Why Is Resultant Wave Energy Proportional to Amplitude Difference?

[somecelxis]

Homework Statement

The energy transferred per second by a progressive waves is directly proportional to the square of amplitude . If two different amplitude waves superpose , the energy per second transferred by the resulatant waves us direcly proportional to (ANS: the diffrence of amplitude)

why the ans shouldn't be sum of amplitude?

Homework Equations

The Attempt at a Solution

in my opinion , when 2 waves superposed , the resultant intensity is I1 +I2 ...but intensity is directly proportional to amplitude square , so the resultant amplitude should be sqrt root of ( (I1)^2 + (I1)^2 )

 

[cwasdqwe]

Superposition principle isn't valid for power (energy per second transferred), for it is not a linear operation.

 

[rude man]

First, the given answer as well as your answer, are wrong.

Second, the energy transferred per unit time depends on the phasing of the two waves.

Third, if the phase happens to be zero the intensity is proportional to the square of the sum of the two amplitudes.

 

[cwasdqwe]

Of course it dependes on phase, and let's assume they are in phase with the purpose of understanding his question, and not adding more questions to the initial problem. Your third claim doesn't contradict my answer, though...

 

[rude man]

Of course it dependes on phase, and let's assume they are in phase with the purpose of understanding his question, and not adding more questions to the initial problem. Your third claim doesn't contradict my answer, though...

I was addressing the OP, not you. Should have made that clearer.

 

[cwasdqwe]

It's ok, I did misunderstand it too. My excuses.

 

[somecelxis]

so the ans should be intensity is proportional to the square of the sum of the two amplitudes
?

 

[rude man]

so the ans should be intensity is proportional to the square of the sum of the two amplitudes
?

Yes, if they're in phase.