One dimensional Wave Equation

(Vibrations of a stretched string)

Y

T2

Q β δs P α y T1 δx

0 x x + δx A X

Consider a uniform elastic string of length l stretched tightly between points O and A and displaced slightly from its equilibrium position OA. Taking the end O as the origin, OA as the axis and a perpendicular line through O as the y-axis, we shall find the displacement y as a function of the distance x and time t.

Assumptions(i) Motion takes places in the XY plane and each particle of the string moves

perpendicular to the equilibrium position OA of the string.(ii) String is perfectly flexible and does not offer resistance to bending.(iii) Tension in the string is so large that the forces due to weight of the string can be

neglected.

(iv) Displacement y and the slope are small, so that their higher powers can be neglected.

Let m be the mass per unit length of the string. Consider the motion of an element PQ of length δs. Since the string does not offer resistance to bending(by assumption), the tensions T1and T2 at P and Q respectively are tangential to the curve.Since there is no motion in the horizontal direction, we have

T1cosα = T2cosβ = T(constant) …..(1)Mass of element PQ is mδs. By Newton’s second law of motion, the equation of motion in the vertical direction is

This is the partial differential equation giving the transverse vibrations of the string. It is also called the one-dimensional wave equation.

Boundary conditionsFor every value of t, y = 0 when x = 0 y = 0 when x = l Initial conditionsIf the string is made to vibrate by pulling it into a curve y=f(x) and then releasing it, the initial conditions are

(i) y = f(x) when t = 0

(ii) when t = 0

Solution of one-dimensional Wave Equation: -

The vibrations of the elastic string, such as a violin string, are governed by the one-

dimensional Wave equation

----------------------------

------(1)

Let

-----------------------

--------(2)

be a solution of (1), where F(x) is only a function of x and G(t) is only a function of t only. On

differentiating equation (5), we obtain

Where dot denotes the partial derivatives with respect to t and primes derivatives with respect to

x. By inserting this into differential equation (1), we have

FG. .

=c2F} } G} {¿¿¿. Dividing c2F} } G} {¿ ¿¿we find

The left-hand side and Right-hand side of the above equation are function of t and x respectively

and both x & t are independent variables & hence both sides must be equal an they must be equal

to a some constant k.

∴

y(x, t)=F(x)G(t)

This yields immediately two ordinary differential equations, namely

---------------------------------------------------------------

(6)

and

---------------------------------------------------------------(7)

Satisfying the boundary conditions: -

We shall now determine solution F & G so that u=FG satisfies the boundary conditions.

u(0, t)=F(0).G(t)=0 ; u(l, t)=F(L)G(t)=0

Solving (6) If G=0 then u=0 which is of no interest and then

-------------------------------------------(8)

So, here three cases arise.

Case-I (k=0) then the general solution of (6) is F=ax+b. Now applying (8) we obtain a=b=0, and

hence F0, which is of no interest because then u0.

Case-II (k>0 and k=2) then general solution of (6) is F=Aex + Be-x and again applying (8), we

get F0 & hence u0. So we discard the case k=2.

Case-III (k<0 and k=-p2) Then the differential equation (6) takes the form

F”+p2F=0.

F”-kF=0

(a) F(0)=0 (b) F(L)=0

Its general solution is

F(x)=Acos px +B sin px

Applying (8), we obtain

F(0)=0=A(1) +B(0)

A=0.

and

F(L)=0BsinpL=0.

We must take B0 since otherwise F0. Hence, sinpL=0pL=n

-----------------------------------------------(9)

We thus obtain infinitely many solutions F(x)=Fn(x) n=1,2,3--------

----------------------------------------------------------(10)

These solutions satisfy the equation (8).

Solving (7), the constant k is now restricted to the values k=-p2

k=−( nπL )2

Inserting to the equation (7), we get G. .

+λn2G ( t )=0 where λn=

nπcL,

A general solution is Gn ( t )=ancos λn t+bn sin λn t

p=nπ

L (n integer)

Fn( x )=Bnsin( nπxL )

And hence

---------(11)

Using principle of superposition the general solution is given by

-------------(12)

Now applying the initial conditions y = f(x) and , when t = 0 , we have

Since eqn.(13) represents Fourier series for f(x), we have

From (14), bn = 0 for all n.

Hence (12) reduces to

Where an is given by (15)

Examples

1. A string is stretched and fastened to two points l apart. Motion is started by displacing the

string in the form from which it is released at time t=0. Show that the

displacement of any point at a distance x from one end at time t is given by

2. The points of trisection of a string are pulled aside through the same distance on opposite

sides of the position of equilibrium and the string is released from rest. Derive an

expression for the displacement of the string at subsequent time and show that the mid-

point of the string always remains at rest.

Ans.

3. A tightly stretched string with fixed end points x=0 and x=l is initially at rest in its

equilibrium position. If it is set vibrating by giving to each of its points a velocity λx(1-x), find

the displacement of the string at any distance x from one end at any time t.

Ans.