UPSC 2017 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

Given the one-dimensional wave equation $\dfrac{\partial^2 y}{\partial t^2}=c^2\dfrac{\partial^2 y}{\partial x^2};\ t>0$ , where $c^2=\dfrac{T}{m}$ , $T$ is the constant tension in the string and $m$ is the mass per unit length of the string.

(i) Find the appropriate solution of the above wave equation.
(ii) Find also the solution under the conditions $y(0,t)=0,\ y(l,t)=0$ for all $t$ and $\left[\dfrac{\partial y}{\partial t}\right]_{t=0}=0,\ y(x,0)=a\sin\dfrac{\pi x}{l},\ 0<x<l,\ a>0$ .

Technique

Separation of variables with negative separation constant $-k^2$ (oscillatory); fixed-end BCs quantise $k=n\pi/l$ ; zero initial velocity kills all $\sin(n\pi ct/l)$ time-terms; the single-mode initial shape selects only $n=1$ .

Solution

Part (i) — Appropriate solution by separation of variables

Seek $y(x,t)=X(x)\,G(t)$ . Substituting into $y_{tt}=c^2y_{xx}$ :

X\ddot G=c^2 X'' G\ \Longrightarrow\ \frac{\ddot G}{c^2 G}=\frac{X''}{X}=\text{const}.

For a vibrating string (bounded, oscillatory) the separation constant must be negative, say $-k^2$ :

X''+k^2X=0,\qquad \ddot G+c^2k^2G=0.

Hence

X(x)=A\cos kx+B\sin kx,\qquad G(t)=C\cos ckt+E\sin ckt.

The appropriate (physically relevant, periodic-in-time) solution is

\boxed{\;y(x,t)=(A\cos kx+B\sin kx)(C\cos ckt+E\sin ckt).\;}

(The choices " $+k^2$ " → exponentials in $x$ and " $0$ " → linear are rejected as they cannot satisfy fixed-end oscillation.)

Part (ii) — Apply the boundary and initial conditions

BC1: $y(0,t)=0$ for all $t$ . Then $X(0)=0\Rightarrow A=0$ . So $y=B\sin kx\,(C\cos ckt+E\sin ckt)$ .

BC2: $y(l,t)=0$ for all $t$ . Then $\sin kl=0\Rightarrow kl=n\pi\Rightarrow k=\dfrac{n\pi}{l}$ , $n=1,2,3,\ldots$ . The normal modes are

y_n(x,t)=\sin\frac{n\pi x}{l}\Big(C_n\cos\frac{n\pi c t}{l}+E_n\sin\frac{n\pi ct}{l}\Big),

and the general solution is the superposition $y=\sum_{n\ge1}y_n$ .

IC1: $\left.\dfrac{\partial y}{\partial t}\right|_{t=0}=0$ . Differentiate in $t$ and set $t=0$ :

\frac{\partial y}{\partial t}\Big|_{t=0}=\sum_n\sin\frac{n\pi x}{l}\cdot\frac{n\pi c}{l}E_n=0\quad\forall x\ \Longrightarrow\ E_n=0\ \forall n.

So $y=\sum_n C_n\sin\dfrac{n\pi x}{l}\cos\dfrac{n\pi ct}{l}$ .

IC2: $y(x,0)=a\sin\dfrac{\pi x}{l}$ . At $t=0$ :

\sum_n C_n\sin\frac{n\pi x}{l}=a\sin\frac{\pi x}{l}.

Matching coefficients of the orthogonal modes: $C_1=a$ and $C_n=0$ for $n\ge2$ . Only the fundamental ( $n=1$ ) survives.

Result

Answer

\boxed{\;y(x,t)=a\sin\frac{\pi x}{l}\cos\frac{\pi c t}{l}.\;}