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Question

A uniform horizontal rod of length 0.40 m and mass 1.2 kg is supported by two identical wires as shown in figure(g=10 m/sec2), then the distance x (in cm) from left wire, where a mass of 4.8 kg should 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 is


Solution

Let n1 be frequency of vibration of left wire and n2 that of right wire. It is given that n2=n1. If T1 and T2 are tensions in left and right wire, then
n1=12lT1m and n2=22lT2m
(where l is the length of identical rods)
  n2n1=2T2T1

As n1=n2 or T1=2T2
T1T2=2 or T1=4T2  ..........(i)
For vertical equilibrium of rod
T1+T2=4.8+1.2=6kgwt....(ii)From Eqs. (i) and (ii)T1=4.8kg,T2=1.2kg



Taking moments about 
T1x+W(0.2x)=T2(0.4x)4.8x+0.241.2x=0.481.2x or 4.8x=0.24x=0.244.8=0.05m=5cm

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