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whatisreality

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## Homework Statement

A 1D wave function ψ(x,t) satisfies these initial conditions:

ψ(x,0) = 0 for all x

∂ψ/∂t (x,0) is v for -a≤x≤a

0 otherwise

Plot ψ(x,t) as a function of x at time t=a/v.

## Homework Equations

## The Attempt at a Solution

I know the 1D wave equation is given by d'Alembert's:

##\psi(x,t) = 0.5[\psi(x+vt,0) + \psi(x-vt,0)] + \frac{1}{2v} \int_{x-vt}^{x+vt} \frac{\partial \psi}{\partial t}(x,0) \mathrm dx##

But for this function because ψ(x, 0) = 0 for all x, that simplifies to

##\psi(x,t) = \frac{1}{2v} \int_{x-vt}^{x+vt} \frac{\partial \psi}{\partial t}(x,0) \mathrm dx##

I wanted to sketch ψ(x,t) first but I'm not sure how to evaluate the integral. For -a≤x≤a,

##\frac{\partial \psi}{\partial t}(x,0)## = v,

But do I still integrate between x+vt and x-vt? Do I sub x=a or x=-a into the limits, or maybe t=0? I;m fairly sure you just integrate v between x+vt and x-vt (w.r.t x).

Then I have to plot the graph at t=a/v. And I'm not given what

##\frac{\partial \psi}{\partial t}(x,\frac{a}{v})## is, and there's no easy relationship I can spot between the two graphs in terms of for example ψ(x,0) is ψ(x, a/v) translated or rotated etc.

In fact, I can't see how it being t=a/v instead of t=0 would affect the graph at all!

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