---
title: "Spring-mass-damper instrument parameters"
subject: "Instrumentation and Measurement"
chapter: 4
key: "semester_4/instrumentation_and_measurement:spring-mass-damper-instrument-numerical"
source: https://www.bctengineer.com/q/semester_4/instrumentation_and_measurement/spring-mass-damper-instrument-numerical
site: BCT Engineering
---

# Spring-mass-damper instrument parameters

Instrumentation and Measurement · Chapter 4: Dynamic Response of Measurement System

- Asked in: year not given, 6 marks
- Video: [Understanding Vibration and Resonance](https://www.youtube.com/watch?v=vLaFAKnaRJU) — The Efficient Engineer
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/instrumentation_and_measurement/spring-mass-damper-instrument-numerical
- Whole chapter: https://www.bctengineer.com/md/semester_4/instrumentation_and_measurement/chapter-4

## Question

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 $\zeta = 0.7$.

## 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.

## Answer

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

#### (a) Natural frequency and damping ratio
$$
\omega_n = \sqrt{\frac{k}{m}} = \sqrt{\frac{2000}{0.05}} = 200\ \text{rad/s}
$$

$$
\zeta = \frac{c}{2\sqrt{km}} = \frac{4}{2\sqrt{2000 \times 0.05}} = \frac{4}{2(10)} = 0.2
$$

The system is underdamped.

#### (b) Damped frequency
$$
\omega_d = \omega_n\sqrt{1-\zeta^2} = 200\sqrt{0.96} = 195.96\ \text{rad/s}, \qquad f_d = \frac{195.96}{2\pi} = 31.19\ \text{Hz}
$$

#### (c) Step response
$$
M_p = \exp\left(\frac{-\pi(0.2)}{\sqrt{0.96}}\right) = \exp(-0.6413) = 0.527 = 52.7\%
$$

$$
t_p = \frac{\pi}{\omega_d} = \frac{\pi}{195.96} = 16.0\ \text{ms}
$$

$$
t_s \approx \frac{4}{\zeta\omega_n} = \frac{4}{0.2 \times 200} = 0.100\ \text{s}
$$

#### (d) Damping for $\zeta = 0.7$
$$
c = 2\zeta\sqrt{km} = 2(0.7)(10) = 14\ \text{N s/m}
$$

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

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. For m = 0.05 kg, k = 2000 N/m, ωn is
   A. 200 rad/s
   B. 40000 rad/s
   C. 63.2 rad/s
   D. 20 rad/s
   Answer: A. ωn = sqrt(2000/0.05) = 200 rad/s. (Key point: Natural frequency)

2. With c = 4 N s/m, ζ = c/(2 sqrt(km)) =
   A. 0.4
   B. 0.1
   C. 0.2
   D. 2
   Answer: C. sqrt(km) = 10, so ζ = 4/20 = 0.2. (Key point: Damping ratio)

3. The step-response overshoot with ζ = 0.2 is about
   A. 5.3%
   B. 52.7%
   C. 20%
   D. 100%
   Answer: B. exp(-π·0.2/sqrt(0.96)) = 0.527. (Key point: Overshoot)

4. Damping coefficient needed for ζ = 0.7 is
   A. 14 N s/m
   B. 2.8 N s/m
   C. 7 N s/m
   D. 28 N s/m
   Answer: A. c = 2ζ sqrt(km) = 2(0.7)(10) = 14 N s/m. (Key point: Required damping)

## Related in Instrumentation and Measurement

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- [Second-order step response and damping](https://www.bctengineer.com/question.md?key=semester_4%2Finstrumentation_and_measurement%3Asecond-order-step-response-damping) — Instrumentation and Measurement, chapter 4

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