A massless rod BD is suspended by two identical massless strings AB and CD equal lengths. A block of mass, ‘m’ is suspended at point P such that BP is equal to ‘x’ . If the fundamental frequency of left wire is twice the fundamental frequency of the right wire, then find the value of ‘x’ .

Question

Arun · Answer

Dear student [1/(2* lambda)] *sqrt(T1/mu) = [1/ lambda] * sqrt (T2/mu) T2 = T1/4 Now T1 * x = T2 * (L-x) 4 T2 * x = T2 * (L-x) x = L/5 Regards Arun (askIITians forum expert)

Kushagra Madhukar · Answer

Dear student, Please fidn the answer to your question. [1/(2* lambda)] *sqrt(T1/mu) = [1/ lambda] * sqrt (T2/mu) T2 = T1/4 Now T1 * x = T2 * (L-x) 4 T2 * x = T2 * (L-x) x = L/5 Thanks and regards, Kushagra

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