Back to the main pageThe spring pendulum is characterized by the spring constant D, the mass m and the constant of attenuation G. (G 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 (wt).
yE means the exciter's elongation compared with the mid-position; AE is the amplitude of the exciter's oscillation, w 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 w0 = (D/m)1/2 this problem is described by the following differential equation:
|
y''(t) = w02
(AE cos (wt) – y(t))
– G 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: G < 2 w0 |
| Case 1.1: G < 2 w0; G ¹ 0 or w ¹ w0 |
y(t) = Aabs sin (wt)
+ Ael cos (wt)
+ e–Gt/2
[A1 sin (w1t)
+ B1 cos (w1t)]
w1 =
(w02
– G2/4)1/2
Aabs = AE
w02
G w
/ [(w02
– w2)2
+ G2 w2]
Ael = AE
w02
(w02
– w2)
/ [(w02
– w2)2
+ G2 w2]
A1 = – (Aabs w
+ (G/2) Ael)
/ w1
B1 = – Ael
| Case 1.2: G < 2 w0; G = 0 and w = w0 |
y(t) = (AE w t / 2) sin (wt)
| Case 2: G = 2 w0 |
y(t) = Aabs sin (wt)
+ Ael cos (wt)
+ e–Gt/2
(A1 t + B1)
Aabs = AE
w02
G w
/ (w02
+ w2)2
Ael = AE
w02
(w02
– w2)
/ (w02
+ w2)2
A1 = – (Aabs w
+ (G/2) Ael)
B1 = – Ael
| Case 3: G > 2 w0 |
y(t) = Aabs sin (wt)
+ Ael cos (wt)
+ e–Gt/2
[A1 sinh (w1t)
+ B1 cosh (w1t)]
w1 =
(G2/4
– w02)1/2
Aabs = AE
w02
G w
/ [(w02
– w2)2
+ G2 w2]
Ael = AE
w02
(w02
– w2)
/ [(w02
– w2)2
+ G2 w2]
A1 = – (Aabs w
+ (G/2) Ael)
/ w1
B1 = – Ael
URL: http://www.walter-fendt.de/ph14e/resmath_e.htm
© Walter Fendt, September 9, 1998
Last modification: January 18, 2003