Two wires made of the same material, one thick and the other thin, are connected to form a composite wire. The composite wire is subjected to some tension. A wave travelling along the wire crosses the junction point. The characteristic that doesn't undergoes a change at the junction point is
A. Frequency
B. Speed of propagation only
C. Wavelength only
D. The speed of propagation as well as the wavelength.

Hint: One should refer to the definition of frequency and wavelength. The tension for the two strings is the same. We have to differentiate between the two strings based on their diameters and find the velocity and frequency based on this.
Formula used:
For a stretched string with tension T and mass per unit length m, the velocity of the wave is:
$v = \sqrt{ \dfrac{T}{m} }$.
The frequency of fundamental mode for stationary waves on a string of length L:
$\nu = \dfrac{1}{L} \sqrt{ \dfrac{T}{m} }$

Complete answer:
For the propagation of sound waves on a stretched string, we have a formula:
$v = \sqrt{ \dfrac{T}{m} } = \sqrt{ \dfrac{T}{A \rho} } $.
where A is the area of cross section of a string and $\rho$ is mass per unit volume of the string which can be written as
$\rho = \dfrac{M}{Al} = \dfrac{m}{A} $.
The relation for velocity clearly suggests that more the cross section area of the string less will be wave velocity. Thus, in going from one string to another, the velocity definitely changes.

When we consider the entire composite system, we can write the frequency for the entire system as
$\nu = \dfrac{1}{L} \sqrt{ \dfrac{T}{\mu} }$
Here, L is the length of the composite system and $\mu$ is the mass per unit length for the composite system.
This frequency is the same for the entire system as it will be produced when the two ends of the composite string will be tied to form stationary waves.

Now as the frequency is same for the two elements of the composite string and velocity is different, in the light of the following relation
$v = \nu \lambda$,
we can say that wavelength will also change for the two strings.
Therefore, across the junction point frequency is the characteristic of the wave which will not change.

So, the correct answer is “Option A”.

Note:
The laws of refraction are based on this so one can also apply that logic here to get the right answer. In all the formulas used here we have denoted mass per unit length of the string to be m.