An aluminium wire, of length d1 = 60.0 cm, cross-sectional area 1 times 10 cm, a

Question

An aluminium wire, of length d1 = 60.0 cm, cross-sectional area 1 times 10 cm, and density 2.60g cm3, is connected to a steel wire, of density 7.80g/cm3 and the same cross-sectional area. The compound wire, loaded with a block of mass m = 10.0kg, is arranged as displayed below so that the distance d2 = 86.6 cm. Transverse waves are set up in the wire by using an external source of variable frequency, (a) Find the lowest frequency of excitation for which standing waves are observed such that the joint in the wire is a node, (b) How many nodes are observed at this frequency?

Explanation / Answer

Tension in the two cables is T = (10kg)(9.8m/s2) =98N If there is a node at the joint, there should be an inegralmultiples of waves must be on both the wires. let there are n loops of standing waves on thealuminum wire and m loops of standing waves on the steelwire. then we should have n = d1/(Al/4)= 4d1/Al or n = 4 d1/vAl then = nvAl /4 d1 Similarly we can have = mvSteel /4d2 ( since frequency remain same on both thewires) Now speed of wave in the Aluminum wire is vAl =[T/A] A is area of cross section and is thedensity of aluminum wire. therefore vAl =[(98N)/{(1*10-6m2)(2600kg/m3)}]                     =194m/s And speed of wave in the Steel wire is vSteel= [T/A] A is area of cross section and is thedensity of steel wire. therefore vSteel =[(98N)/{(1*10-6m2)(7800kg/m3)}]                        =112m/s Therefore we can have = n(194m/s) /[(4)(0.6m)]                                      = 80.83n and = m(112m/s) /[(4)(0.866m)]                                      = 32.33m then 80.83n= 32.33m or m = 2.5n for both m and n to be integers the minimum values of m and nare m = 5 and n = 2 then the required frequency is = (32.33)(5)                                                 =161.65Hz                                               ˜ 162Hz and teh number of waves on the string is n + m = 2 + 5 =7 Similarly we can have = mvSteel /4d2 ( since frequency remain same on both thewires) Now speed of wave in the Aluminum wire is vAl =[T/A] A is area of cross section and is thedensity of aluminum wire. therefore vAl =[(98N)/{(1*10-6m2)(2600kg/m3)}]                     =194m/s And speed of wave in the Steel wire is vSteel= [T/A] A is area of cross section and is thedensity of steel wire. therefore vSteel =[(98N)/{(1*10-6m2)(7800kg/m3)}]                        =112m/s And speed of wave in the Steel wire is vSteel= [T/A] A is area of cross section and is thedensity of steel wire. therefore vSteel =[(98N)/{(1*10-6m2)(7800kg/m3)}]                        =112m/s Therefore we can have = n(194m/s) /[(4)(0.6m)]                                      = 80.83n and = m(112m/s) /[(4)(0.866m)]                                      = 32.33m then 80.83n= 32.33m or m = 2.5n for both m and n to be integers the minimum values of m and nare m = 5 and n = 2 then the required frequency is = (32.33)(5)                                                 =161.65Hz                                               ˜ 162Hz and teh number of waves on the string is n + m = 2 + 5 =7                                      = 32.33m then 80.83n= 32.33m or m = 2.5n for both m and n to be integers the minimum values of m and nare m = 5 and n = 2 then the required frequency is = (32.33)(5)                                                 =161.65Hz                                               ˜ 162Hz and teh number of waves on the string is n + m = 2 + 5 =7

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