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, from left end (in cm), 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 $g = 10 m / s 2$

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Detailed Solution

Question Image

Let mass 4.8 kg be placed at distance x from A
Since rod is in equilibrium
$T 1 + T 2 = 4.8 g + 1.2 g = 60$
Taking torque about A
$4.8 g x + 1.2 g × 20 = T 2 × 40 48 x + 240 = 40 T 2 … (ii)$
Question Image $n = 1 2 L T 1 μ = 2 2 L T 2 μ$
$T 1 = 4 T 2 … (iii)$
Solving (i) and (iii) , $T 2 = 12 N, T 1 = 48 Nin (ii)$
$48 x + 240 = 40 × 12 = 480 48 x = 240 x = 5 cm$

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