# Forced OscillationsMathematical Appendix

The spring pendulum is characterized by the spring constant D, the mass m and the constant of attenuation Γ. (Γ is a measure of the friction force assumed as proportional to the velocity.)
The top of the spring pendulum is moved to and fro according to the formula yE   =   AE cos (ωt).
yE means the exciter's elongation compared with the mid-position; AE is the amplitude of the exciter's oscillation, ω means the corresponding angular frequency and t the time.

It is a question of finding the size of the resonator's elongation y (compared with its mid-position) at the time t. Using ω0   =   (D/m)1/2 this problem is described by the following differential equation:

 y''(t)   =   ω02 (AE cos (ωt) − y(t))   −   Γ y'(t) Initial conditions:     y(0) = 0;   y'(0) = 0

If you want to solve this differential equation, you have to distinguish between several cases:

 Case 1: Γ < 2 ω0

 Case 1.1: Γ < 2 ω0; Γ ≠ 0 or ω ≠ ω0

y(t)   =   Aabs sin (ωt) + Ael cos (ωt)   +   e−Γt/2 [A1 sin (ω1t) + B1 cos (ω1t)]
ω1   =   (ω02 − Γ2/4)1/2
Aabs   =   AE ω02 Γ ω / [(ω02 − ω2)2 + Γ2 ω2]
Ael   =   AE ω0202 − ω2) / [(ω02 − ω2)2 + Γ2 ω2]
A1   =   − (Aabs ω + (Γ/2) Ael) / ω1
B1   =   − Ael

 Case 1.2: Γ < 2 ω0; Γ = 0 and ω = ω0

y(t)   =   (AE ω t / 2) sin (ωt)

 Case 2: Γ = 2 ω0

y(t)   =   Aabs sin (ωt) + Ael cos (ωt)   +   e−Γt/2 (A1 t + B1)
Aabs   =   AE ω02 Γ ω / (ω02 + ω2)2
Ael   =   AE ω0202 − ω2) / (ω02 + ω2)2
A1   =   − (Aabs ω + (Γ/2) Ael)
B1   =   − Ael

 Case 3: Γ > 2 ω0

y(t)   =   Aabs sin (ωt) + Ael cos (ωt)   +   e−Γt/2 [A1 sinh (ω1t) + B1 cosh (ω1t)]
ω1   =   (Γ2/4 − ω02)1/2
Aabs   =   AE ω02 Γ ω / [(ω02 − ω2)2 + Γ2 ω2]
Ael   =   AE ω0202 − ω2) / [(ω02 − ω2)2 + Γ2 ω2]
A1   =   − (Aabs ω + (Γ/2) Ael) / ω1
B1   =   − Ael

URL: https://www.walter-fendt.de/html5/phen/resonance_math_en.htm
Walter Fendt, September 9, 1998
Last modification: February 3, 2010