---
title: "Mass between two springs, period"
subject: "Engineering Physics"
chapter: 1
key: "semester_1/engineering_physics:mass-spring-two-springs"
source: https://www.bctengineer.com/q/semester_1/engineering_physics/mass-spring-two-springs
site: BCT Engineering
---

# Mass between two springs, period

Engineering Physics · Chapter 1: Oscillation

- Asked in: 2068 Shrawan, 5 marks
- Video: [Physics - 16.5  Simple Harmonic Motion - 2 Springs (1 of 5) 2 Equal Springs, 1 Mass](https://www.youtube.com/watch?v=l0BphtEX8Ts) — Michel van Biezen
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/engineering_physics/mass-spring-two-springs
- Whole chapter: https://www.bctengineer.com/md/semester_1/engineering_physics/chapter-1

## Question

Explain the theory of a simple mass spring system. Develop the relation for time period and frequency of two springs having spring constants K1 and K2 supporting a mass 'M' between them on a frictionless horizontal table.

## Short answer

For a simple mass-spring, m d²x/dt² = –kx so T = 2π√(m/k). With mass M between two springs K1, K2 on a frictionless table, displacing x stretches one and compresses the other, net restoring force = –(K1+K2)x. Thus T = 2π√(M/(K1+K2)).

## Answer

#### Simple mass-spring system
A mass m attached to a spring of constant k on a frictionless surface. When displaced by x, the spring exerts the restoring force F = −kx (Hooke's law).
```text
m d²x/dt² = −kx   →   d²x/dt² + (k/m) x = 0
x = a sin(ωt + φ) ,  ω = √(k/m) ,  T = 2π√(m/k) ,  f = (1/2π)√(k/m)
```
The motion is SHM; the period depends on m and k only.

#### Mass between two springs
A mass M rests on a smooth horizontal table between two fixed walls, joined to them by springs of constants K1 and K2.
```text
 |W|~~~K1~~~[ M ]~~~K2~~~|W|
              <-x->
```
1. At equilibrium both springs are unstretched (or equally pre-stretched). Displace M by x to the right.
2. Spring 1 is stretched by x and pulls the mass back with force K1 x. Spring 2 is compressed by x and pushes the mass back with force K2 x. Both act in the same direction (towards equilibrium).
3. Net restoring force:
```text
F = −(K1 + K2) x
```
4. Equation of motion: M d²x/dt² = −(K1 + K2) x, i.e. d²x/dt² + [(K1 + K2)/M] x = 0.
5. This is SHM with equivalent spring constant K = K1 + K2:
```text
ω = √((K1 + K2)/M)
T = 2π √(M/(K1 + K2))
f = (1/2π) √((K1 + K2)/M)
```
The springs in this arrangement behave as springs in parallel, so the effective stiffness is higher and the period is shorter than for either spring alone.

## Check yourself

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

1. A mass M lies between two springs K1 and K2 and is displaced by x. What is the net restoring force?
   A. (K1 - K2)x
   B. K1K2 x/(K1+K2)
   C. (K1 + K2)x
   D. K1 x only
   Answer: C. One spring is stretched and the other compressed, and both push the mass back toward equilibrium, so the forces add: (K1 + K2)x. (Key point: Net restoring force)

2. What is the time period of the mass between the two springs?
   A. 2π√(M/(K1+K2))
   B. 2π√(M(K1+K2))
   C. 2π√((K1+K2)/M)
   D. 2π√(M/(K1 - K2))
   Answer: A. With equivalent constant K = K1 + K2, T = 2π√(M/K). (Key point: Time period)

3. Compared with either spring alone, the two-spring arrangement has
   A. lower stiffness and longer period
   B. the same period
   C. higher stiffness and shorter period
   D. higher stiffness and longer period
   Answer: C. The springs act like springs in parallel, so K = K1 + K2 is larger and T is smaller. (Key point: Effective stiffness)

4. For a mass m on a spring k, the equation m d²x/dt² = -kx leads to angular frequency
   A. √(m/k)
   B. k/m
   C. √(k/m)
   D. m/k
   Answer: C. Writing d²x/dt² + (k/m)x = 0 gives ω² = k/m, so ω = √(k/m). (Key point: Single spring SHM)

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