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))

A piece of wood of mass m is floating erect in a liquid whose density

1.	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))
Answer» Solution :SPRING factor of liquid (K) = `Arhog` INERTIA factor of PIECE of wood = m m Time period T = `2pi SQRT(("Intetia factor")/("spring factor"))` T = `2pi sqrt((m)/(Arhog))`