Are Kinetic and Potential Energy Densities Equal in Stationary Waves?

[bananabandana]

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!

 

[Vipho]

I think the equation you used that is the general solution for the wave equation. In my opinion, the form of the standing wave is give as follows
Ψ(x,t)=g(x)f(t)

 

[bananabandana]

Yes, but a stationary wave is by definition formed from two traveling waves moving in opposite directions which means it can then be rewritten in the form
that you suggest with the time and space separated?
Or is that not right?

 

[haruspex]

Yes, but a stationary wave is by definition formed from two traveling waves moving in opposite directions

No, the definition is as Vipho posted. It is a matter of deduction that two traveling waves of the same amplitude, frequency and speed moving in opposite directions, and satisfying the wave equation, form a standing wave.

 

[HalfLostAndSearching]

Your approach is correct, but there are a few small mistakes in your calculations. Here's a corrected version:

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:

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

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

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

iv) Simplify and rearrange:

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

$$ \implies v^{2}f'^{2}(z_{0}) + 2v^{2}f'(z_{0})f'(z_{1}) + v^{2}f'^{2}(z_{1}) = f'^{2}(z_{0}) - 2f'(z_{0})f'(z_{1}) + f'^{2}(z_{1}) $$

$$ \implies (v^{2}-1)f'^{2}(z_{0}) + (v^{2}-1)f'^{2}(z_{1}) + 2v^{2}f'(z_{0})f'(z_{1}) = 0 $$

v) Since the above equation must hold for all values of x and t, it implies