A uniform rod AB of length 0 . 4 m and mass 1 .2 kg is suspended by two identical wires from a ceiling as shown in fig. (3). Where should a mass of 4.8 kg be suspended from the rod so that a tuning fork of frequency 256 Hz is in resonance with the first wire at Awhile its first overtone resonates with second wire at B?
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a
5 cm
b
10 cm
c
15 cm
d
20 cm
answer is B.
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Detailed Solution
Here n1=12lT1m=256 and n2=12lT2m=2×256 From these equations, we have T1=4T2 Let the mass be suspended at a distance x ftom A T1+T2=4⋅8+1⋅2=6 Taking moments aboutA for equilibrium 1⋅2×0⋅2+4⋅8×x=T2×0⋅4 From eqs. (1) and (2), T2=1.2 1⋅2×0⋅2+4⋅8×x=1⋅2×0⋅4 x=5cm