Two wires $$W_1$$ and $$W_2$$ have the same radius $$r$$ and respective densities $$\rho_1$$ and $$\rho_2$$, such that $$\rho_2 = 4\rho_1$$. They are joined together at the point $$O$$, as shown in the figure. The combination is used as a sonometer wire and kept under tension $$T$$. The point $$O$$ is midway between the two bridges. When a stationary wave is set up in the composite wire, the joint is found to be a node. The ratio of the number of antinodes formed in $$W_1$$ to $$W_2$$ is:

Wave speed in a stretched string:

$$v=\sqrt{\frac{T}{\mu\ }}$$

Since radius is same, linear mass density:

$$\mu\ ∝\ ρ$$

Thus, $$\frac{v_1}{v_2}=\sqrt{\frac{\mu_2}{\mu_1\ }}=\sqrt{\frac{ρ_2}{ρ_1}}=\sqrt{4}=2$$

$$v_1=2\cdot v_2$$

Since the junction is a node, each segment behaves like an independent string fixed at both ends.

Number of antinodes:

$$n=\frac{2L}{λ}$$

Using $$v=fλ$$,

$$λ=\frac{v}{f}$$

$$n=\frac{2Lf}{v}$$

$$L_1=L_2\ \&\ f_1=f_2$$

Thus, the ratio of antinodes:

$$\frac{n_1}{n_2}=\frac{v_2}{v_1}=\frac{1}{2}$$