Prujina–massa (2-tartibli ODE) — tahlil

m x'' + c x' + k x = 0  |  (vertical bo‘lsa: muvozanat xeq=mg/k)
← Interaktiv masalalarga qaytish
Parametrlar
Vertical bo‘lsa x_eq = m·g/k, grafiklar x - x_eq bo‘yicha tahlil qilinadi.
c=0 bo‘lsa undamped (so‘nishsiz) tebranish.
Faqat vertical rejimda ishlatiladi.
Tahlil (mass_spring2 uchun)
ω₀ = √(k/m)
4.47214
ζ = c / (2√(mk))
0
Rejim (damping)
undamped (so‘nishsiz)
x_max / x_min
0,15 / -0,14999970740973306
v_max / v_min
0,6708128983326315 / -0,6708194904555465
Natijalar
x(t), v(t) va 3D fazoviy trayektoriya
x(t)
v(t)
a(t)
X: x (m), Y: v (m/s), Z: t (s)
Xulosa

Berilgan tizim m x'' + c x' + k x = 0 (muvozanatga nisbatan) ko‘rinishida yechildi. Tabiiy chastota ω0 = √(k/m) = 4.472 rad/s. So‘nish koeffitsiyenti bo‘yicha damping ratio ζ = c/(2√(mk)) = 0, rejim: undamped (so‘nishsiz). Boshlang‘ich shartlar: x(0)=0.15 m, v(0)=0 m/s; hisoblash oraliği t∈[0, 10] s. Gorizontal holatda (oddiy Hooke modeli) x_eq = 0 deb olinadi.

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