Free vibration: natural frequency and damping from a tap
引入| English | 释义 | 拼音 |
|---|---|---|
| natural frequency/ˈnætʃərəl ˈfriːkwənsi/ | 固有频率 | gù yǒu pín lǜ |
| damping ratio | 阻尼比 | zǔ ní bǐ |
| phase plane | 相平面 | xiāng píng miàn |
| logarithmic decrement | 对数衰减率 | duì shù shuāi jiǎn lǜ |
| resonance/ˈrezənəns/ | 共振 | gòng zhèn |
| critical damping/ˈkrɪtɪkl ˈdæmpɪŋ/ | 临界阻尼 | lín jiè zǔ ní |
| vibration | 振动 | zhèn dòng |
Tap it and listen
- The pump sits on rubber mounts. Its natural frequency decides whether running speed will excite it, and its damping decides how quickly a knock dies away.
- $m\ddot x + c\dot x + kx = 0$: write $\omega_n = \sqrt{k/m}$ and the damping ratio $\zeta = c/(2\sqrt{km})$.
Three kinds of return
- Below $\zeta = 1$ the mass oscillates at $\omega_d = \omega_n\sqrt{1 - \zeta^2}$ inside a decaying envelope. At $\zeta = 1$ it returns fastest without overshoot; above 1 it creeps back.
- In the phase plane the motion is an inward spiral; without damping it would be a circle.
How does it return?
Classify each system by its damping.
Light damping changes the oscillation frequency:
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omega_d = omega_n root(1 - zeta^2).
A critically damped system overshoots once.
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It returns fastest without overshoot.
The pump on its mounts
- $m = 120$ kg on 300 kN/m: $\omega_n = \mathbf{50.0}$ rad/s, $f_n = \mathbf{7.96}$ Hz.
- A tap test gives peaks of 5.0 and 3.1 mm one period apart. The logarithmic decrement $\delta = \ln(5.0/3.1) = \mathbf{0.478}$ gives $\zeta = \delta/\sqrt{4\pi^2 + \delta^2} = \mathbf{0.076}$.
- $\omega_d = 49.86$ rad/s: light damping barely changes the frequency.
Natural frequency of 120 kg on 300 kN/m (one decimal, rad/s)?
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Root of k over m.
The same in hertz (two decimals)?
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omega over 2 pi.
Log decrement from peaks of 5.0 and 3.1 mm (three decimals)?
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ln(5.0/3.1).
Damping ratio from that decrement (three decimals)?
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delta over root(4 pi^2 + delta^2).
A torsional system
- The output shaft joins the gear ($J_1 = 0.64$) to the impeller ($J_2 = 0.12$ kg·m²) with $k_t = GJ_p/L = \mathbf{10.6}$ kN·m/rad.
- Two discs vibrate against each other at $\sqrt{k_t(J_1 + J_2)/J_1J_2} = \mathbf{324}$ rad/s, 51.6 Hz. The shaft runs at 180 rpm, far below: no torsional resonance.
Two discs, J1 = 0.64, J2 = 0.12 kg m2, k_t = 10 600 N m/rad (rad/s, whole number)?
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Root of k_t (J1 + J2)/(J1 J2).
The damping in newtons
- The critical damping 临界阻尼 is $c_c = 2\sqrt{km} = 2\sqrt{300\,000 \times 120} = 12\,000$ N·s/m. The mounts' $\zeta = 0.076$ means $c = 0.076 \times 12\,000 = \mathbf{912}$ N·s/m.
- That is the number a damper supplier needs; $\zeta$ alone is not something you can buy.
Mounts 300 kN/m under 120 kg with $\zeta = 0.076$: damping coefficient (N·s/m, nearest whole)?
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$c = \zeta \cdot 2\sqrt{km} = 0.076 \times 12\,000 = 912$ N·s/m.
Doubling the damping ratio halves the natural frequency.
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Light damping barely moves the frequency ($\omega_d = \omega_n\sqrt{1 - \zeta^2}$); it halves the ringing time.
Worked example: how long does a knock ring?
- Each cycle the amplitude falls by $e^{-\delta}$, with $\delta = 0.478$. To fall to 1% takes $\ln 100/0.478 = \mathbf{9.6}$ cycles.
- At 7.96 Hz that is 1.2 s. A pump knocked at start-up rings for about a second; doubling $\zeta$ would halve that.
Log decrement 0.478: cycles for a knock to fall to 1% (one decimal)?
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$\ln 100/0.478 = 4.605/0.478 = 9.6$.
Three free-vibration slips
- Peaks not one period apart. The decrement needs consecutive peaks.
- $f_n$ quoted in rad/s. Divide $\omega_n$ by $2\pi$.
- Two discs treated as one disc on a fixed shaft. Both ends move.
Recap
- $\omega_n = \sqrt{k/m} = 50$ rad/s, 7.96 Hz.
- $\delta = 0.478$, $\zeta = 0.076$.
- Two-disc torsion: 324 rad/s, far above running speed.