A mass-spring system has b/m = ω0/2, where b is the damping constant and ω0 the natural frequency. How does its amplitude A when driven at a frequency 12% above ω0 compare with its amplitude at resonance Ares at ω0?

Answer 1

The amplitude at resonance (ω₁ = ω₀) ia Ar =F₀/b•ω₀,
A/Ar=A/( F₀/b•ω₀)= { b•ω₀/m}/sqrt{( ω₁²- ω₀²)²+(b²•ω₁²/m²)}=
=1/sqrt{(m•ω₀/b)²•[(ω₁²/ω₀²)-1]²+(ω₁²/ω₀²)}.
If m•ω₀/b =2, and ω₁/ω₀ =1.12 (12% above resonance),
and ω₁²/ω₀²= 1.2544, then
A/Ar =1/sqrt{4•(1.2544 -1)²+1.2544} =
=0.81.