How to Prove AI = AT + AR for Wave Amplitudes?

[Aaron8153]

I know that:
yI(x,t) = AI sin [k1( x - v1t )]
yR(x,t) = AR sin [k1( x - v1t )]
yT(x,t) = AT sin [k1( x - v2t )]

W1 = W2
k1v1 = k2v2

I am unsure about how to prove that AI = AT + AR
Where AI, AT, and AR are the different amplitudes of the incident, transmitted, and reflected waves.

 

[HallsofIvy]

Basically, it's that "W1 = W2" which, I take it, is "conservation of energy". Do you know how to calculate the energy in a wave?

 

[GSXR750]

To prove that AI = AT + AR, we can use the fact that the total energy of the system must be conserved. In other words, the sum of the energies of the incident, reflected, and transmitted waves must be equal to the total energy of the system.

Let's start by finding the energy of each wave. The energy of a wave is given by the equation:

E = 1/2 * ρ * A^2 * ω^2 * V

Where ρ is the density of the string, A is the amplitude, ω is the angular frequency, and V is the volume of the string.

For the incident wave, we can rewrite the equation as:

EI = 1/2 * ρ * AI^2 * ω^2 * V

Similarly, for the reflected and transmitted waves, we have:

ER = 1/2 * ρ * AR^2 * ω^2 * V
ET = 1/2 * ρ * AT^2 * ω^2 * V

Now, since the total energy of the system is conserved, we can write:

EI + ER = ET

Substituting the equations for EI, ER, and ET, we get:

1/2 * ρ * AI^2 * ω^2 * V + 1/2 * ρ * AR^2 * ω^2 * V = 1/2 * ρ * AT^2 * ω^2 * V

Dividing both sides by 1/2 * ρ * ω^2 * V, we get:

AI^2 + AR^2 = AT^2

Taking the square root of both sides, we get:

AI + AR = AT

Therefore, we have proved that AI = AT + AR. This means that the amplitude of the incident wave must be equal to the sum of the amplitudes of the reflected and transmitted waves. This makes intuitive sense as the incident wave is the initial energy input into the system, which is then divided into the reflected and transmitted waves.