Question

The amplitude of a lightly damped oscillator decreases by 4.2% during each cycle. What percentage of the mechanical energy of the oscillator is lost in each cycle?

Answer 1

V= A ω maximum KE of object in SHM

V2 / V1 = .958 ratio of amplitudes since ω is constant

KE2 / KE1 = 1/2 m V2^2 / (1/2 m V1^2) = (V2 / V1)^2

KE2 / KE1 = .958^2 = .918

So KE2 = .918 KE1 and .082 = 8.2% of the energy is lost in one cycle