Question 6

Let $v_1$ and $v_2$ be the wave speeds in the two strings.

$v_1 = \sqrt{\frac{T}{\mu}}$ and $v_2 = \sqrt{\frac{T}{4\mu}} = \frac{1}{2}v_1$.

Given $\nu_0 = \frac{1}{2L}\sqrt{\frac{T}{\mu}} = \frac{v_1}{2L}$.

For a node at O:

Both segments must have an integer number of half-wavelengths.

$f = \frac{n_1 v_1}{2L}$ and $f = \frac{n_2 v_2}{2(2L)} = \frac{n_2 (v_1/2)}{4L} = \frac{n_2 v_1}{8L}$.

Equating the frequencies: $\frac{n_1 v_1}{2L} = \frac{n_2 v_1}{8L} \Rightarrow 4n_1 = n_2$.

For minimum frequency, $n_1 = 1$, which gives $n_2 = 4$.

$f_{min} = 1 \cdot \frac{v_1}{2L} = \nu_0$. Thus (A) is correct.

Number of nodes: String 1 has $n_1+1 = 2$ nodes (P and O). String 2 has $n_2+1 = 5$ nodes. Total nodes = $2 + 5 - 1$ (overlapping at O) = 6. Thus (C) is correct.

For an antinode at O:

Each segment must have an odd multiple of quarter-wavelengths.

$f = \frac{(2n_1-1)v_1}{4L}$ and $f = \frac{(2n_2-1)v_2}{4(2L)} = \frac{(2n_2-1)v_1}{16L}$.

Equating frequencies: $\frac{2n_1-1}{4} = \frac{2n_2-1}{16} \Rightarrow 4(2n_1-1) = 2n_2-1 \Rightarrow 8n_1 - 4 = 2n_2 - 1 \Rightarrow 8n_1 - 2n_2 = 3$.

Since the left side is even and the right side is odd, no integer solution for $n_1, n_2$ exists. Thus (D) is correct and (B) is incorrect.

Final Answer: A, C, D.