A cylindrical wooden block of cross-section 15.0 `cm^(2)` and mass 230g is floated over water with an extra weight of 50 g attached to its bottom. The cylinder floats vertically. From the state of equilibrium, it is slightly depressed and released. If the specific gravity of wood is 0.30 and `g=9.8ms^(-2)` deduce the frequency of the block, `A=15.0cm^(2)=15xx10^(-4)m^(2)`

Answer

`1.15`

Explanation

Area of cross -section of the block
`A = 15.0 cm^(2) = 15 xx 10^(-4) m^(2)`
Total mass of the block
` m = 230 + 50 = 280 g = 0.28 kg`
Density of water
`sigma = 10^(3) kg m^(-3)`
Density of wood
`rho = 0.30 xx 10^(3) kg m^(-3) = 300 kg m^(-3)`
Let the cylinder the depressed through a small distance y . Then
Restoring force = weight of water displaced
or ` F = A y sigma g `
Force constant ,
`K = (F)/(y) = A sigma g = 15 xx 10^(-4) xx 10^(3) xx 9.8 = 14.7 Nm^(-1)`
Frequency ` v = (1)/(2 pi) sqrt((K)/(m)) = (1)/(2 pi) sqrt(( 14.7)/(0.28))`
`= (7)/(44) xx sqrt(52.5) =1.15` Hz