Forced Oscillations
Mathematical 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 y_E   =   A_E cos (ωt).
y_E means the exciter's elongation compared with the mid-position; A_E 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:

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)   =   A_abs sin (ωt) + A_el cos (ωt)   +   e^−Γt/2 [A₁ sin (ω₁t) + B₁ cos (ω₁t)]
ω₁   =   (ω₀² − Γ²/4)^1/2
A_abs   =   A_E ω₀² Γ ω / [(ω₀² − ω²)² + Γ² ω²]
A_el   =   A_E ω₀² (ω₀² − ω²) / [(ω₀² − ω²)² + Γ² ω²]
A₁   =   − (A_abs ω + (Γ/2) A_el) / ω₁
B₁   =   − A_el

Case 1.2: Γ < 2 ω₀; Γ = 0 and ω = ω₀

y(t)   =   (A_E ω t / 2) sin (ωt)

Case 2: Γ = 2 ω₀

y(t)   =   A_abs sin (ωt) + A_el cos (ωt)   +   e^−Γt/2 (A₁ t + B₁)
A_abs   =   A_E ω₀² Γ ω / (ω₀² + ω²)²
A_el   =   A_E ω₀² (ω₀² − ω²) / (ω₀² + ω²)²
A₁   =   − (A_abs ω + (Γ/2) A_el)
B₁   =   − A_el

Case 3: Γ > 2 ω₀

y(t)   =   A_abs sin (ωt) + A_el cos (ωt)   +   e^−Γt/2 [A₁ sinh (ω₁t) + B₁ cosh (ω₁t)]
ω₁   =   (Γ²/4 − ω₀²)^1/2
A_abs   =   A_E ω₀² Γ ω / [(ω₀² − ω²)² + Γ² ω²]
A_el   =   A_E ω₀² (ω₀² − ω²) / [(ω₀² − ω²)² + Γ² ω²]
A₁   =   − (A_abs ω + (Γ/2) A_el) / ω₁
B₁   =   − A_el

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

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