A massless rod of kngth L is suspended by two identical strings AB and CD of equal length. A block of mass m is suspended from point O such that BO is equal to 'x'. Further it is observed that the frequency of `1_(st)` harmonic in AB is equal to `2_(nd)` harmonic frequency in CD, 'x' is

Options

A. `(L)/(5)`
B. `(4 L)/(5)`
C. `(3 L)/(4)`
D. `(L)/(4)`

Answer

A

Explanation

Frequency of the first harmonic of
`AB = (1)/(2l) sqrt((T_(AB))/(m))`
Frequency of the `2nd` harmonic of
`CD=(1)/(l) sqrt((T_(CD))/(m))`
Given that the two frequencies are equal.
`:. (1)/(2l) sqrt((T_(AB))/(m)) = (1)/(l) sqrt((T_(CD))/(m)) rArr (T_(AB))/(4) = T_(CD)`
`rArr T_(AB) =4T_(CD)`

For rotational of massless rod taking torque about point `O`,
`T_(AB) xx x = T_(CD)(L -x)`
Solve to get `x = (L)/(5)`.