Timeline for Energy Conservation of waves at a boundary
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
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| Oct 31, 2016 at 12:36 | comment | added | freecharly | The conservation of energy is the conservation of energy fluxes of the waves. The energy densities of the waves is not conserved. | |
| Oct 31, 2016 at 12:32 | history | edited | freecharly | CC BY-SA 3.0 |
added 30 characters in body
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| Oct 30, 2016 at 23:56 | comment | added | freecharly | If in the right string you assume $\rho_2 → 0$ then $Z_2=\sqrt{T_2\rho} → 0$ and the reflection coefficient becomes $R=1$ and the transmission coefficient becomes $T=2$. Thus the transmitted wave amplitude is, indeed, twice the amplitude of the incident wave. However, the wave velocity of the right string $v_2 → ∞$ meaning that the infinitely light string has infinite wavelength. Due to $\rho v_y^2 → 0$ the energy flux becomes zero meaning that there is zero energy transport in this wave. | |
| Oct 30, 2016 at 19:32 | comment | added | Sundesh | And for the energy conservation, why will they have same velocity? Shouldn't it be different because they have different amplitudes? And my question exactly is why energy flux is conserved and not energy? | |
| Oct 30, 2016 at 19:24 | comment | added | Sundesh | And for the whip, the right side of the boundary is taken to have zero impedance which implies free end and also that the linear mass density is zero to the right. But the left side of the boundary is still present and when you do the math, it tells us that the amplitude of the transmitted wave is twice that of the incident wave. But my question is, the physics that we have built these concepts on till now tells us that no particles in the medium implies wave cannot travel in the medium (because wave consists of infinitely many simple harmonic oscillators). How can one explain this? | |
| Oct 30, 2016 at 19:24 | comment | added | Sundesh | I understood your point. | |
| Oct 30, 2016 at 19:15 | comment | added | freecharly | Because at $x=0$ you have an incident wave with amplitude $A_1$ and a reflected wave with amplitude $B_1$, the total amplitude of the oscillation of the left string at $x=0$ is $A_1+B_2$. This has to be equal to the amplitude $A_2$ of the transmitted wave at $x=0$. | |
| Oct 30, 2016 at 19:03 | comment | added | Sundesh | If the y-coordinates are the same then, shouldn't it be $A_1$=$A_2$ only. | |
| Oct 30, 2016 at 18:22 | history | answered | freecharly | CC BY-SA 3.0 |