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 ti 'x'. Further it is obsreved that the frequency of `1st` harmonic in `AB` is equal to `2nd` harmonic frequency in `CD`. 'x' is

Options

A. (a) `(L)/(5)`
B. (b) `(4L)/(5)`
C. ( c ) `(3L)/(4)`
D. (d) `(L)/(4)`

Answer

A

Explanation

(a) Frequency of `Ist` harmonic of `AB = (1)/(2l)sqrt((T_(AB))/(m)`
Frequency of `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)` ...(i)
For rotational equilibrium rod, taking torque about point `O`.
`T_(AB) xx x = T_(CD)(L - x)` ...(ii)
For trabslational equilibrium,
`T_(AB) + T_(CD) = mg` ...(iii)
On solving, (i) and (iii), we get
`T_(CD) = (mg)/(5)` :. T_(AB) = (4mg)/(5)`
Substituting these values in (ii),we get
`(4mg)/(5) xx x = (mg)/(5)(L - x)` rArr `x = (L)/(5)`