# Cauchy Boundary Conditions on a Wave

1. May 3, 2010

### bon

1. The problem statement, all variables and given/known data

So using the D'Alembert solution, I know the solution of the wave equation is of the form:

y(x,t) = f(x-ct) + g(x+ct)

I'm told that at t=0 the displacement of an infinitely long string is defined as y(x,t) = sin (pi x/a) in the range -a<= x <= a

and y =0 otherwise.

The string is initially at rest.

I'm told that the waves move along the string with speed c and told to sketch the displacement of the string at t=0, t=a/2c and t=a/c

2. Relevant equations

3. The attempt at a solution

So substituting t=0 into the d'alembert solution gives

f(x) + g(x) = sin pix/a

similarly since the string is initially at rest, we can calculate that f(x) - g(x) = const. therefore f(x) = 1/2sin pix/a + k where k is some const. and g(x) = 1/2 sin pix/a - k

So is the full solution y(x,t) = 1/2 [ sin (pi(x-ct)/a) + sin(pi(x+ct)/a) ] ? Isn't this a stationary wave..? Im not sure how to sketch for t = a/2c etc...thanks :)

2. May 3, 2010

### ideasrule

I think you're overcomplicating this. You can easily sketch the wave at t=0; the equation's given to you. For t=a/2c, the wave travels at speed c, so the pattern at t=0 would just be shifted by a/2c*c=a/2.

3. May 3, 2010

### bon

Thanks..I do see that, but I'm trying to understand the general method so that I can apply it to situations where the boundary conditions aren't so simple...

any guidance would be great
thanks

4. May 4, 2010

### ehild

No, you do not get standing waves on an infinite string.
Remember that f(x)=g(x)= sin(pix/a) is valid only in the interval between [-a, a ], and both functions are 0 everywhere else.
Letting time run, the argument of f(x) is replaced by x-ct and that of g(x) by x+ct. The condition above means that

f(x) = 1/2 sin(pi/a(x-ct)) if -a<=x-ct <=a that is -a+ct<=x<=a+ct

and g(x)= 1/2 sin(pi/a(x+ct)) if -a<=x+ct <=a that is -a-ct<=x<=a-ct ,

and y=0 everywhere else.

Can you sketch y(x,t) now?

ehild

Last edited: May 4, 2010
5. May 4, 2010

### bon

Sorry I'm not quite sure I agree with what you've written here :S

I thought it was that f(x) + g(x) = sin pix/a (rather than f(x) = g(x) = ...)

and f(x) - g(x) = constant.

So f(x) = 1/2 sin pix/a + k
and g(x) = 1/2 sin pix/a - k

so f(x) + g(x) = 1/2 sin pi(x-ct)/a + 1/2 sin pi (x+ct)/a...? Over the limits on x that you have spelt out..

which seem to be two waves of equal magnitude, phase etc moving in opposite directions? i.e. a standing wave?

6. May 4, 2010

### ehild

OK, you are right, but that constant is arbitrary, so choose it zero for start.
The sum of two waves of the same frequency and amplitude moving in opposite direction in a closed interval would produce a standing wave, but it is not the case here.
Your function is not y(x.t)=1/2 sin (pi(x-ct)/a)+ 1/2 sin (pi(x+ct)/a) in the whole interval from minus infinity to infinity, but you need a piece-wise definition:

y(x,t) =f((x-ct) + g(x+ct)

where f(x) = 1/2 sin(pi/a(x-ct)) if ct-a<=x<=ct +a , zero otherwise

and g(x)= 1/2 sin(pi/a(x+ct)) if -ct-a<=x<=-ct +a, zero otherwise.

y(x,t) = 0 outside [-ct-a, -ct+a] U[ct-a, ct+a].

Try to sketch y for t = 4a/c, you will understand what I mean.

If you want to include an arbitrary constant, you can do it, by subtracting it form g and adding to f, it is just shifting them up or down.

ehild

7. May 4, 2010

### bon

Ahh great help. I see now. Thanks ehild.