A wire is made of two parts joined end to end: a part 0.50 m long with a linear mass density of 2.5…

Step 1. Speeds (a). $v_1 = \sqrt{\dfrac{100\ \text{N}}{2.5 \times 10^{-3}\ \text{kg/m}}} = 200\ \text{m/s}$ and $v_2 = \sqrt{\dfrac{100\ \text{N}}{10.0 \times 10^{-3}\ \text{kg/m}}} = 100\ \text{m/s}$.

Step 2. A node at the joint (b). Part 1 can vibrate at $p_1\dfrac{v_1}{2L_1} = p_1 \times 200\ \text{Hz}$ and part 2 at $p_2\dfrac{v_2}{2L_2} = p_2 \times 66.7\ \text{Hz}$, for whole numbers $p_1$ and $p_2$. They agree first when $p_1 = 1$ and $p_2 = 3$, at $f = 200\ \text{Hz}$.

Step 3. Nodes (c). There are $1 + 3 = 4$ loops in all, so $5$ nodes, the ends and the joint included.

Step 4. Wavelength (d). In the heavier part $\lambda_2 = \dfrac{v_2}{f} = \dfrac{100\ \text{m/s}}{200\ \text{Hz}} = 50.0\ \text{cm}$: three half-wavelengths fill its $0.75\ \text{m}$.

Step 5. The lowest mode of the whole wire (e). Take $y_1 = A\sin k_1x$ from the left support and $y_2 = B\sin k_2x'$ from the right support, with $k = \omega/v$. At the joint the displacements are equal and the transverse forces $T\,\partial y/\partial x$ balance, which leaves

With $\omega = 400\,\theta$ (in $\text{rad/s}$) this is $\cot\theta + 2\cot 3\theta = 0$. Its smallest root is $\theta = 0.7017$, so $\omega = 280.7\ \text{rad/s}$ and $f = 44.7\ \text{Hz}$.