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

Question

Infinity Learn · Accepted Answer

given nAB $$=2nCD$$ $$12lT1$$μ $$=212lT2$$μ $$T1T2=4$$      $$T1$$ $$=4T2$$ as it is in rotational equilibrium   $$T1x=T2l-x$$ $$4x$$ $$=l-x$$     $$x=l5$$

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