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The 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

Fall 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/ph11e/resmathengl.htm
© Walter Fendt, September 9, 1998
Last modification: December 27,2002

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