Differential equations: the rig’s free and forced response
مُقدَّم| English | المعنى |
|---|---|
| characteristic equation/ˌkærɪktəˈrɪstɪk ɪˈkweɪʒn/ | An algebraic equation whose roots determine the response modes of a linear differential equation. |
| underdamped | Having positive damping below the critical value, producing decaying oscillations in the passive second-order model. |
| resonance/ˈrezənəns/ | A large forced response when excitation is near a system's natural frequency, with the response depending on damping and the measured quantity. |
The rig is a differential equation
- $\ddot{x} + 2\dot{x} + 25x = 0$: mass 1 kg, damping 2 N s/m, stiffness 25 N/m. Reading its response is reading whether the till mount survives the festival.
- The characteristic equation 特征方程 $ms^2 + cs + k = 0$ decides the personality.
Free response, checked by substitution
- $\omega_n = 5$, zeta $= 0.2$ (underdamped 欠阻尼), $\omega_d = \sqrt{24} = 4.899$.
- From $x_0 = 1$, $v_0 = 0$: $x = e^{-t}(\cos\omega_d t + \frac{1}{\omega_d}\sin\omega_d t)$ - residual $\sim 10^{-6}$ on substitution. First minimum $-0.527$ at $t = 0.641 = -e^{-\zeta\pi/\sqrt{1-\zeta^2}}$ - two routes, one number.
Which personality is this?
Route each second-order system to its response family.
m = 1, k = 25. The natural frequency (rad/s)?
الصعوبة 2/5 · 140 عملة + مكافأة صغيرة عند الإنجاز الأول اليوم
sqrt(25) = 5.
With c = 2: the damped frequency (three decimals)?
الصعوبة 2/5 · 140 عملة + مكافأة صغيرة عند الإنجاز الأول اليوم
5 x sqrt(1 - 0.04) = sqrt(24).
Match each parameter to its role.
الصعوبة 2/5 · 140 عملة + مكافأة صغيرة عند الإنجاز الأول اليوم
Four numbers, the whole story.
Forced response and the worst case
- Magnification $M(r) = 1/\sqrt{(1-r^2)^2 + (2\zeta r)^2}$: $M(1) = 2.5$, peak 2.552 at $r = 0.959$ - just below resonance.
- The floor vibrates at $r = 0.959$: 1 mm becomes 2.55 mm against a 2 mm clearance. Fix by stiffness, damping or isolation - one curve, three strategies.
First minimum from x0 = 1, v0 = 0 (three decimals)?
الصعوبة 2/5 · 140 عملة + مكافأة صغيرة عند الإنجاز الأول اليوم
-exp(-zeta pi/sqrt(1-zeta^2)) = -e^-0.641.
Magnification at r = 1 with zeta = 0.2?
الصعوبة 2/5 · 140 عملة + مكافأة صغيرة عند الإنجاز الأول اليوم
1/(2 zeta) = 2.5.
The true peak of M(r) for zeta = 0.2 sits at:
الصعوبة 1/5 · 100 عملة + مكافأة صغيرة عند الإنجاز الأول اليوم
sqrt(1 - 2 zeta^2) = 0.959 - just below resonance.
Three ODE slips
- Using omega_n where omega_d belongs. The correction factor is not a rounding detail at zeta 0.2.
- Quoting M(1) as the peak. The peak is below r = 1 and 2% higher.
- Solving without a substitution check. One line of code converts a claim into a fact.
A supplier’s claimed solution can be audited by substituting x, x-dot and x-double-dot into the ODE.
الصعوبة 1/5 · 100 عملة + مكافأة صغيرة عند الإنجاز الأول اليوم
Residual ~1e-6 makes it a fact, not a claim.
Raising the stiffness always reduces the amplification at a fixed drive frequency.
الصعوبة 1/5 · 100 عملة + مكافأة صغيرة عند الإنجاز الأول اليوم
It moves omega n - the point slides along the curve; direction depends on where you started.
Recap
- Roots of $ms^2 + cs + k = 0$ give the personality; $\omega_d = \omega_n\sqrt{1-\zeta^2}$.
- First undershoot $-0.527$; peak 2.552 at r = 0.959; $M(1) = 1/(2\zeta) = 2.5$.
- Check by substitution; fix by stiffness, damping or isolation.