How to correlate the frequency of source to The frequency of particle just between the two medium while transmission and reflection of wave? When a pulse travels through rarer to denser medium,two new waves (reflected and transmitted) are formed from incident wave,While the wavelength of the Transmitted wave differs from wavelength of Incident wave,also velocity is decreased,But the frequency of incident,Reflected and transmitted ray remains same
I cannot figure out why does this happen,
There are many answers on physics stack exchange forum whose essense is as follows(or they explain it as follows)
1.(Why doesn't the frequency of light change during refraction?)
An answer here(for light wave) states that wave has to follow boundry conditions, i.e to remain in phase to match electric fields
2.similiarly
Why does frequency remain the same when waves travel from one medium to another?
An answer here states example of swimming pool,Which itself Referrs to boundry condition that particle just above and just below water surface must oscillate with same frequency so that rope would not break
Similarly,Other answers at such question indirectly refer to a boundry condition to be satisfied
My question is that,during the transition process of incident wave to reflected and transmitted wave,we can call the Particle at common interface of two medium of rope as source of these new reflected and Transmitted wave so we can deduce that reflected and transmitted wave must have a common frequency,
But why should this common frequency must equal the Frequency of incident wave??As you can see in the animation attached below,During transition,In first medium,The final wave formed(during transition) is superposition of Incident and reflected wave,Final wave has to follow the frequency of Interface particle(we called it new source earlier)

On a note,I understand that superpositioned wave in first medium,Reflected wave and transmitted wave must have common frequency,But I can't understand why this common frequency must equal frequency of incident wave
This is the link of animation stated in question
https://surendranath.org/GPA/Waves/TWRT/TWRT.html (please select thinner to thicker option in website above)
 A: Because you have assumed a linear time invariant medium. A linear time invariant medium is described by a linear time-invariant operator whose eigenfunctions are pure sinusoids that are characterized by their amplitude, frequency and phase. Only amplitude and phase may be changed by such system, meanwhile frequency stays the same between input and output. This is true for lumped as well as for distributed systems, such as RLC circuits or dielectric crystals, resp. This is not really physics. The physics question is rather why a material medium through which an EM wave propagates, for example glass or crystalline dielectric, is linear and time invariant? In practice, nothing is truly linear nor time invariant. For large enough EM intensities in the nonlinear regime you get as  minimum harmonics and may also unplanned for nonlinear modulation on the incoming wave and thus shift or spread its fundamental frequency.
If the medium is nonlinear and/or time varying the frequency of the transmitted wave will change. For example,  a linear time varying medium is a crystal in which acoustic waves (sound) modulate the density and thus the dielectric permittivity of the medium and when light is scattered off the acoustic wave the the scattered light suffers frequency shift that can be used to measure the the acoustic frequency (an instantaneous spectrum analyzer, see 1.
A: As hyportnex mentioned, it is a linear system and in that case the frequency won’t change. If it a nonlinear material you can have all kinds of things happen like frequency doubling or tripling.
Or if you look closely at other situations you can have other types of scattering like Raman or Brillioun scattering where an interaction with the lattice vibrations (phonons) in material can result in a change in frequency and direction.
So a more general way to think about your question is that energy and momentum need to be conserved. This is always true. With photons you have to be a little careful since they are massless and in the special cases like Raman scattering it is essentially keeping track of the energy of the photon the energy that is lost to the molecule vibration and the momentum conservation is taken care of by change and of direction and phonon vibration.
But for your case you can also calculate the energy of your incident wave and the momentum of that wave and the energy and momentum of the transmitted and reflected waves. And the Total Energy would be equal to the the sun of the reflected energy and transmitted energy.
If there was a change in frequency of one of the waves then energy would not be conserved.
If you come in at an angle to the surface of some refractive index then you then match up the vector components to determine the angle of refraction, and the angle of reflection being equal to the angle of incidence. This is essentially ensuring the momentum is conserved.
In the medium with the refractive index, the wavelength does change, but the frequency remains constant. The medium doesn’t add or subtract any energy to the wave since in the linear system and energy is conserved.  The change in wavelength is also important in keeping the phase correct.
When you start dealing with nonlinear systems, you still have to have conservation of energy and momentum…but it is more complicated to do the math.
A: Short answer
It follows from the algebraic conditions governing interface condition at the interface between the two media, with linear behavior.
Details
You can think at the problem as the union of the problems in two neighboring media and a common boundary. Let's call medium 1, the medium where the original incident wave is traveling and reflection occurs, and medium 2 where the transmission occurs. Medium 1 is in the region of space $M_1: x \in [x_1, x_i]$, medium 2 in $M_2: x \in [x_i, x_2]$, where $x_i$ is the position of the interface.
Interface conditions. What happens in the media depends on the common boundary surface, and what happens on both sides of the boundary is governed by interface conditions (coming from integral balance equations, for an elementary volume across the interface surface). As an example, the electric field on a interface with no charge surface density is constant across the surface
$\mathbf{e}_1(x_i,t) = \mathbf{e}_2 (x_i,t)$.
At the medium interface, usually an algebraic condition holds to determine the transmission and reflection ratio, namely
$i_t = 1 + i_r,\qquad$ with $i_t \in (0,1)$,
that quantifies the amount of the signal transmitted in medium 2 and the amount of signal reflected, as fraction of the incident signal.
Solution as superposition of traveling waves. The field in medium 1 is the superposition of the incident and the reflected wave, while field in medium 2 come from the transmitter wave only
$e_1(x, t) = e_i(x,t) + e_r(x,t) = e^{t.w.}_i(x-v_1t) + e^{t.w.}_r(x+v_1t)$
$e_2(x, t) = e_t(x,t) \qquad \qquad \ = e^{t.w.}_t(x-v_2 t)$,
having written them explicitly as traveling waves. Now we can write the three contributions as:

*

*incident wave: its value at the interface is $e^{t.w.}_i(x_i -v_1 t)$.

