In the figure, an aluminum wire, of length L 1 = 46.6 cm, cross-sectional area 1

Question

In the figure, an aluminum wire, of length L1 = 46.6 cm, cross-sectional area 1.01 × 10-2 cm2, and density 2.60 g/cm3, is joined to a steel wire, of density 7.80 g/cm3 and the same cross-sectional area. The compound wire, loaded with a block of mass m = 9.6 kg, is arranged so that the distance L2 from the joint to the supporting pulley is 33.6 cm. Transverse waves are set up in the wire by using an external source of variable frequency; a node is located at the pulley. (a) Find the lowest frequency that generates a standing wave having the joint as one of the nodes. (b) How many nodes are observed at this frequency, including the end nodes?

Explanation / Answer

a.)l=n*lambda/2

f=(n/2l)(sqrt(mg/rho*A))

So n2/n1=(l2/l1)(sqrt(rho2/rho1))=(33.6/46.6)(sqrt(7.8/2.6))=1.248=1.25

The smallest values of n1 and n2 that satisfy are 4 and 5

So f=(4/2*46.6)(sqrt(9.6x10^3x980/2.6x1.01 x10^-2))=812.35 Hz

b.)Draw loops and see for n1=4 and n2=5

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects