Question

If the quality factor Q in the steady state forced vibration is defined as

Q = 2 Average energy stored per cycle/Average energy dissipated per cycle then show that

Show that the quality factor is minimum at resonance p = ω and its minimum value is

Answer 1

Total energy at any instant of time in the steady state is

From problem 5 we know that the average power dissipated

Thus the minimum of Q occurs at resonance when p = ω and the minimum value is ω/2b.Quality factor at resonance = sharpness of resonance.