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(A) Let the distance from $B$ to the first particle be $r$ and the distance between the two particles be $2r$. Let $\omega$ be the angular velocity of

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$4:3$

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(A) Let the distance from $B$ to the first particle be $r$ and the distance between the two particles be $2r$. Let $\omega$ be the angular velocity of the system.The tension in the outer part of the rope (between the two particles) is $T_2$. This tension provides the centripetal force for the particle at end $A$:$T_2 = m \omega^2 (r + 2r) = 3m \omega^2 r$The tension in the inner part of the rope (between $B$ and the middle particle) is $T_1$. This tension provides the centripetal force for both particles:$T_1 = T_2 + m \omega^2 r$$T_1 = 3m \omega^2 r + m \omega^2 r = 4m \omega^2 r$The ratio of the tension in the smaller part $(T_1)$ to the tension in the larger part $(T_2)$ is:$\frac{T_1}{T_2} = \frac{4m \omega^2 r}{3m \omega^2 r} = \frac{4}{3}$

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