---
title: "Damped block-spring oscillation period"
subject: "Engineering Physics"
chapter: 1
key: "semester_1/engineering_physics:damped-block-spring-period"
source: https://www.bctengineer.com/q/semester_1/engineering_physics/damped-block-spring-period
site: BCT Engineering
---

# Damped block-spring oscillation period

Engineering Physics · Chapter 1: Oscillation

- Asked in: 2068 Chaitra, 5 marks
- Video: [Free damped mechanical oscillations](https://www.youtube.com/watch?v=veUgjspHEKM) — Everest Physics
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/engineering_physics/damped-block-spring-period
- Whole chapter: https://www.bctengineer.com/md/semester_1/engineering_physics/chapter-1

## Question

A 750g block oscillates on the end of a spring whose force constant is k = 56 N/m. The mass moves in a fluid which offers a resistive force F = −bv, where b = 0.162 Ns/m. What is the period of the oscillation?

## Short answer

The damped angular frequency is ω′ = √(k/m − (b/(2m))²). With m=0.750 kg, k=56 N/m, b=0.162 N·s/m, ω′ ≈ 8.640 rad/s, giving period T = 2π/ω′ ≈ 0.727 s.

## Answer

Given: m = 750 g = 0.750 kg, k = 56 N/m, b = 0.162 N·s/m.

Equation of motion: m x'' + b x' + k x = 0. For weak damping the angular frequency of the oscillation is
```text
ω′ = √(k/m − (b/2m)²)
```

#### Steps
1. k/m = 56 / 0.750 = 74.667 s⁻² (ω0 = 8.641 rad/s).
2. b/2m = 0.162 / (2 × 0.750) = 0.108 s⁻¹, and (b/2m)² = 0.01166 s⁻².
3. ω′² = 74.667 − 0.0117 = 74.655, so ω′ = 8.6403 rad/s.
4. Period:
```text
T = 2π/ω′ = 6.2832 / 8.6403 = 0.7272 s
```
Compare undamped: T0 = 2π√(m/k) = 0.7271 s. The damping is very small (b² << 4mk since 0.026 << 168), so the period barely changes.

**Answer: T ≈ 0.727 s.**

## Check yourself

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

1. For weak damping, the angular frequency of the oscillating block is
   A. √(k/m + (b/2m)²)
   B. √(k/m) + b/2m
   C. √(k/m - (b/2m)²)
   D. √(k/m - b/m)
   Answer: C. Damping reduces the frequency: ω' = √(k/m - (b/2m)²). (Key point: Damped angular frequency)

2. With m = 0.750 kg and b = 0.162 N s/m, what is b/2m?
   A. 0.216 s^-1
   B. 0.108 s^-1
   C. 0.0117 s^-1
   D. 0.324 s^-1
   Answer: B. b/2m = 0.162/(2 × 0.750) = 0.108 s^-1. (Key point: Calculating b/2m)

3. What is the period of oscillation?
   A. 0.727 s
   B. 1.45 s
   C. 0.363 s
   D. 8.64 s
   Answer: A. T = 2π/ω' = 6.2832/8.6403 ≈ 0.727 s. (Key point: Period result)

4. Why is the damped period almost equal to the undamped period T0 = 2π√(m/k)?
   A. Because the mass is large
   B. Because damping makes b = 0
   C. Because the spring constant changes
   D. Because b² is much smaller than 4mk, so (b/2m)² is negligible next to k/m
   Answer: D. Here (b/2m)² = 0.0117 compared with k/m = 74.7, so the correction to ω is tiny. (Key point: Effect of light damping)

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