The two strings have different mass densities:
The wave speeds in the strings are:
\( v_1 = \sqrt{\frac{T}{\mu_1}}, \quad v_2 = \sqrt{\frac{T}{\mu_2}} = \sqrt{\frac{T}{4\mu}} = \frac{v_1}{2} \)
The ratio of wavelengths is given by:
\( \frac{\lambda_1}{\lambda_2} = \frac{v_1}{v_2} = \frac{\mu_2}{\mu_1} = \frac{4\mu}{\mu} = 2 \)
At the minimum frequency, the lengths of the two strings must fit an integral number of half-wavelengths:
\( L_1 = \frac{p_1 \lambda_1}{2}, \quad L_2 = \frac{p_2 \lambda_2}{2} \)
Substituting \( \lambda_2 = \frac{\lambda_1}{2} \):
\( \frac{p_1}{p_2} = 4 \)
Let \( p_1 = 4 \) and \( p_2 = 1 \), then:
\( f = \frac{v_1}{2L} = v_0 \)
This is the fundamental frequency, with 6 nodes (including P, O, and Q).
For an antinode at O, the lengths of the strings must satisfy the condition for continuity of displacement:
\( L_1 = \frac{p_1 \lambda_1}{4}, \quad L_2 = \frac{p_2 \lambda_2}{4} \)
Substituting \( \lambda_2 = \frac{\lambda_1}{2} \):
\( \frac{p_1}{p_2} = 2 \)
The condition for continuity of vibration cannot be satisfied with integer values of \( p_1 \) and \( p_2 \), so no vibration mode with an antinode at O is possible.
(A), (C), (D)