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 = 10 m/s2.

Question

Arun · Answer

There is no figure given. But I assume the figure as the case for which I have solved thia type of question. Frequency of right wire is two times the frequency of left wire. f= T^1/2 Thus tension in right wire is four times the tension in left wire. Thus T+ 4T = 60 T= 12. & 4T = 48 Taking moment, 12* 40 = 12* 20 + 4.8* x x = 5 cm

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