# Transverse waves propagating on a stretched string encounter another string of different characteristic impedance. i) Write down the equations of particle displacement due to the incident, reflected and transmitted waves. ii) Specify the boundary conditions and iii) use these to obtain expressions for reflection and transmission amplitude coefficients.

1
by abhinavassivns

2015-09-23T15:15:52+05:30

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If you want quick solution, then
Ai = - Ar + At      from boundary condition related to phase angles and displacement
Z₁ Ai² = Z₁ Ar²  + Z₂ At²    from Energy/sec. Conservation
or  k₁ Ai² = k₁ Ar² + k₂ At²

Solving them we get
Ar/Ai = (k₁ - k₂)/(k₁+k₂)   and    At / Ai = 2 k₁ / (k₁+k₂)
Ar/Ai = (Z₁ - Z₂)/(Z₁+Z₂)   and    At / Ai = 2 Z₁/ (Z₁+Z₂)
Ar/Ai = (v₂ - v₁)/(v₁+v₂)   and    At / Ai = 2 v₂ / (v₁+v₂)

This is the same formula for all Transverse waves,  mechanical, light and electromagnetic waves too.
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Transverse waves oscillate in (y) perpendicular to the direction (positive x) of propagation.  X-axis is along the string.  Let F be the tension force in the string at any x.

y(x, t) = Ai  Sin (ω t – k₁ x)    with an initial phase of  0,  at  t = x = 0
ω = angular freq.  wavelength = λ₁,  wave number = k₁ = 2π/λ₁ , T = time period
velocity in x direction = v₁ = λ₁/T        k₁ = ω/v₁
let μ₁ = mass per unit length of the string.   We can derive that :  v₁ = √(F/ μ₁)
So  F = μ₁ v₁² = v₁ Z₁,  where Characteristic impedance of the string = Z₁ = μ₁ * v₁

k₁ is the wave number specific to the string and it may depend on the impedance of the string, mass per unit length and temperature.   Suppose the wave encounters a heavier string of higher wave number k₂, and impedance Z₂, then most of the energy in the wave is reflected.  A little is transmitted.   The reflected wave has a phase difference of π with the incident wave.  The transmitted wave has the same  phase as the incident wave.   Frequency of the wave remains same in both strings.

Let  k₂, λ₂, v₂, Z₂  be the wave number, wavelength, velocity and characteristic impedance of the wave on the second string.   Let both strings meet at  x = L  and at   t = t₁.
So   v₂ = √(F/ μ₂)         and     F = μ₂ v₂² = v₂  Z₂
Since, F = v₁ * Z₁ = v₂ *Z₂,   we get   v₁ / v₂ = Z₂/Z₁ = k₂/k₁

Yr (x, t) = Ar  Sin (ω (t-t₁) - k₁ (L- x) + θ)   where Ar = amplitude of reflected wave.
As the phase difference with Yi(x,t) is  π at x =L and t= t₁,  we get:
ω (t-t₁) - k₁(L-x) + θ = ω t₁ - k₁ L - π       =>    θ = ω t₁ - k₁ L – π
=>       Yr(x,t) = Ar Sin [ω t - k₁(L- x) + k₁ L - π]

Yt (x, t) = At  Sin [ω (t-t₁) - k₂ (x-L) + Ф)  , where At = amplitude of transmitted wave
as phase angle is same as Yi(x,t) at  x = L and t = t₁ , we get
ω (t-t₁) - k₂ (x-L) + Ф = ω t – k₁ x     => Ф = (k₂ – k₁) L
=>  Yt(x, t) = At Sin [[ω t - k₂ x + (k₂  - k₁) L]

Boundary conditions :

1) Displacement of the initial wave is the algebraic sum of the other two displacements at the boundary of the two strings.  It is like vector or phasor addition.  Let δ be the phase angle of incident wave at the boundary.
Ai Sin δ  =  Ar Sin (δ - π) + At Sin δ
Ai   = At – Ar       =>  Ai + Ar = At   --- (1)

2)   The energy incident at the boundary (per unit time) is split into two components: reflected and transmitted.  Using the conservation of energy principle, we get:

1/2  μ1 v1 ω^2  Ar^2 + 1/2 μ2 v2 ω^2 At^2 =  1/2  μ1 v1 ω^2  Ai^2

=>    Z1 (Ai^2 – Ar^2) = Z2  At^2
=>   Z1 (Ai - Ar ) = Z2 At      --- (2)

Solving (1) and (2) we get:
Ar = (Z1 – Z2) Ai  / (Z₁ + Z₂)      And     At = 2 Z₁ Ai /(Z₁ + Z₂ )
Or,  Ar = (k₁ – k₂) Ai / (k₁ + k₂)     and     At = 2 k₁ Ai /(k₁ + k₂)
Or    Ar = (v₂ – v₁) Ai / (v₁ + v₂)     and      At = 2 v₂ Ai /(v₁ + v₂)
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