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 the left end

Explanation

For the given condition we must have
`T_(2) = 4T_(1)`
if `T_(1) & T_(2)` are tensions in left and right strings. So if 4.8 kg mass is supported at a distance x from left we use for equilibrium of rod.
`T_(1) + T_(2) = 60`
`5I_(1) = 60`
`T_(1) = 12N`
For torque about right and of rod we use
`48(x) + 12(0.2) = 12(0.4)`
`48x = 4.8 - 2.4 = 2.4`
`x = (2.4)/(48) = 0.05m`