The interaction of pulse between two ropes of different thickness What occurs when you send a pulse from a;


*

*Thicker string to a thinner string

*Thinner string to a thicker string


And why does such a result occur? I have learnt that the frequency does not change when waves interact with other media, but v = f x lambda does not help me understand why such an interaction happens!
 A: A piece of rope presents a characteristic impedance for waves traveling along it which is set by the elasticity of the rope and its mass per unit length. A connection between two pieces of rope with different impedances represents a discontinuity which causes part of the incoming wave energy in one rope to be reflected backwards off the discontinuity and part of it to be transmitted through the connection into the other rope. 
If the impedance difference is zero, no reflection occurs: transmission is 100%. If it is slight, the reflected wave will be small. If it is big, the reflected wave will be large. If it is infinite, there is no transmitted wave at all.
In the case of a wave pulse in a thick string hitting the thin string, the reflected wave going backwards along the thick string is simply a time-reversed replica of the incoming pulse. In the case of the wave pulse coming in on the thin string and hitting the thick string, the reflected wave pulse is flipped upside-down and sent backwards. 
A: At the transition the string must maintain continuity. This means matching deflection and slope on each side of the transition.
Also on each side the wave speed differs, since speed is defined from the tension $T$ and the unit weight $w$ as $$c = \sqrt{ \frac{T}{w} } \tag{1} $$
On each side the tension is the same (force balance), but the unit weight differs.
Let say there is a known wave going from left to right. Before the transition the wave function is
$$ y_{\rm left} = Y \sin\left( \omega \left( \tfrac{x}{c}-t \right) \right) \tag{2}$$ 
where $Y$ is an arbitrary amplitude, $\omega=2\pi f$ the frequency, and $c$ the wave speed on the left side.
To match deflection and slope, the wave after the transition is
$$ y_{\rm right} = Y \left( \tfrac{c+c'}{2 c} \right) \sin\left( \omega \left( \tfrac{x}{c}-t \right) \right) + Y \left( \tfrac{c'-c}{2 c} \right) \sin\left( \omega \left( \tfrac{x}{c}+t \right) \right) \tag{3}$$
where $c'$ is the wave speed on the right. This has two components, one moving to the right with amplitude $ Y \left( \tfrac{c+c'}{2 c} \right)  $ and one moving to the left with amplitude $ Y \left( \tfrac{c'-c}{2 c} \right) $
But the problem is symmetric, and so you can deduce that

Every incident wave causes two waves on the the side of a transition. The relative amplitudes of these wave depends on the difference between the wave speeds on either side.

By applying symmetry you can find the wave amplitudes on the right side as a function of the wave amplitudes on the left side
$$ y_{\rm left} = Y \sin\left( \omega \left( \tfrac{x}{c}-t \right) \right) +   W \sin\left( \omega \left( \tfrac{x}{c}+t \right) \right)\tag{4}$$ 
leads to
$$ y_{\rm right} = \tfrac{Y (c+c')-W (c-c')}{2c} \sin\left( \omega \left( \tfrac{x}{c}-t \right) \right) + \tfrac{Y (c'-c) + W (c+c')}{2 c} \sin\left( \omega \left( \tfrac{x}{c}+t \right) \right)\tag{5}$$ 
So depending on the situation you can describe each amplitude contribution as "transmitted" wave, or "reflected" wave, etc. 
