From the data of the problem we will first calculate the speeds of wave motion in the aluminium and steel wires. As the speed depends on the tension and the mass per unit length we will first compute these quantities. The tension in the wires that have the same cross-sectional area, , is equal to the weight of the block that is attached to one end of the wire through a pulley. It is

Cross-sectional area of the two wires has been given to be

Density of aluminium, . Therefore, the mass per unit length of the aluminium wire is

Density of steel, . Therefore, the mass per unit length of the steel wire is

Velocity of wave motion in aluminium wire

Velocity of wave motion in steel wire

Length of the aluminium wire,

Length of the steel wire from the joint to the pulley,

Let the number of loops of the standing wave in the aluminium wire when its joint with the steel wire is a node be . Then

Resonant frequency

Let the number of loops of the standing wave in the steel wire from the joint with the aluminium wire and the pulley be . Then

(a)

The resonant frequency for standing wave calculated from the data for the steel wire is

We have to find minimum integers and which satisy the condition

We note that this condition is satisfied for and . The lowest frequency of excitation for which standing waves will be observed with the joint of the two wires being a node is

(b)

The total number of nodes observed at this frequency, excluding the two at the ends of the wire, will be 6.