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

5cm

Explanation

Length of the rod, L=40cm=0.4m
mass of the rod =m=1.2kg
let the 4.8 kg mass be placed at a distance x from the left end.
Given `f_d=2f_2`
`:. 1/(2L)sqrt(T_1/m)=2/(2L)sqrt(T_2/m)`
`rarr sqrt(T_1/T_2)=2`
`rarr T_1/T_2=4`.
From the free body diagram
`T_1+T_2=48+12=60N `
`rarr 4T_r+T_r=5T_r=60N`
`:. T_r=12N`
`and T_1=48N`
now taking moment about point A
`T_rxx(0.4)=48x+12(0.2)`
`rarr 4.8=48x-2.4`
`rarr 4.8x=2.4`
`-x=24/48=1/20m=5cm`
so, the mass should be placed at a distance 5 cm from tehleft end.