---
title: "Slope of string equals speed ratio"
subject: "Engineering Physics"
chapter: 2
key: "semester_1/engineering_physics:string-wave-slope-speed-ratio"
source: https://www.bctengineer.com/q/semester_1/engineering_physics/string-wave-slope-speed-ratio
site: BCT Engineering
---

# Slope of string equals speed ratio

Engineering Physics · Chapter 2: Wave motion

- Asked in: 2069 Chaitra, 5 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/engineering_physics/string-wave-slope-speed-ratio
- Whole chapter: https://www.bctengineer.com/md/semester_1/engineering_physics/chapter-2

## Question

Prove that if a transverse wave is travelling along a string, then the slope at any point of the string is numerically equal to the ratio of the particle speed to the wave speed at that point.

## Short answer

For y = f(x−vt), ∂y/∂x = f' and ∂y/∂t = −v f', so |∂y/∂x| = |∂y/∂t|/v. The slope equals the ratio of particle speed to wave speed.

## Answer

Let a transverse wave travel along a string in the +x direction with speed v. Its displacement can be written as a function of (x − vt):
```text
y = f(x − vt)
```
(for example y = A sin(kx − ωt) = A sin k(x − vt), with v = ω/k).

#### Proof
Let u = x − vt.
1. Slope of the string at a point:
```text
∂y/∂x = f′(u) · ∂u/∂x = f′(u)
```
2. Transverse speed of the particle at that point:
```text
∂y/∂t = f′(u) · ∂u/∂t = −v f′(u)
```
3. Compare:
```text
∂y/∂t = −v (∂y/∂x)
```
4. Taking magnitudes:
```text
| ∂y/∂x | = | ∂y/∂t | / v
```
So the slope at any point of the string is numerically equal to the ratio of the particle speed to the wave speed at that point. For a wave travelling in the −x direction, y = f(x + vt) and ∂y/∂t = +v ∂y/∂x, which gives the same magnitude relation.

Check with y = A sin(kx − ωt): slope = Ak cos(kx − ωt); particle speed = −Aω cos(kx − ωt); ratio = ω/k = v.

## Check yourself

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

1. A wave travelling in the +x direction with speed v can be written as
   A. y = f(x)
   B. y = f(x - vt)
   C. y = f(x + vt)
   D. y = f(vt)
   Answer: B. A function of (x - vt) keeps its shape and moves in +x at speed v. (Key point: Form of travelling wave)

2. With u = x - vt, the slope ∂y/∂x equals
   A. -v f'(u)
   B. f'(u)/v
   C. f'(u)
   D. v f'(u)
   Answer: C. ∂u/∂x = 1, so ∂y/∂x = f'(u). (Key point: Slope)

3. The transverse particle speed ∂y/∂t equals
   A. f'(u)
   B. -v f'(u)
   C. v² f'(u)
   D. f'(u)/v
   Answer: B. ∂u/∂t = -v, so ∂y/∂t = -v f'(u) = -v ∂y/∂x. (Key point: Particle speed)

4. Taking magnitudes, |∂y/∂x| equals
   A. v |∂y/∂t|
   B. |∂y/∂t| / v
   C. |∂y/∂t| × v²
   D. |∂y/∂t|
   Answer: B. From ∂y/∂t = -v ∂y/∂x, slope magnitude = particle speed divided by wave speed. (Key point: Conclusion)

## Related in Engineering Physics

- [Transverse wave equation on a string](https://www.bctengineer.com/question.md?key=semester_1%2Fengineering_physics%3Atransverse-wave-string-velocity) — Engineering Physics, chapter 2
- [Average power of wave on a string](https://www.bctengineer.com/question.md?key=semester_1%2Fengineering_physics%3Astring-wave-average-power-linear-density) — Engineering Physics, chapter 2
- [Maximum wave speed in steel wire](https://www.bctengineer.com/question.md?key=semester_1%2Fengineering_physics%3Asteel-wire-maximum-wave-speed) — Engineering Physics, chapter 2
- [Mass of particle from de Broglie ratio](https://www.bctengineer.com/question.md?key=semester_1%2Fengineering_physics%3Adebroglie-ratio-mass-of-particle) — Engineering Physics, chapter 10

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