A wave pulse travelling along a two-piece string experiences...

When a wave pulse reflects from a junction and the reflected wave is inverted, it implies that the wave is reflecting from a denser medium. Thus, the second string is denser than the first string.

Let $\mu_1$ and $\mu_2$ be the linear mass densities of the first and second strings respectively. Since the reflected wave is inverted, $\mu_2 > \mu_1$.

The speed of a transverse wave on a string is given by $v = \sqrt{\dfrac{T}{\mu}}$. Since the tension $T$ is uniform across the two-piece string, the speed of the wave in the second string is less than that in the first string, so $v_2 < v_1$.

The frequency $f$ of the wave remains constant as it passes from one medium to another.

Using the relation $v = f\lambda$, the wavelengths in the two strings are $\lambda = \dfrac{v_1}{f}$ and $\lambda' = \dfrac{v_2}{f}$.

Since $v_2 < v_1$, it follows that $\lambda' < \lambda$.