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UPSC 2013 Maths Optional Paper 2 Q6c — Step-by-Step Solution

20 marks · Section B

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

Question

A tightly stretched string with fixed end points $x=0$ and $x=l$ is initially at rest in equilibrium position. If it is set vibrating by giving each point a velocity $\lambda\cdot x(l-x)$, find the displacement of the string at any distance $x$ from one end at any time $t$.

Technique

Separation of variables; sine-series expansion of initial velocity profile; $n^{-4}$ decay rate from quadratic IC.

Solution

Setup. Wave equation $u_{tt}=c^{2}u_{xx}$ on $0<x<l$, $t>0$, with boundary conditions $u(0,t)=u(l,t)=0$ and initial conditions $u(x,0)=0$, $u_t(x,0)=\lambda x(l-x)$. (Here $c$ is the wave speed in the string.)

Step 1 — Sine-series general solution

Separation of variables on the boundary value problem $u(0,t)=u(l,t)=0$ gives spatial modes $\sin(n\pi x/l)$, $n\ge 1$, with time evolution $\cos(n\pi ct/l)$ and $\sin(n\pi ct/l)$:

Step 2 — Apply $u(x,0)=0$

Step 3 — Apply $u_t(x,0)=\lambda x(l-x)$

So $B_n\cdot n\pi c/l$ is the sine-series Fourier coefficient of $\lambda x(l-x)$ on $(0,l)$:

Step 4 — Compute $I_n=\int_0^l x(l-x)\sin(n\pi x/l)\,dx$

Integrate by parts twice (both bracket terms vanish at the endpoints):

First IBP: $u=x(l-x)$, $dv=\sin(n\pi x/l)\,dx$. $u=0$ at $x=0,l$ ⇒ bracket vanishes; left with

Second IBP: $u=l-2x$, $dv=\cos(n\pi x/l)\,dx$. Bracket terms again vanish ($\sin(n\pi)=\sin 0=0$); left with

Combining:

For odd $n$: $1-(-1)^{n}=2$, so $I_n=\dfrac{4l^{3}}{n^{3}\pi^{3}}$. For even $n$: $I_n=0$.

Step 5 — Solve for $B_n$

For odd $n$:

For even $n$: $B_n=0$.

Step 6 — Final solution

Answer