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Q.

A mass less rod is suspended by two identical strings AB and CD of equal length. A block of mass m is suspended from point O such that BO is equal to ‘x’. Further, it is observed that the frequency of 1st harmonic (fundamental frequency) in AB is equal to 2nd harmonic frequency in CD. Then, length of BO is

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a

L/5

b

4(L/5)

c

3(L/4)

d

L/4

answer is A.

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Detailed Solution

given nAB =2nCD 12lT1μ =212lT2μ T1T2=4 T1 =4T2 as it is in rotational equilibrium T1x=T2l-x 4x =l-x x=l5

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