A string consists of two parts attached at x=0.The right part of the string (x>0) has mass 'µr' per unit length and the left part of the string (x<0) has mass 'µl' per unit length, the tension in the string is T.If a wave of unit amplitude travels along the left part of the string, what is the amplitude of the wave that is transmitted to the right part of the string.
[Physics] Amplitude change when wave travels from one string to another
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D'Alembert equation reads, for the considered case,
$$\mu\frac{\partial^2 y}{\partial t^2} = T_0 \frac{\partial^2 y}{\partial x^2}\:.\tag{1}$$
It is nothing but $F=ma$ along the vertical direction ($y$). Here $y$ denotes the small deformation of the string along the vertical direction from the stationary (horizontal) configuration. We disregard horizontal deformations. $T_0 \frac{\partial^2 y}{\partial x^2}$ is the $x$-derivative of vertical component of force acting on a infinitesimal segment of string $dx$, when the deformation is small, assuming that the absolute value of the tension $T_0$ is constant. In other words
$$f_y = T_0 \frac{\partial y}{\partial x}\tag{2}$$
is the $y$ component of either the tension or the force acting at the endpoints.
From (1) and the defintion of $e(t,x)$
$$e(t,x) = \frac{1}{2}\left[T_0\left(\frac{\partial y}{\partial x}\right)^2 + \mu\left(\frac{\partial y}{\partial t}\right)^2\right]$$ one easily sees (just computing derivatives) that $$\frac{\partial e}{\partial t} = \frac{\partial}{\partial x} \left(\frac{\partial y}{\partial t} T_0\frac{\partial y}{\partial x}\right)\:.\tag{3}$$ Integrating this identity along the string, from the point (maybe endpoint) $x_1$ to the point (maybe endpoint) endpoint $x_2$, we have $$\frac{d}{dt}\int_{x_1}^{x_2} e(t,x) dx = \left.\left(\frac{\partial y}{\partial t}T_0\frac{\partial y}{\partial x}\right)\right|_{x_2} - \left.\left(\frac{\partial y}{\partial t}T_0\frac{\partial y}{\partial x}\right)\right|_{x_1}\tag{4}$$ As $\frac{\partial y}{\partial t}$ is the vertical velocity of the endpoint, taking (2) into account, we conclude that $\frac{\partial y}{\partial t}T_0\frac{\partial y}{\partial x}$ is the power of the force acting on the endpoint (or at a generic point of the string).
Equations (3) and (4) establish relations, respectively local and global, between the density of energy and the power of the forces acting along the string.
The displacement of each particle on one side of the boundary with be equal to the sum of the displacement due to the incident wave $\vec y_{\rm i}$ and the reflected wave $\vec y_{\rm r}$ and particles on the other side of the boundary will have a displacement due the transmitted wave $\vec y_{\rm t}$.
If at the boundary $\vec y_{\rm i}+\vec y_{\rm r}\ne \vec y_{\rm t}$ there would be a gap between the particle immediately one side of the boundary and the particle on the other side of the boundary which is not possible.
The slope condition is a much more subtle condition and it might help to see how the wave equation is derived for transverse waves on a string.
For a more detailed explanation please have a look at one of the many derivations that you can find on the Internet.
The diagram below shows a section of a string $AB$ of mass per unit length $\mu$ with one end at position $x$ displaced from the equilibrium position by $y_1$ and the other end at position $x +\Delta x$ displaced from the equilibrium position by $y_2$.
$T$ as the tension in the string.
The diagram is greatly exaggerated in that the angles are small and their difference is small.
The net force in the y-direction on the element $AB$ is $T \sin \theta_2 - T \sin \theta_1$.
$T \sin \theta_2$ is the gradient of the string at $x+\delta x$ which I shall write as $\left(\dfrac{dy}{dx}\right)_{x+\Delta x}$ and similarly the gradient of the string at the other end is $\left(\dfrac{dy}{dx}\right)_x$
So the net force on the element is
$$T\left(\dfrac{dy}{dx}\right)_{x+\Delta x} - T\left(\dfrac{dy}{dx}\right)_x = T\Delta x \dfrac {\left(\dfrac{dy}{dx}\right)_{x+\Delta x} - \left(\dfrac{dy}{dx}\right)_x}{\Delta x}$$
Now using Newton's second law the acceleration $a$ in the y-direction can be found.
$$\mu \Delta x \,a = T\Delta x \dfrac {\left(\dfrac{dy}{dx}\right)_{x+\Delta x} - \left(\dfrac{dy}{dx}\right)_x}{\Delta x}$$
$$\Rightarrow a = \dfrac{T}{\mu} \dfrac {\left(\dfrac{dy}{dx}\right)_{x+\Delta x} - \left(\dfrac{dy}{dx}\right)_x}{\Delta x}$$
So the acceleration of the string depends on the rate of change of the gradient of the string.
For successive elements of the string the gradient and hence the acceleration of the string will change.
However imagine that the element was at a boundary and it had a kink in it which is saying that the gradient changes in a discontinuous manner as one crosses the kink.
This would mean that the gradient of the string on one side of the kink would be very different from the gradient of the string on the other side of the kink.
Now the element of the string with the kink in it would be subjected to a very much larger acceleration than elements on either side of it.
This is not possible and so there can be no kink in the string and at a boundary the gradient of the string must be the same on both sides.
Best Answer
As a wave travel across medium a part of it gets reflected back and a part of it gets transmitted and the point $x=0$ will be a junction.Let $A_i$ be amplitude of initial and $A_r$ be denoted as amplitude of reflected wave and $A_t$ the amplitude of transmitted one.
$y_i = A_i sin(kx-\omega t)$
at x=0 $y_i + y_r = y_t \\ A_i + A_r=A_t$ Therefore
$\frac{dy_i}{dx}_{x=0^-} + \frac{dy_r}{dx}_{x=0^-}=\frac{dy_t}{dx}_{x=0^+}$ this gives $-A_i + A_r = A_t \frac{v_1}{v_2}$ Solving the two equations we obtain $A_r=\frac{v_2-v_1}{v_1+v_2}A_i$ and $A_t=2\frac{v_{1}}{v_{1}+v_{2}}A_i$
Here $v_1=\sqrt{\frac{T}{ml}}$ and $v_2=\sqrt{\frac{T}{ mr }}$