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A massless rod BD is suspended by two id...

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

A

`(l)/(5)`

B

`(l)/(4)`

C

`(4l)/(5)`

D

`(3 l)/(4)`

Text Solution

Verified by Experts

The correct Answer is:
A
`f propv propsqrt(T)`
`f_(AB)=2f_(CD)`
`:. T_(AB)=4T_(CD)`
Further `Sigma tau_(p)=0`
`:. F_(AB)(X)=T_(CD)(l-x)`
`rArr 4X=l-X ` (as `T_(AB)=4T_(CD)`)
`rArr X = l//5`
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