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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’ .

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’ .

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Grade:12

2 Answers

Arun
25750 Points
6 years ago
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
askIITians Faculty 628 Points
3 years ago
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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