*transmitted wave: its at the interface is
$e^{t.w.}_t(x_i - v_2 t) = i_t e^{t.w.}_i(x_i - v_1t) $,

*reflected wave: its at the interface is
$e^{t.w.}_r(x_i - v_2 t) = i_r e^{t.w.}_i(x_i - v_1t) = 
(i_t - 1) e^{t.w.}_i(x_i - v_1t)$,

so that both the interface condition $e_1(x_i, t) = e_2(x_i, t)$ and the ratio of the transmission and reflection signal are satisfied.
Properties of traveling waves.
Now, we only need to reconstruct the full signal in the media, knowing that its made of traveling waves and knowing the boundary conditions. For a "backward" traveling wave, we know that
$f(x_0,t_0) = f^{t.w.}(x_0+vt_0) = f^{t.w.}(x_1+vt_1) = f(x_1,t_1)$,
if the argument of the traveling wave is the same, i.e.
$x_0 + v t_0 = x_1 + v t_1$$\qquad \rightarrow t_1 = t_0 + \dfrac{ x_0 - x_1}{v}$,
while for a "forward" traveling wave $t_1 = t_0 - \dfrac{ x_0 - x_1}{v}$.
Solution of the problem.
Putting everything together, we can write the solution of our problem. The field in medium 1 reads
$e_1(x,t) = e_i(x,t) + e_r(x,t) = \\
\qquad \ \ \ = e^{t.w.}_i(x-v_1t) + e^{t.w.}_r(x+v_1t) = \\
\qquad \ \ \ = e^{t.w.}_i(x-v_1t) + e^{t.w.}_r\left(x_i, t - \dfrac{x_i-x}{v_1}\right) = \\
\qquad \ \ \ = e^{t.w.}_i(x-v_1t) - (1-i_t) e^{t.w.}_i\left(x_i, t - \dfrac{x_i-x}{v_1}\right) = \\
\qquad \ \ \ = e^{t.w.}_i(x-v_1t) - (1-i_t) e^{t.w.}_i\left(x, t - 2\dfrac{x_i-x}{v_1}\right) \\
\qquad \ \ \ = e^{t.w.}_i(x-v_1t) - (1-i_t) e^{t.w.}_i\left(x -v_1 \left( t - 2\dfrac{x_i-x}{v_1} \right) \right) \\
\qquad \ \ \ = e^{t.w.}_i(x-v_1t) - (1-i_t) e^{t.w.}_i\left(-x -v_1 \left( t - 2\dfrac{x_i}{v_1} \right) \right) \\$
while the field in medium 2 reads
$e_2(x,t) = e_t(x,t) = \\
\qquad \ \ \ = e^{t.w.}_t(x-v_2t) = \\
\qquad \ \ \ = e^{t.w.}_t\left(x_i, t - \dfrac{x-x_i}{v_2}\right) = \\
\qquad \ \ \ = i_t  e^{t.w.}_i\left(x_i, t - \dfrac{x-x_i}{v_2}\right) .$
Harmonic signals.
Assuming the incident signal is an harmonic signal, $e^{t.w.}_i(x-v_1 t) = A e^{i k_1(x-v_1t)} = A e^{i (k_1x-\omega_1 t)}$, with $\omega_1 = \dfrac{v_1} k_1$,
the solution reads
$e_1(x,t) =  A e^{i (k_1x-\omega_1 t)} - (1-i_t)  A e^{i \left(-k_1 x -\omega_1 \left(t -2\frac{x_i}{v_1} \right) \right)}  =
e^ {i \omega t} \left[ A e^{i k_1 x} - (1-i_t)  A e^{i \left(-k_1 x + 2 \omega_1 \frac{x_i}{v_1} \right) } \right]$
$e_2(x,t) = i_t A e^{i \left(k_1 x_i -\omega_1 \left( t - \frac{x-x_i}{v_2} \right) \right) } =
e^{-i \omega_1 t} \left[ i_t A e^{i \left(k_1 x_i + \omega_1 \frac{x-x_i}{v_2}\right)} \right] 
$,
that clearly shows that, at every point in space, the signal has the same time frequency $f_1 = \dfrac{\omega_1}{2\pi}$.
A: You say :superpositioned wave in first medium,Reflected wave and transmitted wave must have common frequency, but this superpositioned wave IS the incident wave. How do you think the boundary point gets its frequency if not from the incident wave?
A: Just think of yourself as a particle at the boundary.
An incident wave of frequency $f$ forces a nearest neighbour and hence you to oscillate at the same frequency $f$ .
You are connected to nearest neighbours via bonds.
How can you force those nearest neighbours to oscillate at any other frequency other than frequency $f$?
The answer is that you cannot, and so as a result of your oscillations at frequency $f$ your nearest neighbours must also oscillate at a frequency $f$.
