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Spring-mass-damper instrument parameters

Short answer

ω_n = √(k/m) = 200 rad/s, ζ = c/(2√(km)) = 0.2. Damped frequency f_d = ω_n√(1-ζ²)/(2π) = 31.2 Hz. Overshoot = exp(-πζ/√(1-ζ²)) = 52.7%, t_p = π/ω_d =16.0 ms, t_s = 4/(ζω_n) = 0.1 s. For ζ=0.7, c = 2ζ√(km) = 14 N s/m.

A seismic-type measuring element has mass 0.05 kg, spring stiffness 2000 N/m and viscous damping coefficient 4 N s/m. Find (a) the natural frequency and damping ratio, (b) the damped frequency in Hz, (c) the percentage overshoot, peak time and 2% settling time for a step input, and (d) the damping coefficient needed for ζ=0.7\zeta = 0.7.

Given: m=0.05m = 0.05 kg, k=2000k = 2000 N/m, c=4c = 4 N s/m.

(a) Natural frequency and damping ratio

ωn=km=20000.05=200 rad/s\omega_n = \sqrt{\frac{k}{m}} = \sqrt{\frac{2000}{0.05}} = 200\ \text{rad/s} ζ=c2km=422000×0.05=42(10)=0.2\zeta = \frac{c}{2\sqrt{km}} = \frac{4}{2\sqrt{2000 \times 0.05}} = \frac{4}{2(10)} = 0.2

The system is underdamped.

(b) Damped frequency

ωd=ωn1−ζ2=2000.96=195.96 rad/s,fd=195.962π=31.19 Hz\omega_d = \omega_n\sqrt{1-\zeta^2} = 200\sqrt{0.96} = 195.96\ \text{rad/s}, \qquad f_d = \frac{195.96}{2\pi} = 31.19\ \text{Hz}

(c) Step response

Mp=exp⁡(−π(0.2)0.96)=exp⁡(−0.6413)=0.527=52.7%M_p = \exp\left(\frac{-\pi(0.2)}{\sqrt{0.96}}\right) = \exp(-0.6413) = 0.527 = 52.7\% tp=πωd=π195.96=16.0 mst_p = \frac{\pi}{\omega_d} = \frac{\pi}{195.96} = 16.0\ \text{ms} ts≈4ζωn=40.2×200=0.100 st_s \approx \frac{4}{\zeta\omega_n} = \frac{4}{0.2 \times 200} = 0.100\ \text{s}

(d) Damping for ζ=0.7\zeta = 0.7

c=2ζkm=2(0.7)(10)=14 N s/mc = 2\zeta\sqrt{km} = 2(0.7)(10) = 14\ \text{N s/m}

Answer: ωn=200\omega_n = 200 rad/s, ζ=0.2\zeta = 0.2; fd=31.2f_d = 31.2 Hz; overshoot 52.7%52.7\%, tp=16.0t_p = 16.0 ms, ts=0.1t_s = 0.1 s; required c=14c = 14 N s/m.

✓ Answer checked by Claude Opus 5.5 (high effort)

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