Question

A wire cable is 4.00m long and has a mass of .6 kg. A transverse wave pulse is produced by plucking one end of the cable. The pulse is observed to make 5 complete round-trips (down and back) in 1.265 seconds. a.) What is the tension in the cable? b.) In order to make standing waves on the cable with a 60 HZ vibrator where we observe 6 nodes and 5 anti-nodes, what tension do we need in the cable?

Answer 1

A) Total mass of cable = 0.6 kg   , Total length = L = 4 m

The formula for tension in a wire is given as ,

m = linear density of wire = $\frac{total mass }{total length}$ = $\frac{0.6 kg }{4 m}$ = 0.15 kg/m.

frequency = $\frac{Number of Pulse }{Total time}$ = $\frac{5 }{1.265}$ = 3.953 Hz

inserting the values in the formula for T ,

T = 4 (3.953 Hz)² (4)² (0.15) = 150.02 N

B) Number of antinodes = 5

so this is 4th overtone and n = 4

so the formula for tension is

f = 60 Hz , L = 4 m , m = 0.15

T = 4 (60)² (4)² (0.15) (4 +1)²

T = 864000 N