Waves Reflection Transmission
Each medium which a wave travels offers a characteristic impedance to wave motion. When a wave travelling in a particular medium of impedance z_{1} meets a normal boundary beyond which the impedance changes to z_{2}, the incident wave may get partially reflected and partially transmitted. In other words, part of the incident flux is transmitted while the remaining is reflected back into first medium. Let us find out the reflected and transmitted flux.
For example, consider a long string which consists of two pieces joined smoothly at point x = 0. The entire string is stretched having uniform tension T. The part of the string left to x = 0 has linear mass density ρ1, hence impedance ; the right part has density ρ2 and impedance .
Suppose the incident transverse harmonic wave travelling along the string (from left) is given by
y_{i} ( x, t ) = A_{i} cos ω (t – (x/v_{1} ) , x < 0
where subscript i stands for incident wave; is the wave velocity in the left string. [Note that frequency ω = 2 π v of the wave is decided by the frequency of oscillations of the source while wavelength λ = v/v depends upon the medium.]
The incident wave meets the discontinuity in impedance suddenly at x = 0. Let us assume that total incident energy is partially reflected back and partially transmitted in the form of reflected and transmitted harmonic waves given by
y_{r} ( x, t ) = A_{r} cos ω (t + (x/v_{1} ) , x < 0
and y_{t} (x, t) = At cos ω (t – x/v^{2} ) , x > 0
In the above expressions, note the following:
(i) y_{r} is the wave moving towards –ve X-axis; y_{t} is moving towards +ve X-axis.
(ii) Amplitudes have to be in general different; remember that energy flux is proportional to square of amplitudes.
(iii) The frequency ω is same for all the three waves. The particles at x = 0 are made to oscillate with frequency ω by incident wave; in turn, these particles become the source of reflected and transmitted waves.
Now, there are two geometrical boundary conditions that must be satisfied (for all values of t) at the point x = 0.
1. First, the net displacement of string must be equal immediately to the left and right of position x = 0. In other words, at any instant t,
y_{i }+ y_{r} = y_{t} at x = 0.
[Note that net displacement on left of x = 0 is y_{L} = y_{i} + y_{r}, and that on right y_{R} = y_{t}.]
This implies A_{i }+ A_{r }= A_{t}
2. Secondly, the slope of string y/x must be same immediately to the left and right of point x = 0. That is,
/x (y_{i} + y_{r}) = yt/x at x = 0, for any t
This implies, – 1/v_{1} ( A_{i} – A_{r} ) = – 1/v_{2} A_{t}
By solving these eq. we find
where we used the relation T/v = z in order to get the last terms.
We define the reflection and transmission coefficients R_{12} and T_{12} for the displacement y (or amplitude A) as
Services: - Waves Reflection Transmission Homework | Waves Reflection Transmission Homework Help | Waves Reflection Transmission Homework Help Services | Live Waves Reflection Transmission Homework Help | Waves Reflection Transmission Homework Tutors | Online Waves Reflection Transmission Homework Help | Waves Reflection Transmission Tutors | Online Waves Reflection Transmission Tutors | Waves Reflection Transmission Homework Services | Waves Reflection Transmission