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Question

A uniform horizontal rod of length 40 cm and mass 1.2 kg is supported by two identical wires as shown in figure (15 - E9). 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 ms2

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Solution

Length of the rod,

L = 40 cm = 0.4 m

mass of the rod = m = 1.2 kg

Let the 4.8 kg mass be placed at a distance 'x' from the left end.

Given that, f1=2f2

12LT1m=22LT2m

T1T2=2

T1T2=4 ...(1)

From the free body diagram

T1+T2=48+12=60 N

4Tr+Tr=5Tr=60 N

Tr=12 N

and T1=48 N

Now taking moment about point A

Tr×(0.4)=48x+12(0.2)

4.8=48x2.4

4.8x=2.4

x=2448=120m=5 cm

So, the mass should be placed at a distance 5 cm from the left end.


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