A rod of length 'L' and negligible mass is suspended by two identical strings AB and CD as shown in the figure. A mass 'M' is suspended from point 'O' which is at a distance 'x' from B. If the frequency of the first harmonic of AB is equal to the frequency of the second harmonic of CD, then the value of 'x' is

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$\frac{L}{9}$

$\frac{2 L}{7}$

$\frac{L}{5}$

$\frac{3 L}{10}$

answer is A.

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Detailed Solution

According to question,

$\begin{array}{l} \frac{1}{2 l} \sqrt{\frac{T_{1}}{μ}} = \frac{1}{l} \sqrt{\frac{T_{2}}{μ}} \\ T_{2} = T_{1} / 4 \\ F o r r o t a t i o n a l e q u i l i b r i u m, \\ T_{1} x = T_{2} (L - x) \Rightarrow x = \frac{L}{5} \end{array}$

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