A 150-kg electromagnet is at rest and is holding 100 kg of scrap steel when the current is turned off dropping the steel. - 2- Knowing that the cable and the supporting crane have a total stiffness equivalent to a spring of constant k 60 kN/m and a damper c = 300 Ns/m, determine (a) the frequency of oscillation (b) the maximum upward displacement І В

Respuesta :

The frequency of the oscillation is determined to be 0.123 Hz.

The number of waves passing through a fixed point in unit time is said to be frequency.

Given that, Mass of the scrap steel = 100 kg

Spring constant k = 60 kN/m

The formula for time period is as follows,

T = 2π √(m/k)

And the relation between time period and frequency is, T = 1/f

f = 1 / [2π √(m/k)] = 1/ [ 2π √(100/60) = 1/ [ 2π * 1.29] = 1/8.105 = 0.123 Hz

Thus, the frequency of oscillation is 0.123 Hz.

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