Question

A string is stretched and fastened to two points at a distance 0

l

0 apart. Motion is started by

displacing the string in the form y = k sin πx

l

from which it is released at time t = 0. Show

that the displacement of any point on the string at a distance x from one end at time t is

given by k sin πx

l

cos πax

l

Answer 1

Solution

The displacement y(x,t) of the string at a distance X from end 0 at time t is give by the following equation;

$\frac{d^2y}{dt^2}=a^2\frac{d^2y}{dx^2}$ ..........(a)

We establish the boundary conditions using the given information.

Since the ends of the spring ,x=0 and x=l are fixed,they undergo zero displacement at any time t;

y(0,t)=0 for t$\geq$ 0..........(1)

y(l,t)=0 for t$\geq$ 0...........(2)

Since the spring is released from rest,at t =0 ,the velocity of every point on the string in the y-direction is zero,hence;

$\frac{dy}{dx}$ (x,0)=0 for 0$\leq$ x$\leq$ l..............(3)

The initial displacement is in the form of y=ksin$\frac{πx}{l}$ ,where y(x,0) is the initial displacement of any point x in the y-direction ,hence

y(x,0)=ksin$\frac{πx}{l}$ for 0$\leq$ x$\leq$ l ..........(4)

(1),(2),(3) and (4) are the boundary conditions of the problem.

The general solutions of equation (a) with respect to the vibration of the string is

y(x,t)=(Acos px +Bsin px)(C cos pat +Dsin pat) ......(b)

A,B,C,D and p are arbitrary constants which can be found by applying the boundary conditions;

Using condition (1) in (b),we have

A(C cos pat + Dsin pat)=0 for all t$\geq$0

A=0

Using condition (2) in (b) ,we have

Since A=0

B sin pl(C cos pat +Dsin pat)=0,for all

t$\geq$ 0;

Bsin pl=0

Either B or sin pl=0

If B = 0 ,y(x,t)=0 and the equation is meaningless,so

sin pl=0

pl=nπ

p=$\frac{nπ}{l}$

Where n=0,1,2,3...$\infin$

Differentiate both sides of (b) with respect to t:

$\frac{dy}{dt}$ (x,y)=(Bsin px)pa(-Csinpat+Dcos pat) ........(c)

Where p=$\frac{nπ}{l}$

Using condition (3) on (c),we have

Bsin px.paD=0

But B$\ne$ 0 and p$\ne$ 0 ,we get D=0

With the found constants ,the equation (b) reduces to

y(x,t)=BC sin $\frac{nπx}{l}$ Cos$\frac{nπat}{l}$ where n=1,2,3.....$\infin$

Taking BC=k,

The most general solution would be

y(x,t)=$\displaystyle\sum_{n=1}^\infin$ $\lambda_n$Sin$\frac{nπ}{l}$Cos$\frac{nπat}{l}$ ....(d)

$\lambda_n$ can be found using condition (4) in (d)

ksin$\frac{πx}{l}$ =$\lambda_nsin\frac{nπx}{l}$

By comparison,

$\lambda_n$ =k

$\frac{πx}{l}=\frac{nπx}{l}$

n=1

Using this values,the equation of displacement at any time on the string will be

y(x,t)=ksin$\frac{πx}{l}$ Cos$\frac{πax}{l}$