Where does extra energy come from in superposed waves?

  • Thread starter Thread starter Chris 197
  • Start date Start date
  • Tags Tags
    Energy Waves
AI Thread Summary
In the discussion about superposed waves, the focus is on understanding the source of "extra" energy when two identical waves combine in phase, resulting in a higher amplitude and energy. The energy of a wave is calculated using the formula E = A^2, leading to a scenario where two 2m waves combine to create a 4m wave, resulting in a significant increase in energy from 400 J to 1600 J. The confusion arises from the misconception that energy should simply add linearly, while it actually depends on the square of the amplitude. This highlights a misunderstanding of energy conservation in wave interactions, as the energy does not simply appear from nowhere. The discussion emphasizes the importance of correctly applying the principles of wave superposition and energy calculations.
Chris 197
Messages
4
Reaction score
0
Where does "extra" energy come from in superposed waves?

1. I am needing to understand how, when two identical waves are superposed in phase, and the amplitude doubles, how is it that the energy of the resulting wave is greater than the sum of the two superposed waves.



2. Given that E = A^2, and my example waves are measured in meters and use this formula: E=100 J/m^2, then,

a wave with an amplitude of 2m therefore has an energy of E = 100 J/m^2 * 2^2 = 400 J

Two 2m waves are superposed in phase and the amplitude is now 4m. The new wave now has an energy of E = 100 J/m^2 * 4^2 = 1600 J

Where did the extra 800J come from? It seems to have come out of thin air, which violates the conservation of energy.
 
Physics news on Phys.org


You cannot write E1 + E2 = A1^2 + A2^2
It should be E1 + E2 = (A1 + A2)^2
 


Thank you rl.bhat.

It seems to me that I did add the amplitudes in the way you described:

"The new wave now has an energy of E = 100 J/m^2 * 4^2 = 1600 J." To get the "4^2" I added two 2m amplitudes of the original waves that were then superposed, and then squared 4 to get 16.

Thanks for the help to understand where I am missing something.
 
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
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