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Undamped dynamic vibration absorber

Short answer

An undamped absorber tuned to the excitation frequency causes zero main‑mass amplitude by cancelling the exciting force. Absorber spring stiffness is 246.7 kN/m; the two new natural speeds are approximately 1281 rpm and 1756 rpm.

(a) Explain the principle of an undamped dynamic vibration absorber and show that the amplitude of the main mass is zero when the absorber is tuned to the exciting frequency. (b) A machine of mass 100 kg is mounted on springs and is excited by a harmonic force at 1500 rpm, which is also its natural speed. Design an absorber of mass 10 kg to remove the resonance. Find the absorber spring stiffness and the two new natural frequencies of the combined system.

(a) Principle

A small auxiliary spring-mass system (m2m_2, k2k_2) is attached to the main system (m1m_1, k1k_1), which is excited by F0sin⁡ωtF_0\sin\omega t. When k2/m2=ω\sqrt{k_2/m_2} = \omega, the absorber's force on the main mass cancels the exciting force and the main mass stays still. The absorber mass vibrates instead.

  F0 sin(wt)
     |
   [ m1 ]--k2--[ m2 ]
     |
     k1
   ///

Equations of motion:

m1x¨1+(k1+k2)x1−k2x2=F0sin⁡ωt,m2x¨2+k2(x2−x1)=0m_1\ddot{x}_1 + (k_1+k_2)x_1 - k_2x_2 = F_0\sin\omega t, \qquad m_2\ddot{x}_2 + k_2(x_2 - x_1) = 0

Put x=Xsin⁡ωtx = X\sin\omega t:

[k1+k2−m1ω2−k2−k2k2−m2ω2]{X1X2}={F00}\begin{bmatrix} k_1 + k_2 - m_1\omega^2 & -k_2 \\ -k_2 & k_2 - m_2\omega^2 \end{bmatrix}\begin{Bmatrix} X_1 \\ X_2 \end{Bmatrix} = \begin{Bmatrix} F_0 \\ 0 \end{Bmatrix} X1=(k2−m2ω2)F0(k1+k2−m1ω2)(k2−m2ω2)−k22,X2=k2F0(…)X_1 = \frac{(k_2 - m_2\omega^2)F_0}{(k_1+k_2-m_1\omega^2)(k_2 - m_2\omega^2) - k_2^2}, \qquad X_2 = \frac{k_2F_0}{(\ldots)}

X1=0X_1 = 0 when k2−m2ω2=0k_2 - m_2\omega^2 = 0, i.e. ω=k2/m2\omega = \sqrt{k_2/m_2}. Then X2=−F0k2X_2 = -\dfrac{F_0}{k_2}, so the absorber spring force k2X2=−F0k_2X_2 = -F_0 exactly cancels the exciting force. The absorber works only at the one tuned frequency; away from it, the combined system has two resonances.

(b) Design

Operating frequency: Ω=2π×150060=157.08\Omega = \dfrac{2\pi \times 1500}{60} = 157.08 rad/s. The main system is at resonance, so k1=m1Ω2=100×157.082=2467401k_1 = m_1\Omega^2 = 100 \times 157.08^2 = 2467401 N/m.

Tuning: k2=m2Ω2=10×157.082=246740 N/m=246.7 kN/mk_2 = m_2\Omega^2 = 10 \times 157.08^2 = 246740\ \text{N/m} = 246.7\ \text{kN/m}.

New natural frequencies: the mass ratio is μ=m2/m1=0.1\mu = m_2/m_1 = 0.1. The absorber and the main system have the same natural frequency Ω\Omega. With r=ω/Ωr = \omega/\Omega, the frequency equation is

r4−(2+μ)r2+1=0  ⇒  r2=2.1±2.12−42=0.7298, 1.3702r^4 - (2+\mu)r^2 + 1 = 0 \;\Rightarrow\; r^2 = \frac{2.1 \pm \sqrt{2.1^2 - 4}}{2} = 0.7298,\ 1.3702 ωn1=Ω0.7298=134.19 rad/s (1281 rpm),ωn2=Ω1.3702=183.87 rad/s (1756 rpm)\omega_{n1} = \Omega\sqrt{0.7298} = 134.19\ \text{rad/s}\ (1281\ \text{rpm}), \qquad \omega_{n2} = \Omega\sqrt{1.3702} = 183.87\ \text{rad/s}\ (1756\ \text{rpm})

The two new resonant speeds lie about 15% below and 17% above the operating speed, so the machine runs between them, safely away from both.

Answer: Absorber stiffness =246.7= 246.7 kN/m; new natural speeds ≈1281\approx 1281 rpm and 17561756 rpm.

✓ Answer checked by Claude Opus 5.5 (high effort)

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