---
title: "Natural frequencies of a two-mass system"
subject: "Theory of Mechanism and Machine II"
chapter: 7
key: "semester_7/theory_of_mechanism_and_machine_ii:two-dof-natural-frequencies-mode-shapes"
source: https://www.bctengineer.com/q/semester_7/theory_of_mechanism_and_machine_ii/two-dof-natural-frequencies-mode-shapes
site: BCT Engineering
---

# Natural frequencies of a two-mass system

Theory of Mechanism and Machine II · Chapter 7: Vibration of Two Degree of Freedom Systems

- Asked in: year not given, 8 marks
- Video: [Mechanical Vibration: MDOF Calculating the Natural Frequencies (Part II)](https://www.youtube.com/watch?v=otQZnq21mCk) — Azma Putra
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/theory_of_mechanism_and_machine_ii/two-dof-natural-frequencies-mode-shapes
- Whole chapter: https://www.bctengineer.com/md/semester_7/theory_of_mechanism_and_machine_ii/chapter-7

## Question

Two masses $m_1 = 2$ kg and $m_2 = 1$ kg are connected in series to a fixed wall by two springs. The spring between the wall and $m_1$ has stiffness $k_1 = 3000$ N/m and the spring between $m_1$ and $m_2$ has stiffness $k_2 = 2000$ N/m. The masses move along a frictionless horizontal line. Find the natural frequencies and mode shapes, and sketch the modes.

## Short answer

Natural frequencies: ω1 = 28.53 rad/s and ω2 = 60.71 rad/s. Mode shapes (normalised to X1=1): first mode {1, 1.686} (in phase), second mode {1, -1.186} (out of phase).

## Answer

**Data:** $m_1 = 2$ kg, $m_2 = 1$ kg, $k_1 = 3000$ N/m, $k_2 = 2000$ N/m.

```text
 wall |--k1--[ m1 ]--k2--[ m2 ]
          x1       x2
```

#### Equations of motion
$$
m_1\ddot{x}_1 + (k_1+k_2)x_1 - k_2x_2 = 0, \qquad m_2\ddot{x}_2 - k_2x_1 + k_2x_2 = 0
$$

In matrix form with $x = X\sin\omega t$:

$$
\begin{bmatrix} 5000 - 2\omega^2 & -2000 \\ -2000 & 2000 - \omega^2 \end{bmatrix}\begin{Bmatrix} X_1 \\ X_2 \end{Bmatrix} = \begin{Bmatrix} 0 \\ 0 \end{Bmatrix}
$$

#### Frequency equation
$$
(5000 - 2\omega^2)(2000 - \omega^2) - (2000)^2 = 0
$$

$$
2\omega^4 - 9000\omega^2 + 6\times10^6 = 0 \;\Rightarrow\; \omega^4 - 4500\omega^2 + 3\times10^6 = 0
$$

$$
\omega^2 = \frac{4500 \pm \sqrt{4500^2 - 4(3\times10^6)}}{2} = \frac{4500 \pm 2872.28}{2}
$$

$$
\omega_1^2 = 813.86 \Rightarrow \omega_1 = 28.53\ \text{rad/s}\ (f_1 = 4.54\ \text{Hz})
$$

$$
\omega_2^2 = 3686.14 \Rightarrow \omega_2 = 60.71\ \text{rad/s}\ (f_2 = 9.66\ \text{Hz})
$$

#### Mode shapes
From the second row, $\dfrac{X_2}{X_1} = \dfrac{2000}{2000 - \omega^2}$ (equivalently $\dfrac{5000 - 2\omega^2}{2000}$).

| Mode | $\omega$ (rad/s) | $X_2/X_1$ | Shape |
|---|---|---|---|
| 1 | 28.53 | 1.686 | both masses move in phase, $m_2$ moves more |
| 2 | 60.71 | -1.186 | masses move in opposite phase (one node between them) |

```text
 Mode 1: m1 -> , m2 ---> (same direction)
 Mode 2: m1 -> , m2 <-   (node between m1 and m2)
```

**Answer:** $\omega_1 = 28.53$ rad/s, $\omega_2 = 60.71$ rad/s; mode shapes $\{1,\ 1.686\}$ and $\{1,\ -1.186\}$.

## Check yourself

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

1. For the wall-k1-m1-k2-m2 system, the equation of motion of m1 is:
   A. m1 x1'' + k1 x1 + k2 x2 = 0
   B. m1 x1'' + k1 x1 - k2 x2 = 0
   C. m1 x1'' + (k1 + k2) x1 - k2 x2 = 0
   D. m1 x1'' + k2 (x1 - x2) = 0
   Answer: C. m1 is pulled by k1 (extension x1) and k2 (extension x1 - x2), giving (k1 + k2) x1 - k2 x2. (Key point: Equation of motion for m1)

2. How are the natural frequencies found?
   A. Set the determinant of [K] - w^2 [M] to zero
   B. Set the trace of [K] equal to w^2
   C. Set [K] x {X} = {0} directly
   D. Set [M] = w^2 [K]
   Answer: A. A non-trivial solution of ([K] - w^2 [M]){X} = 0 requires det = 0; here w^4 - 4500 w^2 + 3x10^6 = 0. (Key point: Frequency equation)

3. The natural frequencies for this system are approximately:
   A. 14.3 and 30.4 rad/s
   B. 28.53 and 60.71 rad/s
   C. 813.9 and 3686 rad/s
   D. 50 and 100 rad/s
   Answer: B. w^2 = 813.86 and 3686.14 (rad/s)^2, so w = 28.53 and 60.71 rad/s; the values 813.9 and 3686 are w^2. (Key point: Natural frequencies)

4. In the second mode (X2/X1 = -1.186):
   A. the masses move in the same direction with m2 moving more
   B. only m1 moves
   C. the masses move in opposite directions with a node between them
   D. the masses move together as a rigid body
   Answer: C. A negative amplitude ratio means opposite phase, so one node lies between m1 and m2; mode 1 has both in phase (ratio +1.686). (Key point: Mode shape character)

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