A piece of wood of mass m is floating erect in a liquid whose density is `rho` . If it is slightly pressed down and released, then executes simple harmonic motion. Show that its time period of oscillation is `T=2pisqrt((m)/(Arhog))`

Explanation

When a wood is pressed and released,
Change in
force `(F= ma)`
`=(Ax) rho g`
`A=(F)/(m)=((A rho g)x)/(m) " "...(1)`
since `a=- omega^(2) x " "....(2)`
From (1) & (2)
`omega^(2)=(A rho g)/(m)`
`omega= sqrt((A rhog)/(m))`
We know that,
Time period `T=(2pi)/(omega)= 2pi sqrt((m)/(A rho g))`