A uniform horizontal rod of length 40 cm and mass 12 kg is supported by two identical wires as shown in figure (15-E9). Where should a mass of 48 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 s1.


We know that



Now, as per question,


fd =2f2


So,


Now,


12 = 48 + 12 = 60


By replacing the previous relation, we get that,


T1=48 N


T2=12 N


Now we need to take moment about point A,


T2 = 48x + 12(0.2)


So,


So,


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