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UPSC 2020 Maths Optional Paper 2 Q8a — Step-by-Step Solution

20 marks · Section B

Wave equation · PDEs · asked 7× in 13 yrs · Read the full method →

Question

One end of a tightly stretched flexible thin string of length ll is fixed at the origin and the other at x=lx=l. It is plucked at x=l/3x=l/3 so that it assumes initially the shape of a triangle of height hh in the xx-yy plane. Find the displacement yy at any distance xx and at any time tt after the string is released from rest. Take horizontal tensionmass per unit length=c2\frac{\text{horizontal tension}}{\text{mass per unit length}}=c^2.

Technique

d’Alembert/Fourier (separation of variables); cosine-in-time modes for the rest condition; half-range sine coefficients of the triangular initial profile.

Solution

Step 1 — Governing equation and boundary/initial conditions

The transverse displacement satisfies the one-dimensional wave equation

2yt2=c22yx2,0<x<l,\frac{\partial^2 y}{\partial t^2}=c^2\frac{\partial^2 y}{\partial x^2},\qquad 0<x<l,

with

y(0,t)=0,y(l,t)=0(fixed ends),yt(x,0)=0(released from rest),y(0,t)=0,\quad y(l,t)=0\quad(\text{fixed ends}),\qquad \frac{\partial y}{\partial t}(x,0)=0\quad(\text{released from rest}),

and initial shape (triangle, peak hh at x=l/3x=l/3):

y(x,0)=f(x)={3hlx,0xl3,3h2l(lx),l3xl.y(x,0)=f(x)= \begin{cases}\dfrac{3h}{l}\,x, & 0\le x\le \dfrac{l}{3},\\[2mm] \dfrac{3h}{2l}\,(l-x), & \dfrac{l}{3}\le x\le l.\end{cases}

(At x=l/3x=l/3 both branches give hh.)

Step 2 — Separation of variables

The solution satisfying the fixed-end conditions and zero initial velocity is

y(x,t)=n=1bnsinnπxlcosnπctl,y(x,t)=\sum_{n=1}^{\infty} b_n\sin\frac{n\pi x}{l}\cos\frac{n\pi c t}{l},

since each mode sinnπxlcosnπctl\sin\frac{n\pi x}{l}\cos\frac{n\pi ct}{l} has zero velocity at t=0t=0 and vanishes at x=0,lx=0,l.

Step 3 — Fourier coefficients from the initial shape

bn=2l0lf(x)sinnπxldx=2l ⁣[0l/33hlxsinnπxldx+l/3l3h2l(lx)sinnπxldx].b_n=\frac{2}{l}\int_0^l f(x)\sin\frac{n\pi x}{l}\,dx =\frac{2}{l}\!\left[\int_0^{l/3}\frac{3h}{l}x\sin\frac{n\pi x}{l}dx+\int_{l/3}^{l}\frac{3h}{2l}(l-x)\sin\frac{n\pi x}{l}dx\right].

Carrying out the integration (integration by parts),

bn=9hπ2n2sinnπ3.b_n=\frac{9h}{\pi^2 n^2}\sin\frac{n\pi}{3}.

(Equivalently, the standard plucked-string result bn=2hl2π2n2a(la)sinnπalb_n=\dfrac{2hl^2}{\pi^2 n^2\,a(l-a)}\sin\dfrac{n\pi a}{l} with a=l/3a=l/3.) Note sinnπ3=0\sin\frac{n\pi}{3}=0 for n=3,6,9,n=3,6,9,\dots, so every third harmonic is absent.

Step 4 — Displacement

Answer

  y(x,t)=9hπ2n=11n2sinnπ3sinnπxlcosnπctl.  \boxed{\;y(x,t)=\frac{9h}{\pi^2}\sum_{n=1}^{\infty}\frac{1}{n^2}\sin\frac{n\pi}{3}\,\sin\frac{n\pi x}{l}\,\cos\frac{n\pi c t}{l}.\;}
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