---
title: "String released from sin³(πx/l)"
subject: "Applied Mathematics"
chapter: 3
key: "semester_4/applied_mathematics:wave-eq-string-sin3"
source: https://www.bctengineer.com/q/semester_4/applied_mathematics/wave-eq-string-sin3
site: BCT Engineering
---

# String released from sin³(πx/l)

Applied Mathematics · Chapter 3: Partial Differential Equations

- Asked in: 2069 Poush; 2068 Magh, 10 marks (asked 2 times)
- Video: [Problem on wave Equation](https://www.youtube.com/watch?v=TyLBj7yO-LY) — Sreeja's MathWorld
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/applied_mathematics/wave-eq-string-sin3
- Whole chapter: https://www.bctengineer.com/md/semester_4/applied_mathematics/chapter-3

## Question

A tightly stretched string with fixed end points x = 0 and x = l is initially in a position given by y = y₀ sin³(πx/l). If it is released from rest from this position, find the displacement y(x, t).

## Short answer

Express sin³(πx/l) as (3 sin(πx/l) – sin(3πx/l))/4. With zero initial velocity, only cosine terms remain, yielding y(x,t) = (y₀/4)[3 cos(πct/l) sin(πx/l) – cos(3πct/l) sin(3πx/l)].

## Answer

**Problem.** u_tt = c² u_xx (here y is the displacement), with y(0, t) = y(l, t) = 0, y(x, 0) = y₀ sin³(πx/l), and zero initial velocity.

**Step 1: general solution (ends fixed).** Separation of variables gives

y(x, t) = Σ_{n=1}^{∞} [ Bₙ cos(nπct/l) + Bₙ* sin(nπct/l) ] sin(nπx/l)

**Step 2: zero initial velocity** gives ∂y/∂t = 0 at t = 0, so every Bₙ* (nπc/l) = 0, hence Bₙ* = 0.

y(x, t) = Σ Bₙ cos(nπct/l) sin(nπx/l)

**Step 3: initial shape.** Use sin 3θ = 3 sin θ − 4 sin³θ, so sin³θ = (3 sin θ − sin 3θ)/4. With θ = πx/l:

```text
y(x, 0) = Σ Bₙ sin(nπx/l)
        = (y₀/4) [ 3 sin(πx/l) − sin(3πx/l) ]
```

Compare coefficients (the sine functions are orthogonal):

B₁ = 3y₀/4, B₃ = −y₀/4, all other Bₙ = 0.

**Answer:**

y(x, t) = (y₀/4) [ 3 cos(πct/l) sin(πx/l) − cos(3πct/l) sin(3πx/l) ]

Check: at t = 0 this equals (y₀/4)(3 sin θ − sin 3θ) = y₀ sin³θ ✓.

## Check yourself

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

1. Released from rest gives
   A. Bn* = 0
   B. Bn = 0
   C. Bn* = Bn
   D. c = 0
   Answer: A. No sine in t. (Key point: Velocity)

2. sin^3 theta =
   A. (3 sin theta + sin 3 theta)/4
   B. (3 sin theta - sin 3 theta)/4
   C. 3 sin theta - sin 3theta
   D. sin 3 theta/4
   Answer: B. From sin 3theta. (Key point: Identity)

3. B1 and B3 are
   A. y0/4 and 3y0/4
   B. 3y0/4 and y0/4
   C. 3y0/4 and -y0/4
   D. y0 and 0
   Answer: C. Match. (Key point: Coefficients)

4. y(x,t) =
   A. (y0/4)[3 sin(pi c t/l) sin(pi x/l) - sin(3 pi c t/l) sin(3 pi x/l)]
   B. y0 sin^3(pi x/l) cos(pi c t/l)
   C. (y0/4)[3 cos(pi c t/l) sin(pi x/l) + cos(3 pi c t/l) sin(3 pi x/l)]
   D. (y0/4)[3 cos(pi c t/l) sin(pi x/l) - cos(3 pi c t/l) sin(3 pi x/l)]
   Answer: D. At t=0 equals y0 sin^3. (Key point: Result)

## Related in Applied Mathematics

- [String with initial velocity 3(lx-x²)](https://www.bctengineer.com/question.md?key=semester_4%2Fapplied_mathematics%3Awave-eq-string-velocity-3lx-x2) — Applied Mathematics, chapter 3, asked 2×
- [String with initial deflection lx - x²](https://www.bctengineer.com/question.md?key=semester_4%2Fapplied_mathematics%3Awave-eq-string-lx-x2) — Applied Mathematics, chapter 3
- [Bilinear map z = 0,1,∞ to -3,-1,1 with fixed points](https://www.bctengineer.com/question.md?key=semester_4%2Fapplied_mathematics%3Abilinear-0-1-inf-fixed-points) — Applied Mathematics, chapter 1

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- [Rod released from horizontal position](https://www.bctengineer.com/question.md?key=semester_3%2Fengineering_mechanics%3Arod-release-from-horizontal-reaction) — Engineering Mechanics, chapter 6
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