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Find M, D, K from 10 lb response

Short answer

Spring stiffness K = 500 lb/ft from final displacement 10/K = 0.02 ft. Damping ratio ζ = 0.237 from overshoot M_p = 0.0093/0.02 = 0.465; natural frequency ω_n = 1.078 rad/s from peak time 3 s. Mass M = K/ω_n² = 430 slug, damping coefficient D = 2ζω_n M = 220 lb‑s/ft.

Following figure shows a mechanical vibratory system and the response when 10 lb of force is applied to the system. Determine the transfer function and value of M, D and K. The displacement x is measured from the equilibrium position. [Figure: mass M suspended from a fixed support by spring K, with damper D from M to the fixed ground below; a 10 lb step force acts on M; x(t) downward. Response: x(t) settles at 0.02 (ft), with peak overshoot 0.0093 above the final value occurring at t = 3 sec]

The system is a mass-spring-damper driven by a force. Find KK from the final value, ζ\zeta from the overshoot, and ωn\omega_n from the peak time.

Transfer function

Mx¨+Dx˙+Kx=f(t)  ⇒  X(s)F(s)=1Ms2+Ds+KM\ddot{x} + D\dot{x} + Kx = f(t) \;\Rightarrow\; \frac{X(s)}{F(s)} = \frac{1}{Ms^2 + Ds + K}

with ωn2=K/M\omega_n^2 = K/M and 2ζωn=D/M2\zeta\omega_n = D/M.

Step 1: KK from steady state

For F(s)=10/sF(s) = 10/s:

x(∞)=lim⁡s→0s⋅10s⋅1Ms2+Ds+K=10K=0.02  ⇒  K=500 lb/ftx(\infty) = \lim_{s\to 0} s\cdot\frac{10}{s}\cdot\frac{1}{Ms^2+Ds+K} = \frac{10}{K} = 0.02 \;\Rightarrow\; K = 500\ \text{lb/ft}

Step 2: ζ\zeta from overshoot

Mp=0.00930.02=0.465M_p = \frac{0.0093}{0.02} = 0.465 ζ=−ln⁡0.465π2+(ln⁡0.465)2=0.76579.8696+0.5863=0.76573.2336=0.2368\zeta = \frac{-\ln 0.465}{\sqrt{\pi^2 + (\ln 0.465)^2}} = \frac{0.7657}{\sqrt{9.8696 + 0.5863}} = \frac{0.7657}{3.2336} = 0.2368

Step 3: ωn\omega_n from tp=3t_p = 3 s

ωn=πtp1−ζ2=3.141631−0.0561=3.14163(0.9715)=1.078 rad/s\omega_n = \frac{\pi}{t_p\sqrt{1-\zeta^2}} = \frac{3.1416}{3\sqrt{1 - 0.0561}} = \frac{3.1416}{3(0.9715)} = 1.078\ \text{rad/s}

Step 4: MM and DD

M=Kωn2=5001.1618=430.4 slugD=2ζωnM=2(0.2368)(1.078)(430.4)=219.7 lb-s/ft\begin{aligned} M &= \frac{K}{\omega_n^2} = \frac{500}{1.1618} = 430.4\ \text{slug} \\ D &= 2\zeta\omega_n M = 2(0.2368)(1.078)(430.4) = 219.7\ \text{lb-s/ft} \end{aligned}

Transfer function

X(s)F(s)=1430.4s2+219.7s+500\frac{X(s)}{F(s)} = \frac{1}{430.4s^2 + 219.7s + 500}

Answer: M≈430M \approx 430 slug, D≈220D \approx 220 lb-s/ft, K=500K = 500 lb/ft.

✓ Answer checked by Claude Opus 5.5 (high effort)

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