the ceiling. One wire is aluminum and the other is copper. The aluminum wire is attached to the left-hand end of the bar, and the copper wire is attached 0.400 m to the left of the right-hand end. Each wire has length 0.600 m and a circular cross section with radius 0.250 mm.

a. What is the fundamental frequency of transverse standing waves for aluminium wire?
b. What is the fundamental frequency of transverse standing waves for copper wire?

[tex]T_a0+T_c(1.8-0.4)=W\dfrac{1.8}{2}\\\Rightarrow T_c=W\dfrac{0.9}{1.8-0.4}\\\Rightarrow T_c=648\dfrac{0.9}{1.8-0.4}\\\Rightarrow T_c=416.571428571\ Nm[/tex]

The sum of the tensions will be equal to the weight of the bar

[tex]T_a+T_a=W\\\Rightarrow T_a=W-T_c\\\Rightarrow T_a=648-416.571428571\\\Rightarrow T_a=231.428571429\ Nm[/tex]

Fundamental frequency is given by

[tex]f=\dfrac{1}{2L}\sqrt{\dfrac{T_a}{\mu}}\\\Rightarrow f=\dfrac{1}{2L}\sqrt{\dfrac{T_a}{\rho\pi r^2}}\\\Rightarrow f=\dfrac{1}{2\times 1.8}\sqrt{\dfrac{231.428571429}{2700\times \pi (0.25\times 10^{-3})^2}}\\\Rightarrow f=183.530885247\ Hz[/tex]

The frequency for aluminum is 183.530885247 Hz

[tex]f=\dfrac{1}{2L}\sqrt{\dfrac{T_c}{\mu}}\\\Rightarrow f=\dfrac{1}{2L}\sqrt{\dfrac{T_c}{\rho\pi r^2}}\\\Rightarrow f=\dfrac{1}{2\times 1.8}\sqrt{\dfrac{416.571428571}{8960\times \pi (0.25\times 10^{-3})^2}}\\\Rightarrow f=135.16781368\ Hz[/tex]

The frequency for copper is 135.16781368 Hz