---
title: "Natural frequency with series-parallel springs"
subject: "Theory of Mechanism and Machine II"
chapter: 6
key: "semester_7/theory_of_mechanism_and_machine_ii:series-parallel-spring-mass-frequency"
source: https://www.bctengineer.com/q/semester_7/theory_of_mechanism_and_machine_ii/series-parallel-spring-mass-frequency
site: BCT Engineering
---

# Natural frequency with series-parallel springs

Theory of Mechanism and Machine II · Chapter 6: Vibration of Single Degree of Freedom Systems

- Asked in: year not given, 8 marks
- Video: [Vibration - Springs in Series and Parallel Problems in GATE](https://www.youtube.com/watch?v=cSimV5-GRNc) — Rahul Yadav
- 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/series-parallel-spring-mass-frequency
- Whole chapter: https://www.bctengineer.com/md/semester_7/theory_of_mechanism_and_machine_ii/chapter-6

## Question

A mass of 12 kg is supported by two springs of stiffness 15 kN/m and 25 kN/m connected in parallel, which are in series with a third spring of stiffness 20 kN/m. Find the equivalent stiffness, the static deflection, the natural frequency and the period of free vibration. If the mass is displaced 20 mm from equilibrium and given an initial velocity of 0.5 m/s, find the amplitude, the maximum velocity and the maximum acceleration.

## Short answer

Parallel springs give k_p = 40 kN/m; in series with 20 kN/m gives k_eq = 13.33 kN/m. Then δ_st = mg/k_eq = 8.83 mm, ω_n = √(k_eq/m) = 33.33 rad/s, f_n = 5.305 Hz, T = 0.1885 s. With initial displacement 20 mm and velocity 0.5 m/s, amplitude X = 25.0 mm, v_max = 0.833 m/s, a_max = 27.78 m/s².

## Answer

**Data:** $m = 12$ kg, $k_1 = 15000$ N/m, $k_2 = 25000$ N/m (parallel), $k_3 = 20000$ N/m (in series with the pair).

#### Equivalent stiffness
$$
k_p = k_1 + k_2 = 15000 + 25000 = 40000\ \text{N/m}
$$

$$
k_{eq} = \frac{k_p k_3}{k_p + k_3} = \frac{40000 \times 20000}{40000 + 20000} = 13333.3\ \text{N/m}
$$

#### Static deflection
$$
\delta_{st} = \frac{mg}{k_{eq}} = \frac{12 \times 9.81}{13333.3} = 0.00883\ \text{m} = 8.83\ \text{mm}
$$

#### Natural frequency and period
$$
\omega_n = \sqrt{\frac{k_{eq}}{m}} = \sqrt{\frac{13333.3}{12}} = 33.33\ \text{rad/s}
$$

$$
f_n = \frac{\omega_n}{2\pi} = 5.305\ \text{Hz}, \qquad T = \frac{1}{f_n} = 0.1885\ \text{s}
$$

#### Response to the initial conditions
$x(t) = x_0\cos\omega_n t + \dfrac{v_0}{\omega_n}\sin\omega_n t = X\sin(\omega_n t + \phi)$

$$
X = \sqrt{x_0^2 + \left(\frac{v_0}{\omega_n}\right)^2} = \sqrt{0.02^2 + \left(\frac{0.5}{33.33}\right)^2} = 0.0250\ \text{m} = 25.0\ \text{mm}
$$

$$
\phi = \tan^{-1}\left(\frac{\omega_n x_0}{v_0}\right) = \tan^{-1}\left(\frac{33.33 \times 0.02}{0.5}\right) = 53.1^\circ
$$

$$
v_{max} = \omega_n X = 33.33 \times 0.0250 = 0.833\ \text{m/s}, \qquad a_{max} = \omega_n^2 X = 27.78\ \text{m/s}^2
$$

**Answer:** $k_{eq} = 13333.3$ N/m; $\delta_{st} = 8.83$ mm; $f_n = 5.305$ Hz; $T = 0.1885$ s; $X = 25.0$ mm; $v_{max} = 0.833$ m/s; $a_{max} = 27.78$ m/s².

## Check yourself

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

1. The two parallel springs (15 and 25 kN/m) in series with 20 kN/m give:
   A. 13333 N/m
   B. 60000 N/m
   C. 40000 N/m
   D. 6667 N/m
   Answer: A. Parallel: 40000 N/m; then series: 40000 x 20000/60000 = 13333 N/m. (Key point: Equivalent stiffness)

2. Static deflection is calculated as:
   A. m g / k_eq = 8.83 mm
   B. k_eq / (m g)
   C. m / k_eq = 0.9 mm
   D. m g k_eq
   Answer: A. 12 x 9.81/13333 = 0.00883 m. (Key point: Static deflection)

3. With initial displacement x0 and velocity v0, the amplitude is:
   A. sqrt(x0^2 + (v0/w_n)^2)
   B. x0 + v0/w_n
   C. x0 only
   D. sqrt(x0^2 + v0^2)
   Answer: A. X = sqrt(0.02^2 + (0.5/33.33)^2) = 25 mm. (Key point: Amplitude)

4. Maximum velocity and acceleration are:
   A. w_n X and w_n^2 X
   B. w_n^2 X and w_n X
   C. X/w_n and X/w_n^2
   D. X w_n and X w_n
   Answer: A. v_max = 33.33 x 0.025 = 0.833 m/s, a_max = 33.33^2 x 0.025 = 27.78 m/s^2. (Key point: Maximum velocity)

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