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## Homework Statement

Show that the potential and kinetic energy densities for a stationary wave are not equal.

## Homework Equations

A) The 1-D Wave Equation:

$$ \frac{\partial^{2} \psi}{\partial x^{2}} = \frac{1}{v^{2}} \frac{\partial^{2}\psi}{\partial t^{2}}$$

B) The general form of a stationary wave: (?)

$$ \psi(x,t) = f(x+vt) +f(x-vt) $$

C)Formula for

**total**

**energy density**in a stationary wave: (w)

$$ w = \frac{\mu}{2}\bigg[ v^{2}\big(\frac{\partial \psi}{\partial x} \big)^{2}+\big( \frac{\partial \psi}{\partial t} \big)^{2} \bigg] $$

## The Attempt at a Solution

i) Work out the partial derivatives of ## \psi(x,t) ##

Let $$ z_{0} =x+vt , z_{1}=x-vt $$

$$ \implies \psi(x,t) = f(z_{0})+f(z_{1}) $$

$$ \frac{\partial \psi}{\partial x} = f'(z_{0})+f'(z_{1}) $$

$$ \frac{\partial \psi}{\partial t} = v[f'(z_{0}) -f'(z_{1})] $$

ii) If the kinetic and potential energy densities are equal it implies:

$$ v^{2} \big(\frac{\partial \psi}{\partial x} \big)^{2} = \big( \frac{\partial \psi}{\partial t} \big)^{2} $$

iii) So substitute the values from i) into this, get:

$$ [f'(z_{0})+f'(z_{1})]^{2} = [f'(z_{0})-f'(z_{1})]^{2} $$

Which is not, in general true, therefore proved?

But I'm not sure I used the correct form for the equation of the stationary wave, or is this still acceptable?

Thanks!