Back to the main pageThe 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
ω02
(ω02
− 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
ω02
(ω02
− 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
ω02
(ω02
− ω2)
/ [(ω02
− ω2)2
+ Γ2 ω2]
A1 = − (Aabs ω
+ (Γ/2) Ael)
/ ω1
B1 = − Ael
URL: http://www.walter-fendt.de/ph14e/resmath_e.htm
© Walter Fendt, September 9, 1998
Last modification: February 3, 2010