Are Kinetic and Potential Energy Densities Equal in Stationary Waves?

AI Thread Summary
The discussion centers on demonstrating that the potential and kinetic energy densities in a stationary wave are not equal. The participants analyze the wave equation and the general form of a stationary wave, leading to the conclusion that if the energy densities were equal, a specific mathematical relationship would hold true, which does not generally apply. They confirm that a stationary wave is formed by two traveling waves moving in opposite directions, supporting the validity of the wave equation used. The conversation also touches on the proper representation of stationary waves, emphasizing the importance of understanding their formation. Ultimately, the proof is accepted as valid, reinforcing the distinction between kinetic and potential energy densities in stationary waves.
bananabandana
Messages
112
Reaction score
5

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!
 
Physics news on Phys.org
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)
 
  • Like
Likes 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?
 
bananabandana said:
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.
 
  • Like
Likes bananabandana
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top