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 y_E   =   A_Ecos (wt). y_E means the exciter's elongation compared with the mid-position; 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 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)   =   A_abs sin (wt) + A_el cos (wt)   +   e^-Gt/2 [A₁ sin (w₁t) + B₁ cos (w₁t)]
w₁   =   (w₀² - G²/4)^1/2
A_abs   =   A_E w₀² G w / [(w₀² - w²)² + G² w²]
A_el   =   A_E w₀² (w₀² - w²) / [(w₀² - w²)² + G² w²]
A₁   =   - (A_abs w + (G/2) A_el) / w₁
B₁   =   - A_el

Case 1.2: G < 2 w0; G = 0 and w = w0

y(t)   =   (A_E w t / 2) sin (wt)

Case 2: G = 2 w0

y(t)   =   A_abs sin (wt) + A_el cos (wt)   +   e^-Gt/2 (A₁ t + B₁)
A_abs   =   A_E w₀² G w / (w₀² + w²)²
A_el   =   A_E w₀² (w₀² - w²) / (w₀² + w²)²
A₁   =   - (A_abs w + (G/2) A_el)
B₁   =   - A_el

Fall 3: G > 2 w0

y(t)   =   A_abs sin (wt) + A_el cos (wt)   +   e^-Gt/2 [A₁ sinh (w₁t) + B₁ cosh (w₁t)]
w₁   =   (G²/4 - w₀²)^1/2
A_abs   =   A_E w₀² G w / [(w₀² - w²)² + G² w²]
A_el   =   A_E w₀² (w₀² - w²) / [(w₀² - w²)² + G² w²]
A₁   =   - (A_abs w + (G/2) A_el) / w₁
B₁   =   - A_el

Click to Back