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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)` |
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Answer» Correct Answer - 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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