A uniform horizontal rod of length 40 cm and mass 1.2 kg is supported by two identical wires as shown in figure. Where should a mass of 4.8 kg be placed on the rod so that the same tuning fork may excite the wire on left into its fundamental vibrations and that on right into its first overtone ? Take `g=10ms^-2`

Answer

`5 cm` from left wire

Explanation

Let `n_(1)` be frequency of vibration of left wire and `n_(2)` that of second wire . Then given `n_(2) = n_(1)`. If `T_(1) and T_(2)` are tensions in first and second wire , then
`n_(1) = (1)/( 2l) sqrt((T_(1))/(m)) and n_(2) = (2)/( 2l) sqrt((T_(2))/(m))`
Therefore , `(n_(2))/( n_(1)) = 2 sqrt((T_(2))/(T_(1)))`
As `n_(1) = n_(2) or sqrt(T_(1)) = 2 sqrt(T_(2))`
`:. sqrt((T_(1))/(T_(2))) = 2 or T_(1) = 4 T_(2)`
For vertical equilibrium of rod
`T_(1) + T_(2) = 4.8 + 1.2 = 6 kg wt`
From Eqs.(i) and (ii)
`T_(1) = 4.8 kg, T_(2) = 1.2 kg`
Taking moments about `A`
`T_(1) x + W(0.2 - x) = T_(2) ( 0.4 - x)`
`4.8 x + 0.24 - 1.2 x = 0.48 - 1.2 x`
or ` 4.8 x = 0.24`
`x = ( 0.24)/(4.8) = 0.05 m = 5 cm`