
The top of the spring pendulum is moved to and fro according to the formula
y_{E} = A_{E}
cos (wt).
y_{E} means the exciter's elongation compared with the midposition; A_{E} 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 midposition) at the time t. Using w_{0} = (D/m)^{1/2} this problem is described by the following differential equation:
y''(t) = w_{0}^{2}
(A_{E} 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 w_{0} 
Case 1.1: G < 2 w_{0}; G ¹ 0 or w ¹ w_{0} 
y(t) = A_{abs} sin (wt)
+ A_{el} cos (wt)
+ e^{Gt/2}
[A_{1} sin (w_{1}t)
+ B_{1} cos (w_{1}t)]
w_{1} =
(w_{0}^{2}
 G^{2}/4)^{1/2}
A_{abs} = A_{E}
w_{0}^{2}
G w
/ [(w_{0}^{2}
 w^{2})^{2}
+ G^{2} w^{2}]
A_{el} = A_{E}
w_{0}^{2}
(w_{0}^{2}
 w^{2})
/ [(w_{0}^{2}
 w^{2})^{2}
+ G^{2} w^{2}]
A_{1} =  (A_{abs} w
+ (G/2) A_{el})
/ w_{1}
B_{1} =  A_{el}
Case 1.2: G < 2 w_{0}; G = 0 and w = w_{0} 
y(t) = (A_{E} w t / 2) sin (wt)
Case 2: G = 2 w_{0} 
y(t) = A_{abs} sin (wt)
+ A_{el} cos (wt)
+ e^{Gt/2}
(A_{1} t + B_{1})
A_{abs} = A_{E}
w_{0}^{2}
G w
/ (w_{0}^{2}
+ w^{2})^{2}
A_{el} = A_{E}
w_{0}^{2}
(w_{0}^{2}
 w^{2})
/ (w_{0}^{2}
+ w^{2})^{2}
A_{1} =  (A_{abs} w
+ (G/2) A_{el})
B_{1} =  A_{el}
Fall 3: G > 2 w_{0} 
y(t) = A_{abs} sin (wt)
+ A_{el} cos (wt)
+ e^{Gt/2}
[A_{1} sinh (w_{1}t)
+ B_{1} cosh (w_{1}t)]
w_{1} =
(G^{2}/4
 w_{0}^{2})^{1/2}
A_{abs} = A_{E}
w_{0}^{2}
G w
/ [(w_{0}^{2}
 w^{2})^{2}
+ G^{2} w^{2}]
A_{el} = A_{E}
w_{0}^{2}
(w_{0}^{2}
 w^{2})
/ [(w_{0}^{2}
 w^{2})^{2}
+ G^{2} w^{2}]
A_{1} =  (A_{abs} w
+ (G/2) A_{el})
/ w_{1}
B_{1} =  A_{el}
ÊÒÃºÑ 
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