Reflection and Transmission

Figure 1: Four possible cases for the reflection and transmission of a wave on a string.

The first figure shows the 4 possible cases for reflection and transmission at an interface. Lets solve the problem, which is shown in the next figure

Figure 2

since μ 1 ≠ μ 2 μ 1 ≠ μ 2 it must be that v 1 ≠ 1 ≠ v 2 2 .

On the left side of the junction we have y l = y i n c + y r e f = A cos ( k 1 x − ω t ) + B cos ( k 1 x + ω t ) y l = y i n c + y r e f = A cos ( k 1 x − ω t ) + B cos ( k 1 x + ω t ) and on the right side of the junction we have y r = y t r a n s = C cos ( k 2 x − ω t ) . y r = y t r a n s = C cos ( k 2 x − ω t ) . At the boundary x = 0 x = 0 the wave must continuous, that is there are no kinks in it. Thus we must have y l ( 0 , t ) = y r ( 0 , t ) y l ( 0 , t ) = y r ( 0 , t ) ∂ y l ( x , t ) ∂ x | x = 0 = ∂ y r ( x , t ) ∂ x | x = 0 ∂ y l ( x , t ) ∂ x | x = 0 = ∂ y r ( x , t ) ∂ x | x = 0

So from the first equation A cos ( ω t ) + B cos ( ω t ) = C cos ( ω t ) A cos ( ω t ) + B cos ( ω t ) = C cos ( ω t ) A + B = C A + B = C ∂ y l ( x , t ) ∂ x | x = 0 = ∂ y r ( x , t ) ∂ x | x = 0 ∂ y l ( x , t ) ∂ x | x = 0 = ∂ y r ( x , t ) ∂ x | x = 0 − A k 1 sin ( − ω t ) − k 1 B sin ( ω t ) = − k 2 C sin ( − ω t ) − A k 1 sin ( − ω t ) − k 1 B sin ( ω t ) = − k 2 C sin ( − ω t ) ( A − B ) k 1 sin ω t = C k 2 sin ω t ( A − B ) k 1 sin ω t = C k 2 sin ω t A − B = k 2 k 1 C A − B = k 2 k 1 C now solve for B B and C C A + B = C A + B = C A − B = k 2 k 1 C A − B = k 2 k 1 C 2 A = ( 1 + k 2 k 1 ) C 2 A = ( 1 + k 2 k 1 ) C Thus we can define the transmission coefficient t r ≡ C / A = 2 k 1 k 1 + k 2 t r ≡ C / A = 2 k 1 k 1 + k 2 and the refection coefficient r ≡ B / A = C A − 1 = k 1 − k 2 k 1 + k 2 r ≡ B / A = C A − 1 = k 1 − k 2 k 1 + k 2 note how the amplitudes can change at the boundary

If μ 2 < μ 1 μ 2 < μ 1 then we must have k 2 < k 1 k 2 < k 1 since v = ω / k = T / μ v = ω / k = T / μ and ω ω and T T must be fixed. We see that k ∝ μ k ∝ μ In this case we see that the amplitude of the wave gets bigger when it moves into a less dense medium. We have probably all experienced this in real life. As waves come ashore they become bigger. This is because shallower water is effectively less dense. A tsumami in open ocean may have an imperceptable amplitude but when it comes ashore it can be tremendous. This seems almost counter intuitive, but in any closed system the energy and power are conserved but there is no rule saying amplitude has to be conserved.

Lets look at the reflected and transmitted power. Recall Power: P = 1 2 μ ω 2 A 2 v P = 1 2 μ ω 2 A 2 v For the incident and reflected waves μ μ and ν ν are the same so the reflected power coefficient (reflected power / incident power) P R = ( B / A ) 2 = ( k 1 − k 2 k 1 + k 2 ) 2 P R = ( B / A ) 2 = ( k 1 − k 2 k 1 + k 2 ) 2 To do transmitted power lets first rewrite the power equation. Recall v = ν λ = 2 π ν k = ω k v = ν λ = 2 π ν k = ω k Also v = T μ v = T μ so μ = T v 2 = T ( ω k ) 2 = T k 2 ω 2 μ = T v 2 = T ( ω k ) 2 = T k 2 ω 2 so now P = 1 2 μ ω 2 A 2 v P = 1 2 μ ω 2 A 2 v becomes P = 1 2 T k 2 ω 2 ω 2 A 2 ω k P = 1 2 T k 2 ω 2 ω 2 A 2 ω k or P = 1 2 T k ω A 2 . P = 1 2 T k ω A 2 .

Now lets look at the 4 specific cases we have:

Rigid wall

μ → ∞ μ → ∞ so k 2 → ∞ k 2 → ∞ r = k 1 − k 2 k 1 + k 2 = k 1 k 2 − 1 k 1 k 2 + 1 r → − 1 r = k 1 − k 2 k 1 + k 2 = k 1 k 2 − 1 k 1 k 2 + 1 r → − 1 Also P R → + 1 P R → + 1 P T → 0 P T → 0 So wave is reflected and inverted, but has same power

Free end

μ → 0 μ → 0 so k 2 → 0 k 2 → 0 r = k 1 − k 2 k 1 + k 2 = k 1 k 1 r → + 1 r = k 1 − k 2 k 1 + k 2 = k 1 k 1 r → + 1 Also P R → + 1 P R → + 1 P T → 0 P T → 0 So wave is reflected and has same power

Moving to higher density

μ 2 > μ 1  k 2 > k 1 μ 2 > μ 1  k 2 > k 1 so r < 0 r < 0 t r > 0 t r > 0

Moving to lower density

μ 2 < μ 1  k 2 < k 1 μ 2 < μ 1  k 2 < k 1 so r > 0 r > 0 t r > 1 t r > 1 Note the transmitted wave's amplitude is larger than the original.