A massless rod of length `L` is suspened by two identical string `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 `1st` harmonic in `AB` is equal to `2nd` harmonic frequency in `CD`. 'x' is

Options

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

Answer

A

Explanation

Frequency of the first harmonic of
`AB=1/21 sqrt((T_(AB))/(m))` Frequency of the `2^(nd)` harmonic of
`CD=1/1 sqrt((T_(CD))/(m))`
Given that the two frequency are equal
`therefore 1/21 sqrt((T_(AB))/(m))=1/1sqrt(T_(CD)/(m)) implies (T_(AB))/4=T_(CD)`
`implies T_(AB)=4 T_(CD)`...(i)
For rotational of massless rod taking torque about point O,
`T_(AB)xxx=T_(CD)(L-x)`...(ii)
solve to get `x=L/5`