- plaatje 772 keer bekeken
We gaan ervan uit dat de veren op t = 0 nog in rust zijn en dat x
1(0) = 0. Dan hebben we:
\( x_1(0) = 0 \,\,\,\,\,\,\,\,\,\,\, (1) \)
\( x_2(0) = \mbox{L} \,\,\,\,\,\,\,\,\,\,\, (2) \)
\( x_3(0) = 2 \mbox{L} \,\,\,\,\,\,\,\, (3) \)
\( \dot{x}_1(0) = 0 \,\,\,\,\,\,\,\,\,\,\, (4) \)
\( \dot{x}_2(0) = 0 \,\,\,\,\,\,\,\,\,\,\, (5) \)
\( \dot{x}_3(0) = 0 \,\,\,\,\,\,\,\,\,\,\, (6) \)
\( F_A(t) = \mbox{k} \cdot \{\mbox{L} - (x_2(t) - x_1(t)) \} \,\,\,\,\,\,\,\, (7) \)
\( F_B(t) = \mbox{k} \cdot \{\mbox{L} - (x_3(t) - x_2(t)) \} \,\,\,\,\,\,\,\, (8) \)
\( F(t) - F_A(t) = \mbox{m} \cdot \ddot{x}_1(t) \,\,\,\,\,\,\,\,\,\,\, (9) \)
\( F_A(t) - F_B(t) = \mbox{m} \cdot \ddot{x}_2(t) \,\,\,\,\,\,\,\, (10) \)
\( F_B(t) = \mbox{m} \cdot \ddot{x}_3(t) \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, (11) \)