A 100-m long high-voltage cable is suspended between two towers.The mass of the 100-m cable is 200 kg.If the tension in the cable is 20,000 N,what is the lowest frequency at which this cable can oscillate?

Question

Quizplus · Accepted Answer

The lowest frequency at which the cable can oscillate is given by:f = (1/2π)√(T/m)where T is the tension in the cable and m is the mass of the cable.Substituting the given values, we get:f = (1/2π)√(20,000 N / 200 kg) = 0.50 Hz

A 100-M Long High-Voltage Cable Is Suspended Between Two Towers