---
title: "Find M, D, K from 10 lb response"
subject: "Control Systems"
chapter: 3
key: "semester_3/control_systems:mdk-from-response-10lb"
source: https://www.bctengineer.com/q/semester_3/control_systems/mdk-from-response-10lb
site: BCT Engineering
---

# Find M, D, K from 10 lb response

Control Systems · Chapter 3: System Transfer Function and Responses

- Asked in: 2071 Chaitra, 8 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/control_systems/mdk-from-response-10lb
- Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-3

## Question

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]

## 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.

## Answer

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

#### Transfer function
$$
M\ddot{x} + D\dot{x} + Kx = f(t)
\;\Rightarrow\;
\frac{X(s)}{F(s)} = \frac{1}{Ms^2 + Ds + K}
$$
with $\omega_n^2 = K/M$ and $2\zeta\omega_n = D/M$.

#### Step 1: $K$ from steady state
For $F(s) = 10/s$:
$$
x(\infty) = \lim_{s\to 0} s\cdot\frac{10}{s}\cdot\frac{1}{Ms^2+Ds+K} = \frac{10}{K} = 0.02
\;\Rightarrow\; K = 500\ \text{lb/ft}
$$

#### Step 2: $\zeta$ from overshoot
$$
M_p = \frac{0.0093}{0.02} = 0.465
$$
$$
\zeta = \frac{-\ln 0.465}{\sqrt{\pi^2 + (\ln 0.465)^2}} = \frac{0.7657}{\sqrt{9.8696 + 0.5863}} = \frac{0.7657}{3.2336} = 0.2368
$$

#### Step 3: $\omega_n$ from $t_p = 3$ 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: $M$ and $D$
$$
\begin{aligned}
M &= \frac{K}{\omega_n^2} = \frac{500}{1.1618} = 430.4\ \text{slug} \\
D &= 2\zeta\omega_n M = 2(0.2368)(1.078)(430.4) = 219.7\ \text{lb-s/ft}
\end{aligned}
$$

#### Transfer function
$$
\frac{X(s)}{F(s)} = \frac{1}{430.4s^2 + 219.7s + 500}
$$

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

## Check yourself

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

1. How is K found from the step response of a 10 lb force?
   A. From the final value: x(inf)=10/K=0.02
   B. From the peak time
   C. From the overshoot ratio
   D. From the damping ratio only
   Answer: A. The final value theorem gives x(inf)=F/K, so K = 10/0.02 = 500 lb/ft. (Key point: Find K)

2. The overshoot ratio Mp = 0.0093/0.02 = 0.465. What does this give?
   A. wn
   B. zeta = about 0.237
   C. M directly
   D. K
   Answer: B. zeta = -ln(Mp)/sqrt(pi^2+ln(Mp)^2) = 0.2368, since overshoot depends only on zeta. (Key point: Damping ratio)

3. Which formula finds wn from the peak time tp = 3 s?
   A. wn = tp/pi
   B. wn = 4/(zeta*tp)
   C. wn = pi/(tp*sqrt(1-zeta^2))
   D. wn = 1/tp
   Answer: C. tp = pi/(wn*sqrt(1-zeta^2)), giving wn = 1.078 rad/s. (Key point: Natural frequency)

4. Given K, wn and zeta, how are M and D obtained?
   A. M = K*wn^2, D = 2*zeta*wn
   B. M = K/wn^2, D = 2*zeta*wn*M
   C. M = wn^2/K, D = zeta*M
   D. M = K/wn, D = 2*zeta*K
   Answer: B. From wn^2=K/M and 2*zeta*wn=D/M: M = 500/1.1618 = 430.4 slug, D = 219.7 lb-s/ft. (Key point: Find M and D)

## Related in Control Systems

- [F-V and F-I analogy of mass-spring-damper system](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Afv-fi-m-k1-k2-b2) — Control Systems, chapter 2
- [State model of a mass-spring-damper system](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Ass-model-mass-spring-damper) — Control Systems, chapter 8
- [Velocity transfer function of two-mass system with analogies](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Atf-velocity-m2-fv-fi) — Control Systems, chapter 2
- [Vertical two-mass system transfer function X1/F](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Atf-vertical-m1-m2-k1-b1) — Control Systems, chapter 2
- [Single-mass system transfer function with F-V and F-I](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Atf-fv-fi-k1-m2-k2b2) — Control Systems, chapter 2
- [Two-mass system transfer function and F-V analogy](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Atf-two-mass-k1k2b1b2) — Control Systems, chapter 2

## Related in other subjects

- [Spring-mass-damper instrument parameters](https://www.bctengineer.com/question.md?key=semester_4%2Finstrumentation_and_measurement%3Aspring-mass-damper-instrument-numerical) — Instrumentation and Measurement, chapter 4