From two identical wires, a rod XY of length l is hung. A wooden block of mass m is hung at point O of the rod at a distance a from wire A. Find the value of a in the case when a tuning fork excites the fundamental note in wire A and third harmonic in wire B.

Explanation

m' = mass per unit length of each wire.
`T_(1), T_(2)` are tensions produced in wires A and B.
According to given condition
For wire 1
v = frequency `= (1)/(2l) sqrt((T_(1))/(m))" "...(i)`
For wire 2
`v = (3)/(2l) sqrt((T_(2))/(m))" "...(ii)`
From (i) and (ii)
`(1)/(2l) sqrt((T_(1))/(m)) = (3)/(2l) sqrt((T_(2))/(m))`
`rArr" "(T_(1))/(T_(2)) = 9`
Applying rotational equilibrium of rod about O. We have,
`T_(1) (a) = T_(2) (l - a)`
`(T_(1))/(T_(2)) = (l - a)/(a)`
But `(T_(1))/(T_(2)) = 9`
`therefore" "(l-a)/(a) = 9`
l - a = 9a
`rArr" "10a = l`
`a = (l)/(10)`