A metre stick swinging about its one end oscillates with frequency `f_(0)`. If the bottom half of the stick was cut off, then its new oscillation frequency will be :

Options

A. `f_(0)`
B. `sqrt(2)f_(0)`
C. `2f_(0)`
D. `2sqrt(2)f_(0)`

Answer

B

Explanation

`f_(0) = (1)/(2pi)sqrt((mgl)/(I))`
where, `l` is distance between point of suspension and centre of mass of the body.
`f_(0) = (1)/(2pi)sqrt((m.g.(L)/(2))/((mL^(2)//3))) = (1)/(2pi)sqrt((3g)/(2L))`
when bottom half of the stick is cut of
`f_(0)'=(1)/(2pi)sqrt(((m)/(2).g.(L)/(4))/((m)/(2)(L//2)^(2)/(3)) )=(1)/(2pi)xxsqrt((3g)/(L)) = sqrt(2)f_(0)` Ans.