# Sketching wave equation solutions.

1. Feb 24, 2012

### Lucy Yeats

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

An infinite string obeys the wave equation (d2z/dx2)=(ρ/T)(d2z/dt2) where z is the transverse displacement, and T and ρ are the tension and the linear density of the string. What is the velocity of transverse travelling waves on the string?
The string has an initial displacement
z= (h/L)(L-x) for 0<x<L, (h/L)(L+x) for -L<x<0, 0 otherwise
where h is a constant. The string is initially at rest. Sketch z(x,t) at the times
t=αL√(ρ/T) for α=0, 1/4, 1/2, and 1.

2. Relevant equations

3. The attempt at a solution

v=√(T/ρ), from the wave equation.

Using D'Alembert's solution, I get z=(h/2L)(L-(x+ct))+(h/2L)(L-(x-ct)) for 0<x<L
and z=(h/2L)(L+(x+ct))+(h/2L)(L+(x-ct)) for -L<x<0 and 0 otherwise.

But the ct terms seem to cancel, so I'm guessing I've gone wrong somewhere. :-/

Thanks in advance for any help! :-)

2. Feb 24, 2012

### sunjin09

I assume your boundary condition is z(x=±L, t)=0;
Now image what happens if both ends of the string are free, the initial triangular shaped wave would
separate into two half-amplitude triangles and travel toward ±∞, since the endpoints are fixed, these
two waves are immediately reflected, the reflected wave must be of the same shape, 180° out of phase
and travel in the opposite direction, so that the total displacement at the endpoints are always zero ...
Now you know how to figure out the rest.

Alternatively, if you like doing more math and less thinking, general solutions to the wave equation on
a string must be given by some cosine and sine waves of wavelengths 2L/n ...

3. Mar 5, 2012

### Lucy Yeats

Sorry to be slow, but why am I not allowed to cancel out the ct terms? :-/

Thanks for helping!